L. I. Volkovyskii, G. L. Lunts, I. G. Aramanovich-A Collection of Problems on Complex Analysis-Dover Publications (1991)
Short Description
Problems on Complex Analysis...
Description
A COL LEC TIO N OF PROBLEMS ON IS LYS COM PLE X ANA L. I. Volko vyskii , G. L. Lunts , .and I.G. Aram anovi ch
A Collection of Problems on Complex Analysis
A Collection of Problems on
COMPLEX ANALYSIS L.I. VOLKOVYSKII, G.L. LUNTS, I. G. ARAMANOVICH Translated by
J. BERRY Translation edited by
T. KOVAR!, PH.D.
DOVER PUBLICATIONS, INC. New York
Copyright © 1965 by Pergamon Press Ltd. All rights reserved under Pan American and International Copyright Conventions. Published in Canada by General Publishing Company, Ltd., 30 Lesmill Road, Don Mills, '.Ibronto, Ontario. This Dover edition, first published in 1991, is an unabridged and unaltered republication of the work first published by Pergamon Press, Oxford, in 1965 as Volume 68 in the International Series of Monographs in Pure and Applied Mathematics. This book is an edited translation of the original Russian Sbornik zadach po teoriyi funktsii kompleksnogo peremennogo, published in 1960 by Fizmatgiz, Moscow. The Dover edition is published by special arrangement with Pergamon Press, Headington Hill Hall, Oxford OX3 OBW England. Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N. Y. 11501 Library of Congress Cataloging-in-Publication Data Volkovyskii, L. I. (Lev Izrailevich) [Sbornik zadach po teorii funkfsii kompleksnogo peremennogo. English] A collection of problems on complex analysis IL.I. Volkovyskii, G.L. Lunts, LG. Aramanovich; translated by J. Berry; translation edited by T. Kovari. p. cm. _ Translation of: Sbornik zadach po teorii funktsiI kompleksnogo peremennogo. Originally published: Oxford : Pergamon Press, 1965. (International series of monographs in pure and applied mathematics ; v. 68). Includes bibliographical references. ISBN 0-486-66913-0 1. Functions of complex variables-Problems, exercises, etc. 2. Mathematical analysis-Problems, exercises, etc. I. Lunts, G. L. (Grigorii L'vovich) II. Aramanovich, I. G. (Isaak Genrikhovich) III. Berry, J. IV. Kovari, T. V. Title. QA331. 7. V6513 1991 515' .9'076-dc20 91-40596 CIP
CONTENTS l!ortJWO'l'd,
ix
CHAPTER I
Complex numbers and f1ll1Ctions of a complex variable § 1 Complex numbers (complex numbers; geometrical interpretation; stereographic projection; quaternions) § 2 Elementary transcendental functions § 3 Functions of a complex variable (complex functions of a real variable; functions of a complex variable; limits and continuity) § 4 Analytic and harmonic functions (the Cauchy-Riemann equations; harmonic functions; the geometrical meaning of the modulus and argument of a derivative)
I 7 11 14
CHAPTER II
Conformal mappings connected with elementary functions § 1 Linear functions (linear furiptions; bilinear functions) § 2 Supplementary questions of the theory of linear transformations (canonical forms of linear transformations; some ap· proximate formulae for linear transformations; mappings of simply connected domains; group properties of bilinear transformations; linear transforma~ions and non-Euclidean geometry) § 3 Rational and algebraic functions (some rational functions; mappings of circular lunes and domains with cuts; the func-
tion
!(z+ ~);
21
26
application of the principle of symmetry;
the simplest non-sohlicht mappings) § 4 Elementary transcendental functions (the fundamental tran· scendental functions; mappings leading to mappings of the strip and half-strip; the application of the symmetry principle; the simplest many-sheeted mappings) § 5 Boundaries of univalency, convexity and starlikeness
33
42
CHAPTER III
Supplementary geometrical questions. Generalised analytic functions § 1 Some properties of domains and their boundaries. Mappings of domains § 2 Quasi-conformal mappings. Generalised analytic functions
51 55
CHAPTER IV
Integrals and power series
§ 1 The integration of fw1ctions of a complex variable
v
64
vi
CONTENTS
§ § § §
2 3 4 5
Cauchy's integral. theorem Cauchy's integral formula Numerical series Power series (determination of the radius of convergence; behaviour on the boundary; Abel's theorem) § 6 The Taylor series (the expansion of functions in Taylor series; generating functions of systems of polynomials; the solution of differential equations) § 7 Some applications of Cauchy's integral formula and power series (Cauchy's inequalities; area theorems for univalent functions; the maximum principle; zeros of analytic functions; the uniqueness theorem; the expression of an analytic function in terms of its real or imaginary part)
68 70 72 74 78
83
CHAPTER V
Laurent series, singularities of single-valued functions. Integral functions § 1 Laurent series (the expansion of functions in Laurent series; some properties of univalent functions) § 2 Singular points of single-valued analytic functions (singular points; Picard's theorem; power series with singularities on the boundary of the circle of convergence) \j § 3 Integral functions (order; type; indicator function)
89 92
CHAPTER VI
Various series of functions. Parametric integrals. Infinite products § 1 Series of functions § 2 Dirichlet series § 3 Parametric integrals (convergence of integrals; Laplace's integral) § 4 Infinite products
102 106 108 111
CHAPTER VII
Residues and their applications § 1 The calculus of residues § 2 The evaluation of integrals (the direct application of the theorem ofresidues; definite integrals; integrals connected with the inversion of Laplace's integral; the asymptotic behaviour of integrals) § 3 The distribution of zeros. The inversion of series (Rouch6's theorem; the argument principle; the inversion of series) § 4 Partial fraction and infinite product expansions. The summation of series
116 118 141 147
vii
CONTENTS CHAPTER VIII
Integrals of Cauchy type. The integral formulae of Poisson and Schwarz. Singular integrals § 1 Integrals of Cauchy type § 2 Some integral relations and double integrals § 3 Dirichlet's integral, harmonic functions, the logarithmic potential and Green's function § 4 Poisson's integral, Sohwarz's formula, harmonic measure § 5 Some singular integrals
152 159 163 167 174
CHAPTER IX
Analytic continuation. Singularities of many-valued character. Riemann surfaces § 1 Analytic continuation § 2 Singularities of many-valued character. Riemann surfaces
192 199
§ 3 Some classes of analytic functions with non-isolated singularities
207
CHAPTER X
Conformal mappings (continuation) § 1 The Schwarz--Ohristoffel fotmula § 2 Conformal mappings involving the use of elliptic functions
211 229
CHAPTER XI
Applications to mechanics and physics § 1 Applications to hydrodynamics § 2 Applications to electrostatics § 3 Applications to the plane problem of heat conduction
243 258 271
ANSWERS AND SOLUTIONS
Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter
I
n
m IV V VI VD
vm IX X XI
274 285 307 310 317
324 328 339
349 374 395
FOREWORD A Collection of Problema on Oomple:x: Analysis is intended mainly for students of the physico-mathematical and mechanico-mathematical faculties of universities and also for students of the physicomathema.tical faculties of teachers training colleges. The authors believe also that the collection will be useful for those specialising in the mechanics of continuous media (hydrodynamics, the theory of elasticity) and electrical engineering, since it contains a large number of problems either on the direct application of the theory of functions of a complex variable to the given subject, or on questions forming their mathematical basis (conformal mappings, harmonic functions, potentials, integrals of Cauchy type, etc.). Problems connected with the compulsory university course on the theory of functions of a complex variable are arranged ma.inly in Chapters I, II (§ 1, 3 and 4), IV, V, VII, X (§ 1). The book also contains groups of problems which go beyond the range of the syllabuses. Some of these may be used as the basis of course work and as material for study in seminars on the theory of functions of a complex variable. It seems to us that the book deals with the main sections of the theory of functions of a complex variable approximately in accordance with the curriculum. For the convenience of those using the book, in the contents besides the titles of the chapters and sections, the main groups of problems contained in them have sometimes been enumerated (this refers ma.inly to the basic course material). It is assumed that the users of the collection are acquainted with the corresponding sections of the course in the theory of functions of a complex variable contained in the book by A. I. Markushevich A Short O, w = u+i11 = eelll (:ii, y, u, 11, r, (!, qi and 8 are real numbers, r;;;;,: O, e ~ 0); Re z =:ii, Im z = y, arg z =qi, lzl = r, z= z-iy. In the absence of any additional indications the principal value of the argument arg z is defined by the inequalities -:ii< arg z ~:ii; the complex plane whose points are represented by the complex number z, will be called the z-plane; usually the terms "complex number z" and "point z" a.re used as synonyms.
§ I. Complex numbers I. Perform the operations indicated: 2 (1) l/i; (3) l-3i;
1-i (2) l+i ;
(4) (l+iy3)s.
2. Find the modulus and argument of each of the following complex numbers (a and b are real numbers): (1) 3i;
(6) 2-5i;
(2) -2;
(7) -2+5i;
(3) l+i;
(8) -2-5i; (4) -1-i; (9) bi (b =/: O); (5) 2+5i; (10) a+bi (a=/: 0). 3. Find all the values of the following roots and plot them:
Jf (1-i);
(1) yl;
(6)
(2> jli; (3> V(-1); (4) (-8);
c1> V(3+4i); (8) j/(-2+2i); (9) (-4+3i),
v
fl
(5) j/1; 4. Prove that both values of (z2- l) lie on the straight line passing through the coordinate origin and parallel to the bisector of the
r
I
2
PROBLEMS ON COMPLEX ANALYSIS
internal angle of the triangle, with vertices at the points -1, I and z, which passes through the vertex z. 5. Let m and n be integers. Prove that (ii'z)m has n/(n, m) different values, where (n, m) is the greatest common divisor of the numbers m and n. Verify that the sets of values of (yz )m and j/ (zDI) are identical if and only if (n, m) =I, that is, if n and mare coprimes. 6. Prove the following inequalities from geometrical considerations:
(I) lzt +z2J ~ Jz1l + Jz2i; (2) Jz1-z2I ;?:: ] Jz1J-lz2i I; (3) iz-IJ
< J izi-Il+lzlJarg zl.
In each case explain when the sign of equality applies. 7. Prove the identity
iz1+z2l 2 +iz1-z.l2 = 2(1Zil 2 +lzal2) and explain its geometrical meaning. 8. Prove that if iz11 = iz2 J = IZsl, then Z:J-Z2 I z2 arg---=-arg-. Zs-Z1 2 Z1 9. Prove that if z1+z2 +z3 = 0 and iz11 = iz21= IZsl = I, then the points z1, z2, Zs are the vertices of an equilateral triangle inscribed in the unit circle. 10. Prove that if Zi +z 2 +z3 +z4 = 0 and Jz11 = lz21= !Zs! = jz4 J, then the points z1, zs, Zs· z4 are either the vertices of a rectangle or pairs of them are identical. 11. Find the vertices of the regular n-gon if its centre is at the point z = 0, and one of its vertices z1 is known. 12. The points z1 and Zs are adjacent vertices of a regular n-gon. Find the vertex Zs• adjacent to z2 (Zs #: z1 ). 13. Given three vertices of a parallelogram z1, z2, Zs· find the fourth vertex z4 , opposite to the vertex z2• 14. Under what conditions do three points z1, z2, z3 , no two of which are coincident, lie on a straight line 1 15. Under what conditions do four points zt, z2, Zs• z4 , no two of which are coincident, lie on a circle or straight line 1
COMPLEX NUMBERS
3
16*. The points z1, z8, • • • , z. lie on one side of a certain straight line passing through the coordinate origin. Prove that the points l/zt, 1/z2, ••• , 1/z. possess a similar property (with respect to what straight linen and that Zi +z.+ ... +z. :#: 0; l/z1 + l/z2+ ... +1/z. ¥: O. 17. Prove that if zt + z. + •.• + z. = 0, then any straight line passing through the coordinate origin divides the points zt, z., ... , z., provided only that these points do not lie on this straight line. 18. Prove that any straight line passing through the centre of gravity of the material points zt, z., ... , z. with masses fni, m.i•••• , m., divides these points provided only that they do not lie on this straight line.
Explain the geometrical meaning of the relations given in problems 19-29. 19. lz-z0 1 < R; lz-z01 > R; lz-z0 1 = R. 20. lz-21+1z+21=5. 21. lz-21-lz+21>3. 22. lz-z1I = lz-zal· 23. (1) Rez ~ O; (2) lmz < 0. !4. 0 O; (4) Rez < 0; (2) Imz < O; (5) iz! < l; (3) Rez > O; (6) !z! >I. 46. What corresponds on the sphere to a family of parallel straight lines on the plane 1 47. Prove that in stereographic projection, circles on the sphere project into circles or straight lines on the plane. Which circles on the sphere correspond to straight lines 1 48. Let K be a circle on the plane corresponding to the circle K' on the sphere, let N be the north pole of the sphere, and let S be the vertex of the cone touching the sphere along K' (it is assumed that K' is not a great circle). Prove that the centre of the circle K lies on the ray NS. Consider the case when K' is a great circle. 49. Prove that in stereographic projection, angles between curves on the sphere are equal to the angles between their images on the plane. Prove also that the sense of angles is preserved if on the sphere the positive direction of rotation is defined with respect to the inward normal. liO. Find the length h(z, a) of the chord connecting points of the sphere which correspond to the points z and a. Consider also the case when a is the point at infinity.
6
PROBLEMS ON OOMPLEX ANALYSIS
51. Given two points z1 and z9 (one of them may be at infinity), :find the geometrical locus of the points of the z-plane which correspond to a circle on the sphere equidistant from the images of the given points.
Quaternions Quaternionst are numbers which are linear combinations with real ooefB. cients of four units-one being real and three imaginary t, J, k;
the multiplication of units being defined by the equations
,..... ,.. = k 1 = I.i==i.I==i,
-1;
I.J=J.I=J,
l.k=k.I=k.
Quaternions with identical components (coefficients of the units) are con· sidered to be equal. In the addition of quaternions the corresponding units are added. Multiplication is subject to the associative and distributive laws. Thus, for example, (2i+43) (3J-2k) = (2i) (3j)+ (4J) (3J)+(2i)(-2k)+(4J) (-2k)
= 2.3.i.j+4.3.3.. +2(-2).i.k+4.(-2).J.k
= 6k-12+4J-8i.
The numbers q = a+bi+ci+dk and iJ = a-bi-cj-dk are said to be conjugate to one another. The number N(q) = a•+b•+cl+dl is called the norm of the quaternion q.
52. Prove that in the quaternion system the operation of subtrac. tion is defined uniquely and that for divisors different from zero the operation of division is defined uniquely. 53. Prove that q1q9= ~ · q9. 54. Prove the equality N (q1 q2) = N (q1) N (q2)
and using this equality show that the product of two integers each of which can be represented as the sum of the squares of four integers is also a number which can be represented as the sum of four squares. Obtain this representation. t For hyperoomplex numbers and qua temions, see for example, A Uniwr sity Algebra, D. E. Littlewood. Heinemann's, London (1950).
7
OOMPLEX NUMBERS
55. Using quaternions construct the matrix B1 B2 B3 B4
A1 ( A2 A3 A4
01 02 03 04
D 1) D2 D3 D4
with integer elements which has the same sum of squares of elements along the rows, columns and principal diagonals and satisfies the relations AkA1+BkB1+0k01+DkD1
=0
(k :F l;
k, l
= 1, 2, 3, 4).
§ 2. Elementary transcendental functions By definition expz =es= e" (cosy+isiny); COSZ=
els+e-ls ; 2
es+e-s cosh z = - 2- ; es-e-s sinhz=---; 2
sinz
00s7;
sinhz tanhz=--; coshz
cosz cotz=-.-;
coshz cothz=--· sinhz
tanz =
SlnZ
56. Using the definition of es, prove that: (1) es1 .es1 = es1 +s1 ; (2) e=+ 2" 1 =es. (3) If e=+(I) =el' for every z, then ro = 2nlci
(k = 0,
±
1, ± 2, ... ) .
The relation exp iq, =ell/>= cos tJ>+i sin q, (Euler's formula) makes it possible to use the exponential form of notation z = relt/> for complex numbers instead of the trigonometric form z = r (cos tJ>+i sin I/>). In what follows by 41 is usually understood the principal value of the argument, that is, -n~n.
57. Represent the following numbers in the exponential form: 1, -1, i, -i, l+i, 1-i, -l+i, -1-i.
58. Find e
:1:.!..1 2
et"' (k = 0, ±1, ±2, ... ) .
8
PROBLEMS ON COMPLEX ANALYSIS
59. Find the moduli and principal values of the arguments of the following complex numbers: eS+I;
eB-81;
e8+'1;
e-8-41;
-ae'.P (a>
o, It/>! : : : ; 3t);
e-1.p (It/>! ~ 3t). 80. Find the sums: (1) l+cosx+cos2x+ ... +cosna:; (2) sinx+sin2x+ ... + sinna:; (3) cou+cosSz+ ... +cos (2n-l)x; (4) sinx+sin3x+ ... +sin (2n-l)x; (5) sinx-sin2x+ ... +(-1)•-lsinna;. 61. Find the sums: (1) cou+cos (oc+ P>+ ... +cos (oc+np); (2) sinoc+sin (oc+P>+ ... +sin(oc+np). 62. Starting from the definitions of the respective functions, prove: (1) sinllz+cosBz = 1;
(2) sinz =cos(; -z);
(3) sin (Zi.+z2) = sinz1cosz2+cosz1sinz2; (4) cos (z1 +zs) = COSZ1 COU2-sin2l1 sinz2; 2tanz (5) tan2z= 1 ta 2 ; - n z (6) cosh(21i+z2) = coshz1cosh212+sinhZi,sinhz2. 63. Prove that if cos (z+ro) = cosz for every z, then ro = 23tlc (le= 0, ±1, ±2, ... ). 64. Prove that: (1) siniz = isinhz; (3) taniz = itanhz; (2) cosiz = coshz; (4) cotiz = -icothz. 65. Express in terms of trigonometric and hyperbolic functions with real arguments the real and imaginary parts and also the moduli of the following functions: (1) sinz; (4) sinhz; (2) cosz; (5) coshz; (3) tanz; (6) tanhz. 66. Find the real and imaginary parts of the following expressions: (1) cos(2+i);
(4) cot(:-ilog2);
(2) sin2i;
(5) coth (2+i);
(3) tan (2-i);
(6) tanh(log3+
~i).
COMPLEX NUMBERS
9
67. For each of the functions e", cosz, sinz, tanz, coshz, cothz, find the set of points z, where it assumes: (1) Real values; (2) Purely imaginary values. By definition, I.ogz
= logr+it1>+2nik (k = 0, ±1, ±2, .•. ), logz = logr-1-it/>{-n:o:;;.;n)
(logz is called the principal value of the quantity Logz).
68. Evaluate: (1) Log4, Log (-1), log (-1); (2) Logi, logi; l±i (3) Log V2 ; (4) Log (2-3i), Log (-2+3i). 69*. Find the error in the reasoning leading to I. Bernoulli's paradox: (-z)2 = z2, hence 2 Log (-z) = 2 Log z, consequently, Log(-z) = Logz(!). 70. The initial value of Im f(z) at z = 2 is assumed to be zero. The point z makes a complete rotation anticlockwise along a circle with centre at the point z = 0 and returns to the point z = 2. Assuming that f(z) varies continuously during the motion of the point z, find the value of Im f(z) after the given rotation, if: (1) f(z) = 2Logz; (2) f(z) =Log 1/z; (3) f(z) = Logz-Log (z+l); (4) f(z) = Logz+Log (z+l). By definition, for any complex numbers a =F 0 and a a"= exp {a Log a}
(1)
or, if as usualt by e" we understand exp z, then a"= e"Loga
71. Find all the values of the following powers: (1) 1>'2 ; (5) i 1; 1-i )l+f (2) (-2)>' 2 ; (6) (
y2
21;
(3) (4) 1-1;
(7) (3-4i)l+i
(8) (-3+4i) 1+i.
t By eqn.(l) a== exp {zLoge} =exp {:z:(l+2nik)}. However, if nothing is said to the contrary, we shall consider that k = 0, that is, as usual e" = exp z.
10
PROBLEMS ON COMPLEX ANALYSIS
72. Show that in the case of a rational index (a.= m/n) the general definition of the power z• is identical with the usual definition:
z"n"' =
(Y'z)"'
(see also problem 5). 73. Are the sets of values of the following identical:
a••,
(a•)11, (a2)•!
By definition, the equation w = cos-1 z is equivalent to the equation z = cosw. The functions sin-1 z, tan- 1 z, cot-1 z and the inverse hyperbolic functions cosh-1 z, sinh-1 z, tanh-1 z, coth- 1 z are defined similarly.
74. Prove the following equalities (all the values of the roots a.re ta.ken): (1) cos-1z = - i Log(z+y(zll-1)); (2) sin-1z = -iLogi(z+y(zll-1)); (3) ta.n-lz = ; Log((i+z)/(i-z)) = (4) cot-lz =
:i
Log !+::;
~Log ((z-i)/(z+i));
(5) cosh-1z =Log (z+y(z2 -1)); (6) sinh-1z = Log(z+y(zll+l));
1 l+z (7) ta.nh-1z = - Log--;
2
1-z
1 z+l (8) coth-1z = - Log - - ·
2 z-1 75. Prove that for any value of cos-1z it is possible to choose a value of sin-1z, such that the sum of these values is equal to n/2. Prove a similar assertion for ta.n-lz and cot-1z. REMA.BE. The equalities sin-lz+cos-lz = n/2 and tan- 1 z+cot- 1 z = n/2 are always understood in the sense indicated in the present problem.
76. Show that all the values of cos-1 z a.re contained in the formula. cos-lz = ±iLog((z+y(z11-1)),
where by y(zll - 1) is understood any one of its values. 77. (1) For what values of z a.re all the values of the functions cos- 1 z, sin-1 z, and ta.n-1 z real!
11
FUNCTIONS OF A OOMI'LEX VARIABLE
(2) For what values of z does the function sinh-1 z assume purely imaginary values 1 78. Find all the values of the following functions: (1) sin-11/2; (5) tan-1(1+2i); (2) cos-11/2; (6) cosh-12i; (3) cos-12; (7) tanh-1(1-i) . (4) sin-1i;
§ 3. Functions of a complex variable In problems 79-85 it is required to determine the curves defined by the given equations. 79. z =I-it; 0 ~ t ~ 2. 80. z = t+it2; -oo < t < oo. 81. z = t 2 +it'; - 00 < t < oo. 3n . . ) n 82. z =a (cost+ismt; 2 ~ t ~2; a> 0 · 83. z = t+(i/t); -oo < t < 0. 84. (1) z = t+iy'(I-t2); - I ~ t ~ 1; (2) z = -t+i y'(l-t2); -1 ~ t ~ 0 (the arithmetic value of the root is taken). 85. (1) z =a (t+i-ie-"); -oo < t < oo, a> O; (2) z = ia+at-ibe-1 ; 0 ~ t ~·2n, a> 0, b > 0. 86. For the mapping w = z2 it is required: (1) To find the images of the curves x = 0, y = 0, x = y, lzl = R, argz =ex and explain which of them are transformed one-one; (2) To find the originals (in the z-plane) of the curves u = 0, ti= 0 (w = u+it1). 87. For the mapping w =.!__find:
z
(1) The images of the curves x = 0, y = 0, argz =ex, jz-11 = I; (2) The originals of the curves u = 0, t1 = 0. 88. For the mappings w =
z+_.!_ z
and w =
z-.!_ z
lzl
= R,
find the
images of the circles lzl = R. 89. For the transformation w = z +(I/z) find on the Z·plane the original of the rectangular net (u = 0, t1 = 0) of the w-plane.
12
PROBLEMS ON COMPLEX .ANALYSIS
90. What is the circle lzl = 1 transformed into in the mapping = z/(1-z2) 1 91. For the mapping w = es find: (1) The images of the curves x = 0, y = 0, x = y; (2) The originals of the curves fl=() (0 ~ () < oo). 92. Find the transformations of the rectangular net (x = 0, y = 0) of the z-plane by means of the functions:
w
(1) w
= z2 +z;
(2) w
= cothz;
(3) w
=e
21'.
93. Into what are the segments of the straight lines x = 0 and the straight lines y = 0, which lie in the strip 0 ~ y ~ :n; transformed by means of the function w = e:r: +z 1 94. What corresponds in the z-plane to the polar net lwl = R, arg w = ix in the transformations: 1
(1) w=e-=;
(2) w=e:r:•1
95. Find the limit points of the sets: (1) z = 1+(-l)•n/(n+l) (n = 1, 2, ... ); (2) z = 1/m+i/n (m, n are arbitrary integers); (3) z = p/m+ i q/n (m, n, p, q are arbitrary integers); (4) izl < 1. 96. Prove that from an infinite bounded sequence of points {zn} it is possible to select a convergent subsequence. 97. Prove the following propositions: (1) The convergence of the sequence {zn = Xn + iyn} is equivalent to the simultaneous convergence of the sequences {xn} and {Yn}· (2) For the limit lim Zn '# 0, to exist, it is necessary and n-+OO
sufficient that the limits lim lznl '# 0 and (for a suitable definition n-+OO
of argzn) lim arg 11-+00
Zn
should exist. If rm
zn
is not a negative numbel'
n-+OO
it is possible, for example, to take -:n: < arg Zn ~ n. In which cases is the convergence of the sequence {zn} equivalent only to the convergence of the sequence {lznl} 1 98. On the basis of the statements of problem 97 prove: (1) lim (I+z/n)n = ex(cosy+isiny); n-+0a
(2) lim [n('j/(z)-1)] = logr+it/J+2nik n-+oo
(k = 0, 1, 2, .•• ).
Jl'UNCTIONS OF A COMl'LEX VARIABLE
13
99*. The function f(z) defined in the neighbourhood of the point
z0 is said to be Heine continuous at the point z0 , if for any sequence {z11}, which converges to z0 , the condition lim f(z 11 ) =f(z0 ) is satisfied; 11-+00
the same function is said to be Cauchy continuous, if for any e > O there exists a c5(e) > 0 such that from the inequality Jz-z0 J < c5 it follows that Jf(z)-f(z0 )J < e. Prove that these definitions are equivalent (see, for example, [l, Chapter I, § 3.6]). . Rez z Re(~) zRez 100. The functions -z-' /ZI' lzj2' -Jz_J_ are defined for
z ..P 0. Which of them can be defined at the point z = 0 in such a way that they are continuous at this point 1 101. Are the functions 1 1 (l) 1-z' (2) l+zs continuous inside the unit disk (JzJ continuous 1
<
1)? Are they uniformly
1
102. (1) Prove that the function e -lif is uniformly continuous
in the disk JzJ
~
R with the point z = 0 deleted. 1
(2) Is the function e:O uniformly continuous in the same domain! 1
0
<
(3) Is the function e:O uniformly continuous in the sector Jzl ~ R, JargzJ ~ n/6? 1
103. The function w = e- z is defined everywhere except at
the point z == 0. Prove that: (1) in the semicircle 0 < Jzl ~ 1, Jarg zJ ~ n/2 this function is bounded but not continuous; (2) inside the same semicircle the function is continuous but not uniformly continuous; (3) in the sector 0 < JzJ ~ 1, Jarg zJ ~ct < n/2 the function is uniformly continuous. 104. The function f(z) is uniformly continuous in the disk JzJ < 1. Prove that for any point Con the circle JzJ = 1 and any sequence z11 -+ C, Jz11 J < 1 there exists the limit lim f(z 11 ). Prove also that this 11-+00
limit does not depend on the choice of the sequence and that if f (z) is defined on the boundary circle by means of passing to the limit, it is continuous throughout the whole of the closed disk Jzl ~ 1.
14
PROBLEMS ON OOMPLEX ANALYSIS
§ 4. Analytic and harmonic functions Tke Oaucky-Riemann equations 105. Verify that the Cauchy-Riemann equations are satisfied. for the functions z•, ez, cos z, Logz and prove that (z•)'
= nz•- 1 ,
(ez)' = ez,
(1 ogz) t
(cosz)' = -sinz,
1 =-. z
106. Write down the Cauchy-Riemann equations in polar coor. dinates. 107. Prove that if f(z) = u+iv is an analytic function and s and n are perpendicular vectors, the rotation from the vector s to the vector n through a right angle being anticlockwise, then
au
Ts=
av an
au
av
Tri= -Ts
and
(a fas and a fan are the derivatives of the function of two real variables with respect to the corresponding directions). 108. Prove that the functionf(z) = is not differentiable anywhere. 109. Prove that the function w = z Re z is differentiable only at the point z = 0; find w' (0). 110. Prove that the Cauchy-Riemann equations are satisfied. for the function f(z) = y(JxyJ) at the point z = 0 but the derivative does not exist. 111. Prove the following assertions: (1) If the limit
z
Jim [Re
.dz-+0
exists for the function w = and v1 exist and are equal.
f (z)
(2) If the limit lim [rm .dz....0
u1 and
~w] LJZ
then the partial derivatives
Uic
~w] exists then the partial derivatives LJZ
exist and u 1 = -Vic. (3) If it is assumed in advance that the functions u and v are differentiable then the existence of either of the limits mentioned in parts (1) and (2) ensures the existence of the other and consequently the differentiability of the function f(z). Vic
15
FUNOTIONS OF A OOMPLEX VARIABLE
112. At the point z0 the function w
= f(z)
possesses the following
properties: (1) u, v a.re differentiable; (2) the limit lim
.dz-+0
I~wl exists. LJZ
Prove that either f(z) or f(z) is differentiable at the point z0 • 113. At the point z0 the function w = f(z) possesses the following properties: (1) the functions u, v a.re differentiable; (2) the limit lim arg 4z....0
~w LJZ
exists. Prove that f(z) is differentiable at the point z0 •
114. f(z) = u+iv = efi- 9 is an analytic function. Prove that if one of the functions u, v, (!., (J is identically equal to a constant, then the function f(z) is also a constant.
Harmonie function8 A function u(m, y) which in some domain possesses continuous partial derivatives up to the second order inclusive and satisfies Laplace's equation
aau aau
Liu== a:i;1+ay•=0, is said to be a harmonic function. Two harmonic functions u(m, y) and v(m, y) connected by the Cauchy-Riemann equations
au av -=z-,
am
8y
are said to be conjugate.
115. Prove the following propositions: n
(1) A linear combination of harmonic functions}; c1u1 (x, y) i=l
is a harmonic function. (2) H the inversion transformation
is applied to the argument of the harmonic function u(x, y) the transformed function is harmonic. (3) H the transformation
x = "' (~. 'YJ),
y = 'ljJ(~. 'YJ), where (x2+?f). In problems 140-143 prove the existence of and find the analytic functions f(z) from the given modulus or argument. 140. = (xt+ys)eX. 141. = e""cos8'[>. 142. 0 = ey. 143. 0 = ef>+rsinef>. 144. Prove that for the family of curves cf>(x, y) = 0, where cf> is a twice continuously differentiable function, to be the family of level lines of some harmonic function it is necessary and sufficient that the ratio LJcf>/(gra.dcf>) 2 should depend only on ef>.
e e
HINT. As a preliminary establish that the required harmonic function is of the form u =/[•(a:, y)].
In problems 145-149 find analytic functions of which the real part, the imaginary part, the modulus or the argument preserves a constant value along any line of the corresponding family. 145. x = 0. 146. y = o. 147. y =Ox. 148. x11+y11 = 0. 149. xl+y2 =Ox. Phe geometrical meaning of the modulUB and argument of a derivative 150. Mappings are made using the functions w = z2 and w = z8• For each function find the angle of rotation (0) of the directions issuing from the point z0 , and the magnification (k) at the following points: (1) 210 = 1; (2) z0 = -1/4; (3) z0 = l+i; (4) z0 = -3+4i. 151. Which part of the plane is compressed and which is stretched if the mapping is effected by the functions: (1) w = z2; (4) w = e:r:; (2) w = z8 +2z; (5) w =log (z-1) 1 (3) w = l/z; 152. The domain G is mapped by means of the function f(z) conformally and one-one onto the domain G'. Find formulae for the calculation of the area S of the domain G' and the length L of the arc onto which some arc l belonging to the domain G is mapped. 153. Find the length L of the spiral onto which the function e:I: maps the segment y = x, 0 .:::;; x .:::;; 23t (see problem 91).
20
PROBLEMS ON OOMPLEX ANALYSIS
154. Find the area of the domain onto which the function e" maps the rectangle 1 < z < 2, 0 ,.s;; y < 4. 155. Find the domain D, onto which the function e" maps the rectangle 1 < z < 2, 0 < y < 8. Calculate the area of the domain D by means of the formula obtained in the solution of problem 152 and explain why this formula gives an incorrect result.
OHAPTER II
CONFORMAL MAPPINGS CONNECTED WITH ELEMENTARY FUNCTIONS § 1. Linear functions Linear function!J 156. Find the linear function which maps the triangle with vertices at the points 0, 1, i onto a similar triangle with vertices at 0, 2, l+i. 157. Find the linear transformation with fixed point 1+2i whioh transforms the point i into the point -i. 158. For the transformations given find the finite fixed point z0 (if it exists), the angle of rotation round it, (), and the magnification le. Reduce these transformations to the canonical form w-z0 = A.(z-z0). (1) w = 2z+l-3i; (4) w-w1 = a(z-z1 ) (a =F O); (2) w = iz+4; (5) w = az+b (a =f:: 0). (3) w = z+l-2i; 159. Find the general form of the linear transformation which transforms : (1) The upper ha.If-plane into itself; (2) The upper half-plane into the lower ha.If-plane; (3) The upper half-plane into the right ha.If-plane; (4) The right ha.If-plane into itself. Show that in each case the transformation is uniquely determined by specifying a single pair of corresponding interior points or two pairs of boundary points. 160. Find the general form of the linear transformation which transforms : (1) The strip 0 < z < 1 into itself; (2) The strip -2 < y < 1 into itself; (3) The strip bounded by the straight lines y = z and y = z-1 into itself. Explain which pairs of points can correspond to one another in these mappings and in which case this correspondence determines the mapping uniquely. 21
22
PROBLEMS ON OOMPLEX ANALYSIS
161. Find the linear function w(z), which maps the strip, contained between given straight lines onto the strip 0 < u < 1 with the given normalisation: (1) x =a, x = a+k; w(a) = O; (2) x=a,
x=a+k;
Imw(a+; +i) (3) y ='/ex, y (4) y = kx+bi,
<
w(a+
;)= !
+i;
1.
= kx+b; w(O) = O; y = kx+b 2 ; w(b1 ) =
0.
162. Find the linear function which maps the circle !z! < 1 onto the circle !w-Wo! < R and is such that the centres of the circles correspond to one another and the horizontal diameter passes into a diameter ma.king an angle a with the direction of the real a.xis, Bilinear functiona 163. For the function w = 1/z find the images of the following curves: (1) The family of circles x2+y2 = ax; (2) The family of circles zs+ys =by; (3) The pencil of parallel straight lines y = x+b; (4) The pencil of straight lines y = kx; (5) The pencil of straight lines passing through a given point z0 =F 0; (6) The para.bola. y = z2. 164. Explain into what the function w
forms:
= - 1-+k trans-
(1) The rectangular net x = 0, y = O; (2) The polar net !z-z0 ! = R, a.rg (z-z0 ) =
z-z0 Cl.
Z-Z1
165. Given the function w = - - : Z-Z2
=A (O < A< oo) is a family of circles (the circles of Apollonius). For a given A find the radius and the position of the centre of the
<
(1) Prove that the original of the family !w!
corresponding circle in the z-pla.ne. (2) Find the original of the ray arg w = (). (3) Construct the net in the z-plane which corresponds to the polar net in the w- plane.
23
CONFOBMAL MAPPINGS
(4) Find the domain of the z-plane which corresponds to the semicircle lwl < 1, Im w > 0. In problems 166-170 explain what the domains indicated are transformed into with the given mapping functions. 166. The quadrant :x:
> 0,
y
> O;
< 1,
Imz
z-i z+i
w = -..
w = 22 z-:i. +iz 168. The angle 0 < < :rc/4; w = z/(z-1). 169. The strip 0 < :x: < 1; (1) w = (z-1)/z; (2) w = (z-1)/(z-2). 170. The ring 1 < lzl < 2; w = z/(z-1). 171. Map onto the vertical strip 0 < Rew < 1: (1) The half-plane Re z > 0 with the disk lz-d/21 d/2 deleted; (2) The lune contained between the circles
167. The semicircle lzl
> O;
<
lz-d1 /21 = d1 /2, lz-d2 /21 = d2 /2
(d1
<
d2 );
(3) The exterior of the circles lz+d1 /21 = di/2, lz-d2/21 = d2 /2 so that w(d2 ) = 0. 172. Find the bilinear function which transforms the points -1, i, 1+i respectively into the points: (1) 0, 2i, 1-i; (2) i, oo, 1. 173. Find the bilinear function which transforms the points -1, oo, i respectively into the points : (1) i, 1, l+i; (2) oo, i, 1; (3) 0, oo, 1. 174. Find bilinear functions from the following conditions: (1) The points 1 and i are fixed and the point 0 passes into the point -1; (2) The points 1/2 and 2 are fixed and the point 5/4+3i/4 passes into oo; (3} The point i is a double fixed point and the point 1 passes into oo. 175. Find the bilinear function which transforms the points -1, 0, 1 respectively into the points 1, i, -1, and explain what the upper half-plane becomes in this mapping. 176. Find the general form of the bilinear transformation whioh transforms: (1) The upper half-plane into itself; (2) The upper half-plane into the lower half-plane; (3) The upper half-plane into the right half-plane.
24
PROBLEMS ON COMPLEX ANALYSIS
177. Find the mapping of the upper half-plane into itself \\ith the given normalisation: (1) w(O) = 1, w(l) = 2, w(2) = oo; (2) w(O) = 1, w(i) = 2i. RBMABK. For mappings of the upper half·plane into itself for other normalisations see problem 188.
178. Find the function w(z) which maps the circle lzl < R onto the right half-plane Re w > 0 in such a way that w(R) = 0, w (-R) = oo, w (0) =I. What is the image of the upper semicircle in this mapping 1 Two points P 1, P 1 are said to be symmetrical with respect to the circle K with centre 0 and radius R, if they lie on the same ray issuing
from 0 and OP1 .0P1 = R 8 •
179. Find the points symmetrical to the point 2+i with respeot to the cii-cles: (1) lzl = 1; (2) lz-il = 3. 180. Find the symmetrical image with respect to the unit circle of the following curves: (1) lzl = 1/2; (4) lz-zol = l: 1); (6) the hyperbola. x2-y2 = 1; (3) y = 2; (7) The perimeter of the rectilinear triangle with vertices Zt• Z,, Z:1(Z1 '¥: 0). 181. Prove that for the symmetry of the points P 1 and P 2 with respect to K it is necessary and sufficient that one of the following two conditions should be satisfied: (1) Every circle K 1 passing through the points Pv P 2 , is orthogonal to K; (2) For all points M of the circle K
MP1 --=Const MP2
(that is, K is a circle of Apollonius with respeot to the points P 1 and P 2). 182. The function w =
z-{3
e'•---= z-{3
({3 = a+ib, b
> 0)
CONFORMAL MAPPINGS
25
maps the upper half-plane onto the unit disk. (1) Find arg w(z) = O(z); (2) Find w' ({3); (3) Explain what part of the upper half-plane is compressed in this mapping and what part is stretched. 183. Map the upper half-plane Im z > 0 onto the unit disk lwl < 1 in such a way that: (1) w(i) = 0, arg w' (i) = -7'/2; (2) w(2i) = 0, arg w' (2i) = 0; (3) w(a+bi) = 0, arg w'(a+bi) = 0 (b > 0). 184. Map the upper half-plane Im z > 0 onto the disk lw-w0 1 < R so that the point i passes into the centre of the disk and the derivative at this point is positive. 185. Map the disk izl< 2 onto the half-plane Rew>O in such a way that w(O) = 1, arg w'(O) = 7'/2. 186. Map the disk iz-4il< 2 onto the half-plane v > u so that the centre of the disk passes into the point -4, and the point 2i on the circumference becomes the coordinate origin. 187. Find the general form of the bilinear function w(z) which maps the disk izl O in such a way that w(z1 ) = 0, w(z2 ) = oo, where Zi· z2 , are specified points on the circumference izl = R. Construct the family of curves in the circle lzl < R corresponding to the polar net in the half-plane Rew> 0. HINT. It is possible to use either the general form of the transformation of the half-plane into a disk or the general form of the bilinear transformation for three pairs of corresponding points and the result of problem 8.
188. Find the function which maps the upper half-plane onto itself BO that w(a) = b, argw'(a) =ix (Ima>O, Imb>O). HINT. As a preliminary map both specimens of the half-plane onto the unit disk with the corresponding normalisation.
189. Map the upper half-plane onto the lower one so that w(a) =ii and arg w'(a) = -7'/2 (Ima>O). 190. For the function
z-a
w = e1• - - (lal 0) pass into themselves. 209. Prove that the linear transformation w=
z-a
e"·---1-az
(a=
lale1«, lal <
1),
which transforms the unit circle into itself can only be elliptic, hyperbolic or parabolic. Explain for what values of a each of the given cases holds. Find the fixed points of the transformation and reduce it to the canonical form. Some approximate formul,ae for linear transformations 210. The upper half-plane is mapped onto the unit disk so that the point z = ki (k > 0) passes into the centre of the circle. Find the length I' of the image of the segment [O, a] of the real axis (a> 0) and obtain linear approximate formulae for I' for small a/k and for small k/a. 211. The unit disk is mapped onto itself in such a way that the original of the centre of the disk, the point x 0 , is on the real axis. Find the length I' of the image of the arc 0 ~ cf> ::( y of the unit
CONFORMAL MAPPINGS
29
circle (y::,;: n). How does the quantity I'/r vary depending on the sign of x0 1 212. With the conditions of problem 211 obtain the formulae: l+xo (1) I'= - 1 - r+o (y8) for small 'Y; -Xo
(2)
I'= n-ecot..l- ~cot2 1..+o(e8 ) 2
2
2
for small e, where
l-x0 • 213. The unit disk is mapped onto itselfso that the point z0 = r0e1 • passes into the centre. The points z1 = e1• 1 and z2 = e1•• lie on opposite sides of the diameter passing through z0 (c/> 0 < c/>1 < 2 ::,;: c/> 0 +n). Assuming that the point z0 is situated close to the unit circle, prove that the length I' of the image of the arc c/> 1 ::,;: cf> ::,;: 2 of the unit circle satisfies the formula 8 =
where e = l-r0 •
Mappings of simply connected domains 214. Prove that if the linear mapping of the disk izl < 1 onto itself does not reduce to a rotation, then no concentric ring with centre at the coordinate origin passes into a concentric ring. This proposition is a particular case of the following theorem: For the conformal mapping of the ring r 1 < lzl < r 8 onto the ring R 1 < lwl < R 1 it is necessary and sufficient that the condition R 1 /R1 = r 1 /r1 should be satisfied. In this case the mapping function can only have two forms: w = az or w = a/z. The mapping is uniquely determined by the specification of a single pair of boundary points (see, for example, [3, Chapter II, § 3]). REMARK.
2Ui. (1) Map the ring 2 < jzj < 5 onto the ring 4 < lwl < 10 so that w(5) = -4. (2) Map the ring 1 < jz-2il < 2 onto the ring 2 < jw-3+2il < 4 so that w(O) = -l-2i. The following theorem holdst: Every doubly connected domain, the boundaries of which do not shrink t See, for example, the paper: M. V. KELDYsH, (1939) Conformal representations of multiply connected domains (Konformnyye otobrazheniya mnogosvyaznykh oblastei), Uape'/chi, matem. naulc, vol. VI, 90-119.
30
PROBLEMS ON COMPLEX ANALYSIS
to points, can be conformally mapped onto a concentric ring with well-defined ratio µ of the radii of the outer and inner circles (µ is the modulus of the doubly connected domain).
216. Map the semicircle Rez > 0 with the disk lz-hl < R (h > R) deleted onto the ring (! < lwl < 1 so that the imaginary axis passes into the circle jwj = 1. Find !!· HINT. Construct the circle with centre at the coordinat.e origin and ortho· gonal to the circle lz-hl = R; then find the linear transformation, which trans• forms the real axis and the constructed circle into two int.ersecting (orthogo· nally) straight lines and verify that the given domain is then mapped into the concentric ring. Prove that the centre of this ring is coincident with the coordinate origin if the points of intersection of the constructed circle and the real axis pass into 0 and oo.
217. Map the half-plane Rez > 0 with the disk jz-hj < 1, h > 1 deleted, onto the ring 1 < jwl < 2. Find h. 218. Map the eccentric ring bounded by the circles lz-31 = 9, jz-81 = 16 onto the ring (! < jwj < 1. Find p. 219. Map the doubly connected domain bounded by the circles lz-z1 j = r1 , jz-z 2 1= r 2 (lz2 -z1 j > r1 +r2 or lz2 -ztl < jr2 -r1 1) onto a concentric oircular ring with oentre at the coordinate origin. Find the modulus (µ) of the domain. HINT. Find a pair of points symmetrical with respect to both circles and map one of them onto 0 and the other onto oo. REMARK. It is easily verified that the methods of solution recommended in the hints to problems 216 and 219 a.re the same.
220. "Q'sing the solution of the preoeding problem find the moduli of the doubly connected domains bounded by the given circles: (1) jz-11=2, lz+ll = 5; (2) lz-3il = 1, jz-4J = 2. Group properties of bilinear transformations The transformation T(z) = T 8 [T1 (z)] is said to be the product of T 1 and = T 1 T 1 (the order is important since, generally speaking, T 1T 1 of: T 1 T 1 ). The set G of transformations T forms a group if it contains the product of every two transformations belonging to it and together with the transformation T contains the transformation T-1 inverse to it. A group consisting of the powers T1I and p-n of a single transformation T is said to be cyclical. If the group G is formed from the transformations TI> T 8, ••• , Tn by the construction of all the inverse transformations and of all possible products of the given transformations and their inverses, these transformations are said to be generatora of the group G. The points obtained from the fixed point z by means of all the transformations of the group G are said to be equivalent or congruent with respect to the group G. The fundamental domain of the group is an open set (connected or discon· nected) which does not contain a single pair of points equivalent to one another T 8 and is written in the form T
31
CONFORMAL MAPPINGS
with respect to the given group, and in the neighbourhood of every boundary point of which there are points equivalent to the points of the set.
221. Let T 1 be the linear transformations:
la,
LI,= db11· ¥: 0 (i = 1, 2, ... ). c1z+d1 c, 1 Prove the following assertions: (1) T = T 1T 2 is a linear transformation with determinant LI = Ll 1 L1 2• (2) The product of the transformations is associative, that is, T,(z) = a1z+b1'
(T3 T 2)T1
=
T 3 (T2T 1 ).
(3) Every transformation T 1 has an inverse Tj 1, that is,
T 1Tl 1 = T'j' 1T 1 = J, where J(z) = z is the identical transformation. (4) The product of transformations, generally speaking, is not commutative (give examples). 222. Prove that the transformations 1
1
T 3 = 1-z, T , = - - , z 1-z, z-1 z -T,,=--, T a--z-1 z form a group (the group of anharmonic ratioB). 223. Prove that the set of linear transformations which consists of the rotation of a plane round the coordinate origin by angles which a.re multiples of cc, forms a cyclical group. In which case does this group consists of a finite number of transformations 1 224. (1) Prove that the set of transformations of the form w = (az+b)/(cz+d) where a, b, c, d a.re real integers and ad-be= 1, forms a group (this group is said to be modular). (2) Prove that if a, b, c and d a.re considered as complex integers (that is, numbers of the form m+ni, where m and n a.re real integers), which satisfy the condition ad-be= 1, then the set of transformations of pa.rt (1) also forms a group (Picard'B group). 225. Find the fundamental domains of the groups generated by the transformations :
T2 = -
ll1fl
(1) T(z) =en z (n a natural number);
32
:PROBLEMS ON COMPLEX ANALYSIS 27d
(2) T 1 (z) = enz, T 2 = 1/z; (3) T(z) = z+w; (4) T 1 (z) = z+w, T 2 (z) = -z; (5) T 1 (z) = z+ro1 , T 2(z) = z+ro 2 (the (Imro 2/ro1 =F 0) doubly periodic group) ; (6) T 1 (z) = z+ro1 , T 1 (z) = z+w1 , T 3 (z) = -z; (7) T 1 (z) = z+w; T 2 (z) = iz; (8) T 1 (z) = z+w, T 2 (z) = e9"'18 z; (9) T 1 (z) = z+w, T 2 (z) = a21Cil&z. 226. Find the group of linear transformations corresponding
in stereographic projection to the rotation of the sphere: (1) Round the vertical diameter; (2) Round the diameter parallel to the real axis; (3) Round the diameter parallel to the imaginary axis; (4) Round the diameter, the stereographic projection of one of the ends of which is the point a. HINT. If z1 , z1 are the images of diametricaJly opposite points on the sphere, then z1ii = -1 (see problem 48 of Chapter I).
227. (1) Prove that the group of linear transformations which correspond to a rotation of the sphere, and transform the points with stereographic projections a and b into one another, is defined by the relation w-b z-a ---=e'•-l+bw l+az · (2) Prove that the differential
ldzl ds = 1+1z12 is invariant with respect to the transformations of this group and represents the spherical length of the element of arc dz (that is, the length of the image of this element on the sphere).
Linear transformations and non-Euclidean geometry In the realisation of non.Euclidean geometry in the unit circle the part of straight lines is played by arcs of circles in the unit circle and orthogonal to it; the role of motion is played by linear transformations of the unit circle onto itself, the role of distance between the points Zi and z1 by the quantity !,>(Z1 , Z 2 )
=
1
2log(a, p,
z., z1 ),
where a and p are the points of intersection of the "straight lines" passing through the points Zi and z2 , with the unit circle (the order of the points is
CONFORMAL MAPPINGS
33
a, z1 , z1 , fl), and (a, fl, z9 , z1 ) is the anharmonic ratio of the given points. The angles are measured as in Euclidean geometry (see, for example, [I, Chapter II, § 4, section 8]).
228. Prove that Q(Z1, Z2) > 0 if Z1 i: Z2 and Q(Z, z) = 0. 229. Prove that Q(Z1, Z3) ~ Q(Z1, Z2)+Q(Z2, Z3) where the sign of equality applies only when the point z3 lies on the "segment", connecting the points z1 and z2 • 230. Prove that if one of the points 211 and z2 tends to a point of the unit circle (or both of them to different points of the unit circle), then the non-Euclidean distance e(z1 , z2) tends to infinity (that is, the points of the unit circle correspond to the infinitely distant points of the non-Euclidean plane). 231. Prove that the differential
ds=~ l-lzl2
(lzl 0 is mapped by the function w=z+z•. 238. (1) Find the domain onto which the disk lzl < 1 is mapped by the function w = R(z+(zn/n)), R > 0, w = R(z+mz•),
where n is an integer, n > 1. (2) Find the domain onto which the exterior of the unit disk lzl > 1 is mapped by the function w = R(z+ (1 /nzn)), R > 0, n an integer, n > 1. RE111A.:ax. For mappings involving the function w
= B(z+ (1/z))
(Zhukovskii's function) see problem 261 et 11eq.
239. (1) Explain for what values of m the function w = R(z+mzn), where n is a natural number, effects the conformal mapping of the disk jzj O onto the upper half-plane. 244. Map onto the upper half-plane: (1) The sector lzl ) = In problems 277-280 find the domains obtained from the mapping of the given domains by the functions indicated. 277. The disk lzl < 1; w = z/(z2 +1). 278. The semicircle lzl < 1, Imz > O; w = l/(z2 +1). 279. The angle 0 < a.rg z < :n/n; w = 1/2 (z"+l/z"). 280. The sector - :n/n < a.rg z < :n/n, lzl < 1;
IX.
z
w=---2-
( w(z)
>0
for
z
>
0 )·
(l+z")" IIINT. Represent the mapping function in the form w = F{/[t/>(z)]}, where t F(t) = t/>(t) = t", J(t) = (l+t) 2 ,
j/t.
281. (1) Using the solution of problem 280 and the 'principle of symmetry find the image of the unit circle in the mapping
z
w=---2-·
(l+z")" (2) Find the function which maps the interior exterior) of the unit circle onto the exterior of the "star" :
(and
2:n;lc (I~= 0, 1, 2, ... , n - 1). a.rgw = - n 282. Map onto the exterior of the unit circle: (1) The whole plane with cuts a.long the segments [-1, l] and [ - i, i] (the outside of a. cross); (2) The whole plane with cuts a.long the rays (-ex>, -1 ], [l, + ex>), (- i=, - i] and [i, + i=). 283. (l)* Using the function of problem 279, map the sector lzl < 1, 0 < a.rg z < :n/n (n an integer) onto itself so that the radial segments lzl 0);
43
CONFORM.AL MAPPINGS
(3) The angle 0 < arg z < oi ::;:; 2~; (4) The sector lzl < I, 0 < arg z < oi ~ 2n; (5) The ring r 1 0 and in the lower half. plane for 'Y/o < 0. The line 'Y/ = 0 corresponds to the segment [-a, a]. The arcs corresponding to the values 'YJ = 'Y/o and 'Y/ = 'Y/o-n ('l}o > 0), supplement one another and make up a complete circle (Fig. 7). !I
:r
FIG. 6
!I
z acot'lo
.r
FIG. 7
(5) Find the magnitude of the segments b (see Fig. 6) and l (see Fig. 7). REMA.BE. A coordinate net constructed in this way in the z-plane is called a bipolar nel.
OONFORMAL MAPPINGS
45
303. Explain what the mapping w = cosz transforms the following into: (1) The rectangular net z = 0, 11 = O; (2) The half-strip 0 < z < n, 11 < 0; (3) The half-strip 0 < z < n/2, 11 > O; (4) The half-strip -n/2 < z < n/2, y > O; (5) The strip 0 < z < n; (6) The rectangle 0 < z < n, -k < 11 < k, (k > 0). 304. Explain what the mapping w = sin-1 z transforms the following into; (1) The upper half-plane; (2) The plane with cuts along the real axis along the segments (-, -1], [1, ); (3) The first quadrant; (4) The half-plane z < 0 with a cut along the real axis along the segment (-, -1]. 305. Explain what the mapping w = cosh z transforms the following into: (1) The rectangular net z = 0, 11 = C; (2) The strip 0 0, 0 < 11 < n. • · Explain what the mapping w = sinh-1z transforms the following into: (1) the plane with cuts along the imaginary axis along the rays 1 ~ 11 < and - < 11 ~ -1; (2) the first quadrant. 307. Explain what the mapping w = tanz transforms the following into: (1) The rectangular net z == 0, 11 = O; (2) The half-strip 0 < z < n, 11 > O; (3) The strip 0 < z < n; (4) The strip 0 < z < n/4; (5) The strip -n/4 < z < n/4. 308. Explain what the mapping w = cothz transforms the following into: (1) The strip 0 < 11 < n; (2) The half-strip 0 < 11 < n, z > 0.
46
PROBLEMS ON COMPLEX ANALYSIS
In problems 309-314 map the given domains onto the upper half-plane. 309. The strip bounded by the straight lines 11 = :i:, 11 = :i:+k. 310. The half-strip :i: < 1, 0 < y < k. 311. The circular lune bounded by the circles lzl = 2, lz-11=1. 312. The domain bounded by the circles lzl = 2, lz-31 = 1 (the plane with the disks deleted). 313. The domain defined by the inequalities: lz-11
> 1,
lz+ll
> 1,
Imz
>0
(the upper half-plane with two semicircles deleted). 314. The domain included between the confocal parabolas ya= 4(:i:+l),
y" = 8(:i:+2).
HINT. See problem 234, (2).
315. Find the function w(21), which maps the domain bounded by the circle lzl = 1 and the straight line Imz = 1 (the half-plane Im z < 1 with the disk deleted): (1) On the disk lwl < 1 with the normalisation w(-3i) = 0, argw'(-3i) = n/3; (2) On the disk lwl < 1 with the normalisation w(-3i) = (-l+i)/2, argw'(-3i) = n/2; (3) On the upper half-plane with the normalisation w(-3i) = l+i, arg w'(-3i) = n.
316. Map on the upper half-plane: (1) The strip 0 < :i: < 1 with a cut along the line :i: = f, k ~y < oo; (2) The strip 0 < :i: < 1 with cuts along the lines :i: = f, hi ~ y < oo; and :i: = ·!. -oo < y ~ k2 (k2 < k1 ). HINT. First map the strip 0 < a: < ! onto the upper half-plane. By the principle of symmetry the mapping function will map the given domain on the whole plane with certain cuts.
In problems 317-327 map the domains indicated onto the upper half-plane. 317. The strip 0 < :i: < 1 with a cut along the segment 0 < :i: < k, y = 0 (k < 1). 318. The strip 0 < :i: < 1 with cuts along the segments 0 ~ :i: ~ k1 , 11 = 0 and 1-h.2 ~ :i: ~ 1, y = 0 (k1 +k 2 < 1).
CONFORMAL MAPPINGS
47
319. The half-strip 0 < x < n, y > 0 with a cut along the segment x = n/2, 0 ~ y ~ k. 320. The ha.If-strip 0 < x < n, y > 0 with a cut along the ray x = n/2, k ~ y < oo (k > 0). 321. The half-strip 0 < x < n, y > 0 with cuts along the segment x = n/2, 0 ~ y ~ k1 and along the ray x = n/2, k2 ~ y < oo (k2 > k1)· 322. The domain bounded by the circles lz-11=1, lz+ll = 1 with a cut along the ray 2 ~ x < oo, y = 0. 323. The domain bounded by the circles lz-11=1, lz-21 = 2 with a cut along the segment y - 0, 2 ~ x ~ a (a < 4). 324. The domain bounded by the circles lz-11=1, lz-21=2 with cuts along the segments y = 0, 2 ~ x ~a and y = 0, b ~ x ~ 4
(a< b). 325. The domain bounded by the imaginary axis and the circle 1, with cuts along the segment y = 0, 2 ~ x ~a and along the ray y = 0, b ~ x < oo (a < b). 326. The domain bounded by the circles lz-11=1, lz+ll = 1, with a cut along the segment x = 0, -ix ~ y ~ fl (ix ~ 0, fl ~ 0). 327. The domain lz-11>1, lz+ll > 1, Im z > 0 (the upper
lz-11 =
half-plane with semicircles removed) with a cut along the segment x=O, 0 ~y ~k. 328. Map the interior of the parabola y 2 = 4oi2 (x+oi2) onto the upper half-plane and onto the unit disk. HINT. Make a cut along the axis of symmetry of the parabola, map the upper half of the parabola onto a half-strip (by means of the function yz) and then onto the half-plane, and use the symmetry principle.
329*. Map the upper half-plane with cuts along the segments 0 ~ y ~a, x = n/2+kn (k = 0, ±1, ±2, + ... ) onto the upper half-plane (Fig. 8).
II
Fm. 8
48
PROBLEMS ON COMPLEX ANALYSIS
330. Map the plane with parallel cuts -a ~ x ~ a, y = n/2+kn (k = 0, ±1, ±2, ... ) onto the plane with cuts along the real axis segments [kn-b, kn+b] (k = 0, ±1, ±2, ... ; 0 < b < n/2). HINT. Make supplementary cuts along the imaginary axis, map one of the domains formed onto the upper half-plane and use the principle of symmetry.
331. Map the plane with cuts along the rays (-oo, -n/2], [n/2, +oo) and along the segments -a ~ y ~a, x = n/2+kn (k = 0, ± 1, ± 2, •.. }onto the exterior of the unit circle (Fig. 9).
FIG. 9
HINT. Map the function which gives the solution of problem 329 onto the plane with cuts along the rays ] l ( -co sin-1 sech a ' '
[sin-1 ~ch a' +oo] '
332. Map the plane with cuts along the rays (-oo,p], [q, +oo} (-n/2 ~ p < q ~ n/2) and along the segments -a ~ y ~a, x = n/2+kn (k = 0, ±1, ±2, ... ) onto the upper half-plane (Fig. 10).
FIG. 10
333*. Map the plane with cuts along the rays 0 ~ y x = kn/2 (k = 0, ±1, ±2, ... ) onto the upper half-plane.
< oo,
CONFORMAL MAPPINGS
49
In problems 334-337 the images are many-sheeted domains (see the footnote on p. 41). 334. Find the domains onto which the following are mapped by means of the function w = es: (1) The rectangle 0 < x 0; (3) The strip 0 < x < a. 335. Find the domains onto which the following are mapped by means of the function w = cos z: (1) The strip -n/2 < x < :i/2; (2) The strip 0 < x < 2n. 336. Find the domain onto which the strip 0 < x < 2n is mapped by means of the function w = tanz. 337. Construct the Riemann surface onto which the function .!
w =es maps the z-plane.
§ 5. Boundaries of univalency, convexity and starlikeness In problems 889-345 r 1 denotes the maximum radius of a circle with centre at the coordinate origin within which the function w = f(z) is univalent; r 1 denotes the maximum radius of a disk with centre at the coordinate origin which the function w = f (z) maps schlicht onto a convex domain and r 3 the maximum radius of a disk with centre at the coordinate origin mapped by the function w = f(z) schlicht onto a domain which is sta.rlike with respect to the point w = O. (A domain is atarlike with respect to a given point, if any point of the domain can be joined to the given point by a rectilinear segment lying wholly within the domain.) It is obvious that r 1 .;:;:;; r 3 .;:;:;; r 1•
338. Find r 1 , r 2, r 3 for the function w = z/(1-z) and construct the images of the disks izl < rv izl < r 2, izl < r 3 • 339. Find r 1 for each of the following functions: (1) w = z+z2; (2) w = z+a.z2, (a is a real number);
z (3) w = (l-z>9.
340. Prove that in the mapping w = f(z) the curvature of the image of the circle izl = r is expressed by the formula
f"(z)]
l+Re [z7fz) k=
jzj'(z)j
50
PROBLEMS ON COMPLEX ANALYSIS
341. Prove that an analytic function /(z) maps the circle Jzl = r onto a convex curve when and only when
a [n
]= l+Re [zf"f'(z)(z)] ;;,:: O
a(O) = O, t/>(l) = 1, then the paths z{t) and z[t/>{'r)l are considered identical. Two paths ,,0 , r 1 with common ends a, b are said to be cominuoualy deformable into one another or homotopic in the domain G, if there exists in G a family of paths 0 ~ .il ~ I, with ends
A1(t) - ~A. (t)I -+ 0 as .ili -+ .il1) and such that the given paths ro· r1 correspond to the values .il = 0, .il = I. The closed path r is said to be continuoualy deformable into
a point or homotopic to zero in the domain G if it is homotopic in the domain G to a null path, that is, a path consisting of a single point.
358*. Prove the following assertions: (1) Every path yin an arbitrary domain is homotopio to a polygonal path. (2) Every closed path in a circle is homotopio to zero. (3) Every closed path in a circular ring is either homotopic to zero or homotopio to any boundary circle, traversed in a definite direction one or several times. 359. Let y be an arbitrary path, closed or not. Prove the following assertions : (1) arg (z-a), z e y, a y, are uniquely defined as continuous functions of z, if arg (zti-a) is given for an arbitrary point z0 e y.
e
54
PROBLEMS ON COMPLEX ANALYSIS
(2) The quantity LI,. arg (z-a), equal to the increment of arg (z-a) along y, does not depend on the choice of z0 or arg (z0 -a). (3) H the path r lies within an angle with vertex at the point a of magnitude IX (0 < IX < 2n) then LI.a arg (z-a) < IX. (4) For a smooth (or even only rectifiable) path r the equality Ll 7 arg (z-a) = holds. If y is a closed path and a
ey
n,a (a)
=
J,, darg(z-a).
then the number I
2:n Ll 7 arg (z-a)
is said to be the indez of the point a with respect to y (or the winding number of y with respect to the point a).
360. Prove the following assertions: (1) In the homotopio deformation of the path r taking place outside the point a the index n,.(a) is unchanged. (2) In every closed path r it is possible to inscribe a closed polygonal curve n such that n 7 (a) = nn(a) (a is a given point outside y). (3) If the points cii, a1 can be connected by a path which does not encounter the path y, then n 7 (a1 ) = n 7 (a9). (4) If the point a and the path rare separated by some straight line then n 7 (a) = 0. 361. Prove that for a simple closed polygonal path n the index n,.(a) = ± 1, if a lies within n (the sign depends on the direction of traversal), and n,.(a) = 0 if a lies outside n. 362*. Let the function w = f(z) effect the topological mapping of the domain G on the domain G'. Let Q be a small circle in G, r a closed path in Q and a a point of Q which does not belong to y. Let, finally, r' and a' be the images of r and a in G'. Prove that n 7 ,(a') = n7 (a)c5,
where c5 equals either +1 or -1, independently of r and a. In particular prove that if the mapping is continuously differentiable and the Jacobian J
=
au !!!.. - ~ !!!.. =I= 0 ax oy oy ax '
then the sign of c5 is identical with the sign of J.
SUPPLEMENT.A.BY GEOMETRIC.AL QUESTIONS
55
363*. Let w = /(z) effect the continuous mapping of the closed region G bounded by the Jordan curve y. Prove that if f(z) ¢a on y and for the image y' = f(y) of the curve y the index n 7,(a) ¢ 0, then within G, f(z) assumes the value a. 364. Prove the equivalence of the following three definitions of a. simply connected domain: (1) The domain G is said to be simply connected if every closed pa.th lying in G is homotopic to zero in this domain. (2) A domain G which does not contain the point at infinity is simply connected if together with every simple polygon situated in G it also contains its interior. (3) A domain G different from the extended complex plane is simply connected if it is bounded by a. single continuum or point. RElllABX.
The extended complex plane is simply connected by the dret
definition.
§ 2. Quasi-conformal mappings. Generalised analytic functions If in the relation /(111, fl)
=
1(
z;i •
z~i )
= • (z, i)
z and i are considered as independent variables the derivatives with respect to these variables are
-i- = ! (:Ill -
l
~).
! = ! (~ + i ~ ) .
In what follows the notations
au
8111 =
etc. will be used.
365. Prove the following relations: (1) w11 = (2) w11 =
! !
[(ux+v,)+i(-u,+vx)]; [(ux-v1 )+i (u,-vx)];
(3) dw = w11 dz+WjdZ;
(4) ux =
! (w +wi+w.+w 11
11 );
"'x•
56
PROBLEMS ON COMPLEX ANALYSIS
-} ; (5} u1 = 2i(Wz-Wi-wz+wr: (6)
Vx
i
-
-
= -2(w.z-wi+wz-wr:};
(7) v1
- }. = 2I (wr:+wz--wi-wr:
368. Prove that the Cauchy-Riemann equations are equivalent tow;= 0. 387. Prove that Laplace's equation Liu= 0 can be written in the form fJlu/fJz(}i = 0. 368. Prove that
wr: =wi, iOz =wr:, dw=dw (the long bar over a symbol indicates passage to the conjugate value after differentiation). 369. Prove that for the function z(w}, inverse to w(z}, dz=
lw.l~lw1l 1 dw+ lwr:l;:i~il1 dw,
370. Prove that the Jacobian of the transformation w(z} is given by
J w/r:
=
D(u,v} D(:.i:, y}
' 1• . = IWz 11 -,wi"
371. Prove the following equalities:
dw
(I} dz
=
wr:+wie-•1•,
maxi~: (3) mini~: (2)
I= I=
where
IX=
argdz;
lw.l+lwsl; lwr:l-lwil.
372. Prove that if IX= argdz, rt.'= argdw, then (I} drt.' =
d O!
Jw1r:
(uxcosrt.+u, sinoc}1 + (vxcosrt.+v, sinoc}1 11
=
dz Idw I
I Jr:fw;
Sm'l'LEMENTABY GEOMETRICAL QUESTIONS
da:'
(2) max da: = p,
57
min--=da:' da:
1 p'
maxl*I
where
minl~~I
p=
-
lw.l+lw.I liw.1-lwill
373. The 8et of monogeneity ID?. of the function w = u+iv = f (z) at the point z is the set of derived numbers at this point, that is, the set of all possible limiting values of the ratio LJw/LJz as LJz-+ 0. Prove that if u and v are differentiable at the point z, then IDl. is either a point or the circumference of a circle. HINT. Use the relation 3'71 (1).
374. By the characteristics of an ellipse is meant the ratio p of its semi-axes (p;;;:::: 1) and, if p #: 1, the angle 8 (0 ~ 8 < n), formed by the major axis of the ellipse with the axis Oz. Show that the equation of such an ellipse with centre at the coordinate origin and semi-axis minor h can be written in the form
where
~=
pcos1 8+
~ sin•O,
r
fJ =
(p- ~)cos8
sin8,
= psin18+2-cos11 8.
p
REMARK. The quantities at, (J, y are also 1ialled characteristics of the ellipse. They are connected by the relation
rx.y-(J• 'l'he function w w-plane, if
l
== /(z) effects an affine
=
1.
(linear) mapping of the z-plane on the
bil
where LI= a1 of:.0. a1 b1
375. Prove that in an affine transformation parallel straight lines are transformed into parallel straight lines, and families of circles into families of ellipses.
58
PROBLEMS ON COMPLEX ANALYSIS
What condition is it necessary to impose on the coefficients of the transformation in order that circles should be transformed into circles (see, for example, [3, Chapter II, § I]) 1 376. The characteristics of an affine transformation a.re the characteristics p, 8 or cc, fl, y of the ellipses which a.re transformed into circles. Prove that: (1) - " - -
-fl
__ cc _ _ _!_.
bf+bl - LI '
c4+a; - a1b1+a1b1 (2) p=K+y(K2 -I),
where K=
cc~_y-;
(3) ta.no= y-cc+y[(:pcc)1+4p•] .
377. By the characteristics of the continuously differentiable mapping w = u(x, y)+iv(x, y) with Jacobian J > 0 is meant the characteristics p(z), O(z) or cc(z), fl(z), y(z) of the affine transformation du = uxdx+u1 dy, d v = vxdx+v,.dy. In this to the infinitely small circle
=de•
du11 +dv1 there corresponds the infinitely sma.11 ellipse ydxl-2fldxdy+ccdyl = pdhl (dh is the minor semi-a.xis). Prove the relations:
SUPPLEMENT.ARY GEOMETRICAL QUESTIONS
59
378. A one-one continuously differentiable mapping w = u+itJ with Jacobian J > 0 is called a quasi-conformal mapping with characteristics p(z), O(z) or a(z), {J(z), y(z) if it transforms infinitely small ellipses with these characteristics into infinitely small circles. Prove that such a mapping satisfies the system of equations
au,.+flu1 = tJ1 ,
f3u,.+,,u,
= -v,..
which can also be written in the form q(z) HINT. Use the first relation for Gt, nection between Gt, fJ, y and p, 8.
fJ,
=
p-l - - - e119 • p+l
y given in problem 877 and the con-
379. Prove that a one-one continuously differentiable mapping w = u+itJ with positive Jacobian which transforms infinitely small circles into infinitely small ellipses with the characteristics a(z), {)(z), y(z), satisfies the system of equations
u,.+{Ju1
= av1 ,
-f3u,.+u1
= -ow,.,
which can be written in the form p-l q(z) = p+l e819 • HINT. Use the relation ydu1 -2/Jdud11+ctd111
= J(tlail+dy1).
380. Prove that the equation Ws = A (z)w. is invariant with respect to analytic transformations of w, and the equation w, = B(z)ws is invariant with respect to conformal transformations of z. 381. The one-one continuously differentiable mapping w = u(x, y) +iv(x, y) with positive Jacobian is called a quasi-conforma.l mapping with two pairs of characteristics p, O; Pi• (Ji (or with two triads of characteristics a, {J, y; ~./Ji, /'i), if it transforms infinitely small ellipses with characteristics p, 6 into infinitely sma.11 ellipses with characteristics Pi· Oi. Prove that such a mapping satisfies the system of equations
au,.+(fJ+f3i)u1 = aiv1 ,
(fJ-f3i)u,.+,,u,
which can also be written in the form
w, = qi(z)w.+q (z)w•, 2
=
-~v,.,
60
PROBLEMS ON OOMPLEX ANALYSIS
where () q1 z = -
P1(pl-l) el"' (PP1+l)(p+P1)
HINT. Use the relation y 1du1 -2P1dudt1+0idv1 = J(yda:l-2Pd:i:dy+ocdy1 ).
382. Let Pwt• be the characteristic p of a quasi-conformal mapping w(z). Prove that Pwt111 = P111tw
and that for the compound quasi-conformal mapping w[z(t)] Pwtt ~ Pw1111P111tt
883. Show that for the quasi-conformal mappings
u = f (x), v = y (longitudinal stretching compression), u = x, v = f(y) (transverse stretching compression), e = r, (J = f() (angular stretching compression) the characteristic is
and for the mapping e = f(r), 8 = "' (radial stretching compression)
f)
rf' rf'.
p=max ( j '
384. Construct the quasi-conformal mapping of the disk JzJ < R onto itself, transforming the point z =a (Jal < R) into the origin and leaving fixed the points of the circumference JzJ = R. Evaluate the characteristic p. 385. Construct the quasi-conformal mapping of the oblique half-strip x > 0, xtana < y < x tana+h onto the rectangular half-strip u > 0, 0 < v < h without stretching on the base and with constant stretching on the lateral side. Evaluate the characteristic p. 386*. Construct the quasi-conformal mapping of the domain, consisting of a half-plane and a circular segment with central angle
61
SUPPLEMENTARY GEOMETRICAL QUESTIONS
2{J0 (Fig. 11), onto a half-plane with preservation of lengths on the boundary. Evaluate the characteristic 'P·
FIG. 11
387. Reduce the quasi-linear equation
atu atu atu ( au au) A aXA +2B axay+o ay2 = F x, y, u, ax 'ay of elliptical type (AO-B 2> 0) by mea.ns of the schlicht mapping C= C(z) = E+i'YJ to the canonical form
atu atu ( au au) aE2 + a'YJ2 =F1 E. 'Y/• u, aE , a'Y/ Prove that the mapping C(z) satisfies Beltre.mi's system of equations
AaE/ax+BaE/oy a'Y/ y(AO-B2) =" ay '
BaE/ax+oaE/ay y(AO-B2)
= -
a'Y/ ax
and is a quasi-conformal mapping with the characteristics a, {J, y, determined from the relations
r
(it is assumed
fJ
~
1
7f =Ii= A= y(AO-B2) that A > 0).
A function w = u+iv, which satisfies the equation
wz-+Aw+Bw = F,
(I)
where ..4., B and F are functions of z, is known as a genemliaed analytic funo-
non.
62
PROBLEMS ON COMPLEX ANALYSIS
In problems 388-394 equations and systems of equations are considered which reduce to the form (1) and also some properties of their solutionst. 388. Show that Carlema.n's system of equations
u.¥ -v7 +au+bv =/, u1
-
v.¥+cu+dv = g,
where a, b, c, d, J and g are continuous functions of the variables z and y, can be written in the form (1)
tos+Aw+BW = F. Express A, B and F in terms of the coefficients of the given system. 389. Show that the equation
w.-q2(z)wi+Aw+Bw = F can be reduced to the form (1) by means of the "affine" transfor· ma.tion
(2)
w = a(z)ro+b(z)w.
Find the general form of the transformation (2). 390. Show that the equation w,-q1(z)w:r-i2(z)w1+Aw+.BW
=
F
can be reduced to the form w.-q~ws+A'w+B'w =
F'
by means of the transformation of the preceding problem. HINT. Apply the transformation considered to the given equation and to the equation
w.-q1(z)w:r-f1(z}t.os+.A.W+iJ.w=F; eliminate W:r and then select the coefficients a(z) and b(z).
391. Show that the equation
wi-q1(z)ws-qz(z)wi+Aw+Bw = F t On this group of problems see the monograph: I. N.
VEXUA, G~
analyUo Junctiona (Obobahchennyye analitichukiye Junktayi), Chap. III. Fizmat-
giz (1958), Moscow, English translation published by Pergamon Press (1962).
63
SUl'PLEMENTABY GEOMET.RICAL QUESTIONS
oan be reduced to the form
wc-q:i»f+.Aw+Bw = F by changing the independent variable z to the variable C connected with z by the relation Cs = ~ C111 • Find and and explain the geometrical meaning of the transformation C(z). 39!. Prove that an elliptic system of differential equations of the form
qt
q:
et1v1 =
cxux+(/1+fl1)u,+au+bv+f, } -at:1Vx = (/1-/11)ux+yu,+cu-t-dv+g (the condition of ellipticity is here a{l-y 2 > 0; in addition at: > 0) can be reduced to the form w.-q1(z)w111 -q2 (z)wi+.Aw+BW = F, where lq1 (z)l+lq1 (z)I < 1, if at:1 >0, and jlq1 (z)l-lq2 (z)I\ > 1 if «1 0 by the replacement of w ... u+w by iii = u-w.
393. Prove that if w(z) is a continuous differentiable solution of the equation
o,
ios-q2(z)w• =
where q2 (z) is an analytic function of z and lq2 (z)I =f:. 1 then w(z) = (z)+q2 (z) ~(z) ,
l-lq2(z)! where (z) is an arbitrary analytic function. 394. Prove that if w(z) is a twice continuously differentiable solution of the equation ios-q2(z)w.
=
where q2 (Z') is an analytic function of w(z) = (z)+
o
z and
lq2 (z)I
.P 1, then
Jq2 (z)~sdz,
where (z) is an arbitrary analytic function of z. HINT. It is first necessary to prove that w(z) is the sum of an analytic function of z and an analytic function of z.
OHAPTER IV
INTEGRALS AND POWER SERIES In the problems of this and also of the following chapters if nothing is said to the contrary, simple closed contours (that is, those without points of self intersection) are traversed in the positive direction.
§ I. The integration of functions of a complex variable 395. By direct summation prove the equalities:
f
z,
Z1
(1)
dz = z1
-
z0 ;
(2)
J
zdz =
! M-zU .
z,
Zo
396. Let 0 be a simple closed contour bounding an area S. Prove the following equalities: (1)
fc zdz = iS;
fc ydz = -
(2)
S;
(3)
Ic zdz = 2iS.
397. Evaluate the integrals
11 =
Jzdz,
12 =
Jydz
along the following paths: (1) Along the radius vector of the point z = 2+i; (2) Along the semicircle lzi = 1, 0::::;;; arg z :s:;;;:n; (the commencement of the path is at the point z = 1); (3) Along the circle lz - al = R. 398. Evaluate the integral lzldz along the following paths: (1) Along the radius vector of the point z = 2-i; (2) Along the semicircle lzl = 1, 0 ::::;;; arg z ::::;;; :n; (the commencement of the path is at the point z = 1); (3) Along the semicircle izl = 1, -:n;/2 ::::;;; arg z ::::;;; :n;/2 (the commencement of the path is at the point z = -i); (4) Along the circle lzl = R. 399. Evaluate the integral izlzdz, where 0 is the closed contour c consisting of the upper semicircle izl = 1 and the segment -1 ::::;;; z ::::;;; 1, y=O.
J
J
64
INTEGRALS AND POWER SERIES
J ~ dz,
400. Evaluate the integral
c z
65
where 0 is the boundary of
the ha.If ring represented on Fig. 12. y
FIG. 12
J
401. Evaluate the integral (z-a)ndz (n is an integer): (1) Along the semicircle lz-al = R, 0 ::::;;; a.rg (z-a) ~ :r: (the commencement of the pa.th is at the point z = a+R); (2) Along the circle lz-al = R; (3) Along the perimeter of the square with centre at the point a and sides parallel to the coordinate axes. In problems 402-405 the branch of the many-valued function which occurs as the integrand is selected by the specification of its value at some point of the contour of integration. If the contour is closed the initial point of the pa.th of integration is always considered to be that point at which the value of the integrand is specified (it must be remembered that the value of the integral may depend on the choice of this initial point). 402. Evaluate the integral
J ~:
a.long the following contours:
(1) Along the semicircle lzl = 1, y ~ 0, yl = 1; (2) Along the semicircle lzl = 1, y ~ 0, yl = -1; (3) Along the semicircle lzl = 1, y ~ 0, yl = 1; (4) Along the circle lzl = 1, f.'l = 1; (5) Along the circle izl = 1, y(- 1) = i. 403. Evaluate the integral Log z dz, where:
J
(1) (2) (3) (4)
0 0 0 0
is is is is
the the the the
c
unit circle and Log 1 = 0; unit circle and Logi = ni/2; circle lzl = R and LogR =log R; circle izl = R and LogR =log R+2ni.
66
PROBLEMS ON COMPLEX ANALYSIS
f
404. Evaluate the integral and:
znLogzdz, where n is an integer
1zr=1
(2) Log (-1) = ni.
(1) Logl = O;
405. Evaluate the integral
J
~« dz, where
oi
is an arbitrary
lzl=l
complex number and 1« = 1. 406. Prove that for any choice of the initial value of the function ar
J a•dz = 0.
lzl=l
407. For what values of oi (0 ,,;;; oi exist:
< 2n)
do the following integrals
(pis a natural number), taken along the radius vector of the point
z = e"'1 408. Prove that if lal #: R, then
f lzl~R
jdzl iz-all 21 +al
<
2nR IR2-lal 2 I'
409. Prove the following assertions: (1) If f(z) is continuous in the neighbourhood of the coordinate origin then 2..
lim
Jf(re1~)d4' = 2n/(O).
, r-0 0
(2) If f (z) is continuous in the neighbourhood of the point z=athen lim r-0
J
lz-al=r
f(z) dz = 2nif(a).
z-a
410. Prove the following assertions: (1) If f(~) is continuous in the half-strip z ~ z 0 , 0,,;;; y,,;;; h and the limit lim f(z+iy) =.A, not depending on y exists, then lim
Jf(z)dz = i.Ah,
x-+00 Px
INTEGRALS AND POWER SERIES
67
where Px is a segment of the vertical straight line 0::::;;; y ~ h, traversed from below upwards. (2) If f (z) is continuous in the sector 0 < lz - al ~ r0 , 0 ~ arg (z - a) ~ cc, (0 < cc ~ 2n) and the limit lim [(z-a)f(z)] = A z_,.a
exists, then
lim
Jf(z) dz = iAcc ,
r...O 1,
where i'r is the arc of the circle lz-al < r which is present in the given sector, traversed in the positive direction. (3) If f(z) is continuous in the region !zl;;:?:: R 0 , 0 ~ arg z ~cc (0 < cc ~ 2n) and the limit lim zf(z)
=
A ,
Z-+00
exists, then lim
Jf(z)dz =
iAcc,
R"""oorR
where JR. is the arc of the circle lzl = R, which lies in the given region traversed in the positive direction with respect to the coordinate origin. 411. Prove the following theorems: (1) If f(z) is continuous in the region lzl;;:?:: R 0 , Im z;;:?:: a (a is a fixed real number) and lim /(z) = 0, then for any positive number m
lim
Je',,,.f(z) dz = 0 ,
R....oorR
where I'R is the arc of the circle lzl considered (Jordan's lemma).
= R,
which lies in the region
HINT. In the evaluation of the modulus of the integral along the semicircle Im z > 0 make use of the inequality sin 6;;;;.: 28/n for 0 =s;;; 6 =s;;; n/2, and in the evaluation along the a.res lying in the lower half-plane (in the case a < 0), use the fact that the length of each of them tends to !al as R -+ co. (2) If f(z) is continuous in the half-plane Re 21;;:?:: 0, and for the contour 0 we have: (1) The circle izl = 2; (2) The circle iz-11=1 and the initia.l point of integration is z = 1 i. 430. By Liouville's theorem a function f(z) which is analytic and bounded throughout the whole plane is a constant. Prove this theorem after calculating the integral
+
f (!al
< R,
!bl
< R)
f(z)dz (z-a)(21-b)
lzl=R and taking its limit as R --. oo.
72
PROBLEMS ON COMPLEX: ANALYSIS
431. Let f M be analytic in the closed domain bounded by the contour O; z1 , z2 , ••• ,Zn are arbitrary distinct points within 0 and Wn (z) = (z-z1 )(z-z 2) ••• (z-zn)· Show that the integral P(z)=-1-J f(C)
2:rti
c
wnm
Wn(C)-wn(z)dC C-z
is a polynomial of degree (n - 1) which is equal to f (z) at the points Z1' Z2, • • • ' Zn t •
432. Prove the following theorem (Cauchy's formula for an infinite domain): Let 0 be a simple closed contour bounding the finite domain D. The function f(z) is analytic in the exterior of the domain D and lim f(z) =A. Then 1
2"7 :n:i
J
1-f(z)+A, if the point z belongs to the exterior f(C) of the domain D, -,-de= A, if the point z belongs to the domain D. c -z
The contour 0 is traversed in the positive direction with respect to the domain D. HINT. First consider the oase A = 0. 433. Let the function f (z) and the contour 0 satisfy the conditions of the preceding problem. Prove that if the coordinate origin belongs to the domain D, then 1
2:rti
J
c
zf(C) Cz-C 2
dC = {
0, if zeD, f(z), if zeD.
§ 4. Numerical series 00
434. Prove that if the series
J; en converges and larg cnl ~ oc < :rt/2,
n=l
then the series converges absolutely. 00
00
435. Let the series
2
en and
2
n=l
n-1
c: . converge.
Prove that if
00
Re en ;?: 0, then the series
2
lcnl 2 also converges.
n=l
t The polynomial P(z) is known as Lagrange's interpolation polynomial.
73
INTEGRALS A.ND POWER SERIES 00
436. The series
2 en
possesses the property that the four parts
n=l
of it, each of which consists of the terms contained in one and the same closed quadrant of the plane, converge. Prove that the given series converges absolutely. 437. Prove the formula (Abel's transformation) n
n-1
}; akbk = } ; 8k(bk-bk+ 1 )-S._ 1 b.+Snbn, k-m
where 1 ::::;;; m
k=m
< n,
Sk
= ai+a2 + ... +ak
(le~
=
1), 8 0
0.
00
2 anbn where bn > 0, it is sufficient that the partial sums of the series 2 an should n=l 438. Prove that for the convergence of the series
n-1
00
be bounded and the sequence of numbers {bn} tend monotonically to zero (Dirichlet's rule). HINT. Use Abel's transformation. 00
439. Prove that for the convergence of the series
2
anbn where
n-1 00
the bn are real numbers, it is sufficient that the series
2
an should
n-1
converge and the sequence {bn} be monotonic and bounded (Abel's test). 440. Let the sequence of real numbers {bn} satisfy the conditions: 00
(1) lim (yn)bn = O; (2) the series n...oo
2 (yn)(bn-bn+i) converges.
n-1
Then, if the sequence { 8 n
yn
}
is bounded (Sn=
.f 1it), the series
k=l
00
2 akbk converges.
k=l
00
441. Let lim
Vlcnl
= q. Prove that the series
n-+oo
(absolutely), if q < 1, and diverges, if q 442. By the examples of the series
1 1 1 1+26+a-+7+ ...
2 en n=l
>
I.
(1
1 with centre at the point C= 0. 524. Prove that the function (4-t2)/(4-4tz+t2 ) is the generating function of the Chebyshev polynomials:
T 11 (z) = 2111_ 1 cos (n cos-1 z). Using the integral formulae for the coefficients of the Taylor series establish that for n ;):. 2 4T11 + 1 (z)-4zT11 (z)+T11 _ 1 (z)
= 0.
82
PROBLEMS ON COMPLEX ANALYSIS
525. The Herm.ite-Chebyshev polynomials H.(z) a.re defined by the expansion e2t:r:-i•
"° Hn(z) en. "1
=
L.;
n!
n=O
Prove the following relations: (1) Hn+ 1(z)-2zH.(z)+2nHn_ 1(z) = 0 (2) H~(z) = 2nH._ 1 (z) (n ~ 1); (3) H~'(z)-2zH~(z)+2nHn(z)
(4) H (z) = n
=
0
(n
~
l);
(n ~ O);
(-l)•e=·~·ce-"'') • dzn
526. The Laguerre polynomials can be defined by the equation Ln(z) = e=
d•(z•e-"') dz• •
Find a genera.ting function for the sequence {L.(z)} and by means of it obtain the recurrence formula. connecting Ln- 1 (z), Ln(z)
and
Ln+ 1 (z).
In problems 522-526 only a few of the particular properties of the given systems of polynomials have been considered. For other important properties of them which play an important part in the solution of various problems of mathematical physics, see, for example, [3, Chapter VII, § 2] or R. CoURANT and D. HILBERT, Methods of Mathematical Physica, (Vol. I, Chapters II and VII, Interscience Publishers, New York). REMARE.
In problems 52'7-529 find the solutions of the given differential equations, which satisfy the conditions w(O) = 0, w' (0) = 1. 52'7. w"-z2 w = 3z2 -z4. 528. (1-z2)w"-2zw'+n(n+l)w = 0. 529. (l-z2)w"-4zw'-2w = 0. 530. Expand the function cos (m sin-1 z) (sin-1 0 = 0) in a series 00
of the form
2
CnZn
after finding a differential equation which is
n=O
satisfied by this function. 531. The differential equation d2w dw z(l-z) dz2 +[c-(a+b+l)z] dz -abw = 0 is known as the hypergeometric equation.
INTEGRALS AND POWER SERIES
83
Find the solution w(z) of the hypergeometric equation analytic at the point z = 0 which satisfies the condition w(O) = 1, assuming that c is not equal to zero or a negative integer. 532. Prove that the general solution of the hypergeometric equation is of the form (c not equal to an integer)
w = 0 1 F(a, b, c, z)+09 z1 -cF(a+l-c, b+l-c, 2-c, z), where F(a, b, c, z) is the function defined in the preceding problem (the hypergeometric series). 583*. Prove that if c is not equal to zero or a negative integer, then
dF(adb, c, z)
z
=
ab F(a+I,b+l,c+I,z). c
§ 7. Some applications of Cauchy,s integral formula and power series mu. Let the expansion of the function /(z) in the circle lzl < R be of the form 00
/(z) =
2 .... o
011 z11 •
(l) Prove that 2ir
2~ Jj/(re'•)illd,P = 0
00
,21c 12rfln 11
(r < R).
m=O
(2) Prove that if
max l.rl•r
lf(z)I
= M(r),
then the coefficients 011 satisfy the inequalities (Cauchy's inequalities)
Ic..I ~ ~r)
(r
< R) •
(3) Prove that if juE-t one of Cauchy's inequalities becomes an equality, that is,
lctl =
.M(r) then the function /(z) is of the form r1:
84
PROBLEMS ON COMPLEX ANALYSIS 00
HINT. Use the following inequality from section 1. }; lcn1•r1111~ [M(r)]'. n=O
(4) Prove Liouville's theoremt. 00
535. The function /(z) =
2 n-0
Cnzn
2 c~~n
is analytic for lzl ~ r. Prove
00
that the series (z) =
converges in the whole plane and
n=O
the estimates l(z)I
l!L
< Me r
,
i(z)I
.M .!!!. e r (M is a constant)
R-r then
it is
HINT. Use the fact that a power series always has at least one singular point on the boundary of its circle of convergence.
642. Prove Pringsheim's theorem: 00
If the radius of convergence of the series
2
c.z• = f(z) is equal
n=O
to unity and all the en ;;;::: 0, then the point z = 1 is a singularity of the funotionf(z) (the sum of the series). HINT. Prove by means of the criterion of the preceding problem that if it is assumed that the point z = l is regular then all the remaining points of the unit circle will be regular (for the proof use the fact that lf 0 and lzl = a:). 00
643. Prove that if the radius of convergence of the series
2 c.zn n=O
= f(z) is equal to unity, where all the c. are real and the sequence
{s.} (s. =
i: c1c) tends to an infinity of a definite
sign as n -+ oo
k-0
(that is beginning with a certain n, all the s. have the same sign), then the point z = 1 will be a singular point for the function f(z). Prove, for example, that the assertion may not hold if we only have lim ls.I = oo. ·~ t See [l, Chapter m, § 6]. See also G. P6LYA and G. Szoo6 (1957), Aufgaben und Le1w8'i.tze a'U/1 der Analysia, 2nd ed. Chapter I, sec. 3,Chapter 5, § 3, or E. C. TITCHMABSH (1951), Theory of functiona, Chapter VII.
97
SINGULAR :POINTS OF SINGLE-VALUED FUNCTIONS
HINT. "Estimate lf(z)I for z real and close to I (z identity 00
J; c
<
1), by means of the
00
11 z11
=
(1-z)
n-0
J; 1,,zn. n=O
644. Prove that if on the circumference of the circle of convergence 00
of the series
2 c,,z!' = f(z)
there is just one pole of the function
n=O
f(z), then the series diverges at all the points of this circumference. HINT. Using the fact that from the convergence of the series at just one point of the circumference of the circle of convergence it follows tha.t lim c,,R11 = 0 (R is the radius of convergence), prove that if (z'I = R then n-+O
the radial limit lim (z' -z)/(z)
== O.
z-+-z 1 00
64lS. Prove that if the power series f(z) =
2 c,,z" has on the cirn=O
cumference of the circle of convergence only one singular point z0 , Anm-1 a. pole of order m, then c,, = [I + 0, the number a
= lim log M(r) is known as the flype of the function. r-+oo re
Ifa = 0, the funotion /(z) is said to be a function of minimal flype; if a= oo, it is a function of maa:imal flype; if 0 < a < oo it is a function of normal flype.
649. Prove the following assertions: (1) H (! ::fa oo and 0 it is possible to find a number R(e) such that the inequalities M(r)
<
e«•+•)rll.
hold for r > R. It is also possible to find sequences of numbers {rn} and {r~} converging to infinity for which
,e-•
M(r0 )
> e'n
and
M(r~)
,e
> ern .
(2) If for some natural number le
.
OOO
M(,r) r
It
0). 657. coshz. 652. znasz. 658. escosz. 653. zlel'Z-e3Z 659. cosy'z. 654. e621 -3e8st. o
00
660*.
2 (2m~nn)
I
(m a natural number).
n=O
661. e•s. 1
662* •
JezPdt. 0
663. The integral function/(z) is of order(! and type a (0~0'~ oo). Prove that the function P(z)/(z)+Q(z), where P(z) and Q(z) a.re any polynomials is also of order (! and type a. 864. The integral functions f 1 (z) and f 2 (z) are of orders (!1 and (! 2 respectively, where (!1 ::p (!a· What can be said of the order e• of the functions Ji (z) f 2(z) and f 1 (z) +f2 (z) 1 665. The integral functions f 1 (z) and f 2 (z) are of the same order (! and are respectively of type& equal to a 1 and 0'2 , where 0'1 #: a 2. What can be said of the order (!* and type a• of the functions: (1) f 1(z)f2 (z); (2) / 1 (z)+/2{z)1 666. The integral functions Ji {z) and f 2 (z) a.re of the same order (! and of the same type a. What can be said of the order e• and type a• of the functions : (1) f1{z)f2(z); (2) f1{z)+/2{z}1 667*. Prove that the order and type of an integral function a.re not changed on differentiating the function. Solve problems 668-675 by the use of the following theorem. If the Taylor expansion of an integral f1mction is of the form
100
PROBLEMS ON OOMPLEX ANALYSIS 00
f(z) =
2 onzn, then the order and type of this function a.re given by
n•O
the formulae :
(O'ee)
.!. 11
-
= lim
( .!. 11 n
)
n ylOnl .
D-+00
688. Prove that the integral function 00
J(~) = is of order
I
e = -a:
~
L.J
n-o
(Az)n I'(o:n+I)
and type O'
HINT. Use Stirling's formula
I'(«n+I) = ( :"
(A >0, ct:>O)
.!
= A •.
t
y'(2n«n) [1+0
(~ )].
In problems 68M75 find the orders and types of the given functions.
689. f(z) =
2 (:t n=l
oo
670. f(z) =
2(
n
lo!n ) 0 zn
n=l oo
(a> O).
n
671./(z)= 2(nl!gn)°zn
(a>O).
n=ll
2 00
672. f(z) =
e-n•zn.
n=O
673. f(z) =
~
L.J n-1
zn n,r.+n
(a> 0).
INTEGRAL FUNCTIONS
101
co
674. f(z) =
2
cos!tn zn.
n=O co
~
-v
675. z Jv(z) =
(-l)nzlln
.L.J nlI'(n+,,+l)
(11
>
-1; Jv (z) is a Bessel
n-0
function of the 11th order). If
A(•)
(!
is the order of the integral function /(z), then
= 1im" r-+co
the function
logl/(re'">I is called the indicator funcwm of the function /(z). rO
In problems 676-681 find the indicator functions of the given functions. 676. es. 679. cosh z. 677. sin z. 680. ezn. 678. cosz. 681. e"+z1. 682. The integral function f(z) has the indicator function h(). What is the indicator function h*() of the function /(z)+P(z), where P(z) is a polynomial~
CHAPTER VI
VARIOUS SERIES OF FUNCTIONS. PARAMETRIC INTEGRALS. INFINITE PRODUCTS § I. Series of functions In problems 683-692 find the domains of convergence of the given series. 00
'\1 (-l)" z+n L.J n=O
688.
.
00
'\1
Zn
L,; 1-zn •
690.
n=l
2 2 s~nz.
00
00
686.
es ios n •
'\1
n-o
n-t
2 00
00
687.
.,.n
L,; 1 ~zll" •
691.
692
n-t
•
n-1
n
z
(4+z)(4+z2) ... (4+zn)
00
693*. Prove that if the series
J; a
11
converges then the series
n=l 00
n
.J: a,.zz" ,._1 1
converges everywhere where 102
izi :Fl; if, however, the series
VA.BIOUS SERIES OF FUNCTIONS 00
103
00
}; an diverges then the series }; 1anz;n converges in the circle of
n-1
00
n-1
2
anz" and diverges outside this circle. n-1 694. (1) Expand in a sum of powers of z the sum of the series
convergence of the series 00
~a z" ,,c,,; 1 n z"; find the radius of convergence of the series obtained.
n-1
(2) Prove that for
lzl <
1
00
'1
L.J cf>(n)
n-1
z"
z
1-z" = (l-z) 2
'
where c/>(n) is the number of those positive integers less than and coprime to n.
2
HINT. Use the relation l/>(n) = m where n assumes in succeBBion the value of each of the divisors of the number m, including 1 and m. 00
00
~. e-s 1o1n (Riemann's n=l n-1 Zeta function) in a Taylor series in the neighbourhood of the point z = 2 and find its radius of convergence. 695. Expand the function C(z) =
,2 =};
In problems 696-699 find the sum of the given series. 696.
.f(l~zn -l+~n-1)
(jzj.,t:l).
n-1
00
697 ·
2
n-1
(1-z");;_zn+l)
(izl ¥: 1).
HINT. Multiply numerator and denominator by (1 - z).
2
2"
00
698.
n=l
n Z
fl (l+zs")
k-0
•
104
PROBLEMS ON COMPLEX ANALYSIS 00
The series
2
ft(z) is said to be uniformly co1wergent on the set E if for
k=l
any
B
> 0 it is possible to find a number N(e) such that the inequality
00
12 /rc(z)I < B is satisfied for all n > k=n
N(e) and for all points z of the set E.
700. Prove the propositions: 00
(1) For the uniform convergence of the series
2
In (z) on
n-1
the set E it is necessary and sufficient that for any e > 0 a number N = N(e) should exist such that for all n > N, all z e E and any natural number p the inequality
should be satisfied. 00
(2) From the uniform convergence of the series
2 lfn(z)I n=l
on the set E there follows the uniform convergence on the same 00
set of the series
2 fn(z).
n-1
701. Find the sets on which the given seq\'J.ences converge uniformly: (l) {
l~zn };
(a) {
sn:
nz } .
702. Prove: In order that the sequence of continuous functions {!n(z)} should converge uniformly on the bounded closed set E, it is necessary and sufficient that this sequence should converge at all the points of this set and that it should converge continuously at all the limit points of the set E, that is, that for every sequence of points Zn, belonging to the set E and converging to the point z0 ,
lim In (zn) = l(zo) • n-+00
In problems 703-707 find the sets on which the given series converge uniformly.
VARIOUS SERIES OF FUNCTIONS 00
703.
00
2 ~8 (zn+ ~ ). 2 2
706.
n-1
2 2
(sin nz)/n1 •
n-1
00
704.
105
00
707.
e-ns.
n-0
(sinnz)/n.
n~l
00
705.
e-slosn •
n=l
L :. converges uniformly in the closed n
oo
708. Prove that the series
n-1
circle Iz I : : ; ; 1. Does the series obtained by term-by-term differentiation converge uniformly in the circle Iz I < 1 ~ oo
709. Prove that the series
,l:
(-l)n-1
z+n
converges uniformly in
n=l
any finite pa.rt of the plane, from which there has been removed disks of arbitrarily small radius e with centres at the points z = 0, -1, -2, ... Prove that this series does not converge absolutely at a single point. n
oo
710. Prove that the series }; :
converges uniformly in the
n=l
interval (-1, 0), and the series }; 1 1'
~I
n=l
converges in the same in-
I
.L:-: oo
terval, but not uniformly. (Hence the series
n
cannot be ma.jorised
n=l
in the interval (-1, 0) by a convergent series of numbers.) REMABJC. This example shows that Weierstrass's sufficient test for uniformity of convergence is not necessary. 00
711. (1) The series
,L: (I ~z2f converges absolutely for izl ;;:,: 0, n=O
largzi ::::;;; n/4 (these values of z do not exhaust the whole of the domain of absolute convergence which, as is easily seen, consists of the point z = 0 and the outside of the lemniscate I I+z2 1 =I). Prove that the series converges non-uniformly in the given domain.
106
PROBLEMS ON COMPLEX ANALYSIS
REMA.RX. This shows that uniform convergence does not follow even from the absolute convergence of a series in a closed region. (lO
(2) Prove that in the same region the series ,L; c\-l~:;n con+ n=O
verges uniformly and absolutely, but not absolutely uniformly (that is, the series of moduli does not converge uniformly). (lO
712. Prove that if the series 21/n(z)i converges uniformly in n-1 (lO
every closed region inside the domain G, then the series 21/~(z)i n=l
also possesses the same property.
§ 2. Dirichlet seriest (lO
Series of the form
2
ane -A,,z, where the an are complex coefficients
n-1
and the An are real non-negative numbers, satisfying the conditions
A1 < A1 < ... and lim An
=
oo,
II-co
are known as Diriokke Bef'iea.
713. Prove that if a Dirichlet series converges at the point z0 = :r:0 +iy0 , then it converges at all the points of the half-plane Re z > Re z0 , the convergence being uniform in every angle Iarg(z-z0 )1 ~ () < n/2. HINT.
Apply Abel's transformation to the sum q
~
L.J ane
q
..
)
-A,,z _ ~ -Anzo -AnCz-z0 - L.J ane e
11-=p
n==p
and use the inequality (a< b, te =a:+ iy) b
je-az-e-11•1
= jz J e-l:tdtl~~(e-u-e-llx). a
a;
714. Prove that if a Dirichlet series converges absolutely at the point z = z0 , then it converges absolutely and uniformly in the half-plane Re z ~Re z0 • From the theorems formulated in problems '718-'714 it follows that the domain of convergence of a Dirichlet series (if it exists) is a half.plane Re z > ll:c (a:c ~ - oo), and the domain of absolute convergence (if it exists)
t On the subject of this group of problems see, for example, [l, Chapter IV, § I].
107
VARIOUS SERIES OF FUNCTIONS
is a half-plane Re z > a:0 (a:a;;;a. - oo), the series either converging absolutely on the whole of the straight line Re z = a:0 , or not converging absolutely at a single point of this straight line. The numbers a:c and a:0 are called respectively the abBciaBa of oonoorgenoe and the absciaaa of abaolul.e cont1ergence of the Dirichlet series. In problems 715-721 find the abscissas of convergence (zc) and
the abscissas of absolute convergence (z0 ) of the given series.
2 00
715.
e-n•e-zn•
719.
2(
00
00
nl)n e-"'lOglOg n,
720.
e-slogn.
n=l
n=2
2 (Y~n 2
00
00
717.
(-l)ne-slogn •
n=l
n=O
716.
2 2 ~. 2 00
e-slollocn.
721.
n-2
e"" e-sn•.
n=O
00
718.
(-l)ne-Zlo1101n, n-2 722. Prove that if lim (log n) /An :-- 0, then
723. Prove that if lim (logn)/An
=
Z, then
Z 4 -Zc ~
Z.
In problems 724-728 investigate the convergence of the Dirichlet series on the boundary of the half-plane of convergence.
2 2 ~3 2~ 00
724.
725.
727.
n!)n e-sn •
n=l
n=l
00
00
e-s 101 n •
n=l
00
726.
2( 2( 00
(-l)ne-"' 101fn.
e-zn.
n=l HINT.
See problem 4'79.
728.
n=l
-I)lvnJ e-"'n. n
108
PROBLEMS ON COMPLEX ANALYSIS
The series
0, is satisfied for all q, of [oc, {1], and diverges at a point for which for all of [oc, PJ x cos -y sin -k < 0. 00
730. An arbitrary generalised Dirichlet series
L; a.,. e-Ans is given. n=l
_ log!antl Let k(, oc) = lim - , 1 -and k() = lim k(c/J, oc), where {nt} is 1 k-oo
«~
Anrc
the sequence of all the suffixes for which -oc::::;;; arg Ant::::;;; +a (if no such subsequence {n,,,} exists for which lim arg Anm = , then it m-oo
is necessary to put k() = -oo). Prove that if lim (logn)/An = 0, then the series converges absolutely
-
within the domain G, the points z = x+iy of which satisfy fo1· any the condition zcosef>-ysin-k(c/J) >0, and diverges at every point lying outside G.
§ 3. Parametric integrals 731. Prove the theorem: Let 0 be a simple contour (closed or not), possessing a finite length, and f(T, z) a function, analytic with respect to the variable z and continuous with respect to T for every z of some domain D and for all the points T, belonging to the contour 0. Then the function represented by the integral F(z) =
Jf(r:, z)dr:,
c
is an analytic function of the variable z and F'(z) =
JJ;(r:, z)dr:.
c
INTEGRALS DEFENDING ON A l'ARAMETER
If the integral
Jf (T, z)dT is
c
109
improper, that is, if the integrand has dis·
continuities at some isolated values TE 0 or the contour of integration includes the point at infinity, then the definitions of convergence and uniform conver· gence of such an integral are exactly similar to the corresponding definitions given in books on mathematical analysis.
732. Prove that for the uniform convergence of the integral f(T, z)dT on the set E with respect to some point To#= oo of the
J
c contour 0 it is necessary and sufficient that for any e > 0 there should exist a number o(e) such that
IJf(T, z)dTJ < e Co
for all the points z of the set E and for every arc 0 a of the contour 0, lying in a a-neighbourhood of the point To and not containing this point either as an interior or as an end point. 733. Formulate and prove a similar criterion for uniform convergence of the integral if To = oo. Consider the cases when the contour 0 is not bounded at one end, or at both ends. 734. Prove that if lf(T, z)j ~ IVJ(T)I for all the points z of the set E and if IVJ(T)ldT converges then the integral f(T, z)dT converges
f
f
c
c
uniformly on the set E. 735. Let f(T, z) be a function analytic with respect to z and continuous with respect to T for all the points z, belonging to some domain D, and points T, belonging to the contour 0, with the exception of some isolated points of it where the conditions imposed on the function f(T, z) are not satisfied either for all points z, or only for some of them. Prove that if the improper integral F(z) =
Jf(T, z)dT
c
converges locally uniformly in the domain D (that is, in every closed sub-region of the domain D), the function F(z) is analytic and F' (z) =
Jazaf
c
dT '
the latter integral converging uniformly within D.
110
PROBLEMS ON COMPLEX ANALYSIS
In problems 736-743 find the sets on which the given integrals converge uniformly.
Jts00
736. I'(z) =
w-1 = eRe z0 , the convergence being uniform in the angle la.rg (z-z0)1 :,; ; (J < n/2. (2) If the integral (1) converges absolutely for z = z0 then it converges absolutely and uniformly in the half-plane Re z ~Re z0 • (3) If lim loglf(t)I t-+OO
t
= p, then the integral (1) converges
absolutely in the half-plane Re z >fl and uniformly in every half-plane Rez ~ fl+e (e > 0) (constructa.n example of a La.place integral which converges absolutely in the whole plane and for which fl= oo). (4) If lim log lf(t)I =ex, then the integral (1) does not cont-.oo
t
verge absolutely at a single point of the half-plane Re z < oc.
111
INFINITE PRODUCTS
It follows from the theorem formulated in problem '744 that the domains of convergence and of absolute convergence of a Laplace integral (if they exist) are the half-planes Re z > a:c and Re z > a:0 ; the number a:c is called the absciasa of convergence, and a:0 is called the abacisaa of absolute convergence of the Laplace integral. co
In problems 745-751 find
Xe
and x 0 for the integral [ e_.' /(t) d.t,
where f(t) is the given function. 746. /(t) = e-t•. 747. /(t) = e'". 745. /(t) = 1. 748. f(t) = e-t• for 0 < t < log log 3 and log log 2le t nvergence of the 00
product
n-1 00
series
21 log (!_+an))
it is necessary and sufficient that the series
11= 00
2 an
should converge absolutely.
11=1 00
769. The infinite products
fl p
OCI
11
and
n=l
fl qn converge. Investigate
n=l
the convergence of the infinite products:
n 00
(1)
n 00
(Pn+qn); (2)
n-1
n 00
(pn-qn);
(3)
11=1
00
p 11 q11 ;
(4) nPn. n=l qn
n=l
In problems 770-774 investigate the convergence and absolute convergence of the given infinite products. 710.
fJ[1+ -sin )def>
(n is an integer).
0
848.
tan (x+ia)cb;
(a is a real number).
0 2n
849.
cot (x+a)cb;
(a is a complex number and Im a#: 0).
0
850. Prove that if b
>a> -
1
y
FIG. 15 where 0 is the contour represented in Fig. 15, and let the radii of the arcs of the small circles tend to zero. In calculating the integral along the vertical segment divide it into two and by corresponding substitutions reduce them
122
PROBLEMS ON OOMPLEX ANALYSIS
to Eulerian integrals of the first kind; also use the known relation B(p, q)
= i;![)I'(~) p+q
and the formula I'(p)I'(l-p)
= ~. smnp
In problems 851-855 evaluate the integrals with infinite limits.
00
852.
zldx (zl+a11) 1
f f J (zl+a•~za+bl) 0
(a> 0).
co
853.
(n is a natural number).
(zl:l)"
0
co
854.
(a > 0, b > 0).
-oo 00
855.
f
zl+l x'+l dx.
0
f
856. Prove that 1 dT ilc-n-l(n+lc-2) I 2ni T"r = (2k)n+lc-l(lc-l)! (n-l)! (na.ndlca.rena.tura.lnumbers),
c
where 0 is a straight line parallel to the real a.xis which cuts off on the imaginary a.xis a segment equal to k (k > 0). 857. Evaluate the integral 1 2ni
f
dT T"(T-z)
(le is a natural number),
c
where 0 is the contour of the preceding problem. In problems 858-861, using Jorda.n's lemma. (see problem 411), evaluate the given integrals.
f
co
00
858 • (l)
-oo
zcoszdx z 11_2z+IO '
(2)
f co
zsinzdx zl-2z+10.
123
RESIDUES AND THEIR APPLIOATIONS
f f f
00
859.
-oo
xsinxdx x8+4x+20 ·
00
860.
cosa:x:
(a and b are real numbers).
xB+b• dx
·-00
00
861
.
x sin a:x: d
x8+b1
(a and b are real numbers).
x
-oo
862. Let f(z) = e1m"F(z), where m > 0 and the function F(z) possesses the following properties : (1) In the upper half-plane it has a finite number of singularities a 1 , a8 , ••• , a,. ; (2) It is analytic at all the points of the real axis, except the points x1 , x2 , ••• , Xm, which are simple poles; (3) F(z)-+ 0 if z-+ oo and Imz ~ 0. Prove that oo
J
f(x) dx =
-oo
n
m
k-1
k=l
2:n:i{2 resU(z)]:r=ak + ! }; resU(z)]:r-~t},
where the integrals are understood in the sense of the principal value (with respect to all the points xk and oo). In problems 863-866 find the principal values of the given integrals. co
f f
00
863.
J 1· x -5x+6 e''~
_ 00
- d x (t is a real x number).
865.
-oo
00
864 •
xcosxdx. 2
sinxdx (x1 +4)(x-1)
'
00
866.
-oo
cosax l+x3 dx (a~ 0).
-co
In problems 867-872 evaluate the given integrals (a and b are real numbers).
f
00
868 ·
0
sina:x:dx x(x1 +b8) '
124
PROBLEMS ON OOMPLEX ANALYSIS co
8'71.
f
s:x ch.
0
J co
S'70.
cos2aa:;;cos2bx dx.
0
!I
FIG. 16. BniT. Use the integral Jells-l dz, where the contour 0 is indicated C z• in Fig. 16. co
f fi:8 sin8 x
8'72.
ch .
0
JlniT, Use the integral
f e81s-3e1•+2 zS
dz, where the contour 0 is indi-
C
cated in Fig. 16.
In problems 8'73--8'76 evaluate the integrals, considering that xP > 0 for x > 0 (this condition is retained in all the following problems). 00
8'73. (1)
JzP--
1
cos aa:ch (a
> 0, 0 < 'P <
1);
0 co
(2)
Jzl'- sin aa:ch (a> 0, -1
1).
0
877. Let the rational function f(z) have no poles on the positive part of the real axis and
lim [z"f(z)] = lim [z"f(z)] = 0. z-0
Z-+CO
Prove that CO
11
J
o :where
zP-1f(x)dx
ix1 , ix1 , ... ,
=
~
.2
smpn k~l
res [(-z)"-1 f(z)]z=et1:•
°'n are the poles of the function f(z), pis an integer.
IIINT. Consider the integral
J(-z)P-1/(z)dz
( (-z)P-1
c
= e(P-1)101(-z)),
where 0 is the contour represented in Fig. 18. co
878. Evaluate
J
x"(:l)
(0 0).
1 1-e oz
=
I+ e-oz
+ e-llloz +
...
135
RESIDUES AND TREIB APPLICATIONS
HmT. Replace 0 1 by the contour shown in Fig. 26.
26
FIG.
929 . _1_ 2ni 930• ~ 2m
J
fl' sinh ry21 dz
c,
J c,
~ffl'log(z+l)dz. 2ni z-1
931.
rz sinhayz
c,
(a >r >0).
co
lo~(l+z)as' dz. z
ot
J Je:~• dt
932.
dz (a >0).
c,
0
HINT. Change the order of integration.
2~i c,J~ dz Je-•scosudx (a> 0, b a real number). 00
933.
0
HINT.
Make use of the fact that
e-• J-z
-dz
=
0 for u
> 0.
c,
934. From the series expansion of the Bessel function
~
J,,(z) =
L.J
(-l)t klI'(k+v+l)
k=O
(z2 )v+lt '
deduce the following integral representations ( y is the contour given in problem 920): 1 (I) 2ni
f c•+i.. dC (z) 2 •
eC-it
=
J,(z);
136
PROBLEMS ON COMPLEX .ANALYSIS
r..
2~i c,f ~~~ dC = (:
(2)
J,(z} (Re,, > -1).
HINT. Expand the function e - '' into a power series and use the solu· tions of problems 910 and HI.
-n
n FIG. 27
935. Prove that if Re z > 0
f
J,(z} = __!__ eliratnt-fvtdC, 2:n; II where ll is the contour given in Fig. 27, and hence obtain that for evecy integer n J.(z} =
~
.
J
cos(zsinC-nC}dC.
0
In problems 938-938 find the integrals which involve Bessel functions.
J 00
936.
e-ztJn(t}dt (Rez
> 0, n
is an integer}.
0
HINT. Use the integral representations of the preceding problem and change the order of integration.
J 00
J 0 (at} cosbtdt;
937. (1)
J 00
0
(2)
J 0 (at} sinbtdt
0
(a and b are real numbers}.
J 00
938.
cos bx
0
sin t JI (xi- a2)
V(xD-aB)
dx (t > lbl}.
RESIDUES A.ND THEm APPLIOATIONS
137
HINT. Make use of the fact that sinut u
=
y/(!!!...)J 2u
u•t•
(ut)
l
=
y'(nt) _l_ 2 2ni
Jez-~ dz c,
?!la
(see problem 93'), and change the order of integration.
The asymptotic behaviour of integral,at 939. Let the analytic function (z) have on the left of 0 1 only a finite number of singularities, all of them being poles, and let (z) -+ 0 as z -+ oo and Re z :,;;;; ct. Let us put f(t)
=
2~i c,Je%'(z)dz
Find limf(t). Consider various cases of the distribution of the t-+00
poles with respect to the imaginary axis. HINT. Use Jordan's lemma (see problem 4ll).
940. Let the analytic function (z) have on the left of 0 1 a finite number of singularities, and let (z) -+ 0 as z-+ oo 11.nd Re 21 :,;;;; ct. Prove that the asymptotic equality
J
2~ic, e%'(z)dz"' .2res [e%'(z)], holds for large values of t, the summation extending over all the singularities of (z) with negative real part. REMA.Rx. The functions f (t) and F (t) are asymptotically equal as t (f(t)-F(t)), if lim f(t)/F(t) = 1.
-+
co
f-+00
941. Investigate the asymptotic behaviour as t-+ oo of the function l e:r:'dz (Rea> 0). f(t) = 2ni 2!2(z+a)3
J
c,
t On the subject of this group of problems and also on the question of the application of asymptotic estimates and other methods for obtaining them see, for example, (3, Chapter V, § 3]; B. A. FucHs and V. I. LEvlN, FuootionB of a oomple3: variable and some of their applications (Funktaii komplekmogo peremennogo i nekotoryye ikh prilozheniya), Chap. IV, Gostekhizdat, 1951 English translation published by Pergamon Press (1961); M. A. EvGRAFOV Asymptotic e8timatu and integral Junetions (Asimptoticheskiye otaenki i taelyyl!i Junktaii), Gostekhizdat, (1957).
138
PROBLEMS ON COMPLEX ANALYSIS
942. Find a.n asymptotic expression as t-+ oo for the function I
t =-
f
zen-"(z•+aa)
(z-wi)y(z2+2az)
2ni
f()
dz
(w
> 0, a> 0).
C1
where
y(z +2az) > 0
>
for z
2
0.
HINT. Repla.ce the contour 0 1 by lihe contour represented in Fig. 2·0, and prove that the integrals along the arc of the circle and along the negative part of the real axis tend to zero as I-+ oo. 00
The series
1:
_
qll
z
0.
1:"" c11µ 11 (z) is said to be a.n asymptotic expansion of the function
n=O
00
j(z),....
2 c11µ 11 (z), n=O
if
i; ~µ11(z)] = 0
lim - 1 -[J(z)q1:(z) n=O
z-+oo
(n = O, 1, 2, ... ).
Frequently the sequence { z!n } is chosen, as the sequence {µ 11 (z)} where the
ot,s
are positive real numbers monotonically tending to oo.
943. Prove that for :x:
J
>
0
00
0
e-xt
I+t2
I
2!
4!
dt.....,-;-7+7- ...
+(-l)
11
(2n)! :x:Dn+i
+ ...
139
RESIDUES AND THEIR APPLICATIONS
Make use of the expansion
HINT.
I
-= I+t•
(-I)n+lt2•+2 l-tD+t'- ... +(-I)nt2n+-'---'---I+t•
and estimate the remainder term.
944. Prove that for :x: > 0
f -,-dt,....,-;-7+-;a-- ... 00
1
e"-t
2
1
n!
11
+(-1) z11+1
+ ...
h
Integrate by parts and estimate the remainder.
HINT.
945*. Prove that for :x:
>
0
00
f -t-dt-.i- -;+-;a+7+ ... + e-ir-t
(
1
1
2
(n-1)1
x"
)
+ ... ,
-x
where the integral is understood as its principal value. 946. Prove that for real values of x
f
00
-co
00
e-'"dt
t-x
~ (2n) !
,...,-Ji:n; .L.J
1
22n.n! x2n+l'
n-0
where the integral is understood as its principal value. 947. Prove that for real values of :x:
where for :x: > 0 the integral is understood as its principal value. 948*. Prove that
the signs + or - being taken according as Re z > 0 or Re z < 0. If Re z = 0 the term in front of the brackets must be omitted.
140
PROBLEMS ON COMPLEX ANALYSIS
949. Find the asymptotic expansion of the function f(t) =
2~i c,Je:r~~~
(ro
> 0).
Find also the expansion of f(t) for small t. HINT.
Replace 0 1 by the contour given in Fig. 28. In order to obtain the co
asymptotic expansion of the integral
J e-~tyaicb w+w•
use the hint to problem
0
HS. For small t it is necessary to choose 0 1 such that and expand l/(11 +w8) in a series.
at
is greater than co,
y
FIG.
28
950. Prove that 1
2ni
J
pJt
• (
y(z)(z1 +l)dz-sm
C1
n)
2
~
t-4,..., yn .Lt (-l)
n
(4n)I ( 1 (2n)I 2yt
)'n+l •
n=O
951. Find the asymptotic expansion of the function
1
f(t) = 23ii
J
endz !.
c, z(l+z1 )
8
(i
>0 for z >0).
Obtain also an approximate formula for f(t) for small t. HINT. In order to obtain the asymptotic expansion replace 0 1 by the I iy3 I iy8 contour shown in Fig. 29, where z1 = - 2 + - 2-, z, = -2- - 2- •
141
RESIDUES AND THEm Al'l'LIOATIONS
For small I the abscissa of the straight line 0 1 muat be taken greater than unity.
FIG. 29
§ 3. The distribution of zeros. The inversion of series In problems 952-954 using Rouche's theorem find the number of roots of the given equations lying within the circle lzl < 1. 952. z9 -2z&+z•-Sz-2 = 0. 953. 2z6 -z3 +sz•-z+s = O.
954. z7 -5z'+z2 -2 = 0. 955. Prove that if the inequality
la1:zil > lau+a1 z+ ... +a1:_ 1 zt-1 +a1:+1zi+i+ ... +anz"I, is satisfied at all the points of a contour 0, then the polynomial
a0 +12iz+ ... +anzn has le zeros within the contour 0, if the point z = 0 lies within this contour, and has no zeros if it lies outside the contour 0. 956. How many roots of the equation
z'-5z+l =0 a.re situated inside the circle lzl < 1 ¥ In the ring 1 957. How many roots of the equation
< lzl < 21
z'-Sz+IO = 0 a.re situated inside the circle lzl < 11 In the ring 1 <
lz! <
31
142
PROBLEMS ON COMPLEX ANALYSIS
958. How many roots has the equation z = (z) in the circle lzl < 1, if for izl ~ 1 the function (z) is analytic and satisfies the inequality l(z)I < 1? 959. How many roots has the equation ez = az" in the cii'clti Jzj < R (n a natural number), if !al> eR/R"? 960. Prove that in the right half-plane the equation
z = A.-e-z
(A> 1)
has a unique (and hence real) root. 961*. Prove that no matter how small is great n all the zeros of the function
I f,.(z) =I+ z
1
+ 2!z2
-1 ··•
(!
> 0,
for sufficiently
I
+ n!z".
are situated inside the circle izl < (!· 962. Prove that if (! < I then the polynomial
P,.(z)
= I+2z+3z2+ •.• +nz"-1
for sufficiently great n has no zeros in the circle izi
<
(!.
HINT. Use the method of solution of problem 961.
963. The function (z) is meromorphic in the domain G and analytic on its boundary 0. Prove the following assertions: (I) If l(z)i < 1 on 0, then the number of roots of the equation (z) = I situated inside the domain G is equal to the number of poles of the function (z) in the domain G. (2) If l(z)i >I on 0, then the number of roots of the equation (z) =I situated in the domain G is equal to the number of zeros of the function t/>(z) in the domain G. (3) The assertions (I) and (2) remain true if the equation (z) = I is replaced by the equation (z) =a, where Jal ?- I in case (1) and 1~! < l in case (2). 964. Let none of the zeros of the polynomial
P,.(z)
= z"+a1 z"-1 + ... +a,.
lie on the imaginary axis. Prove that when the point z traverses the imaginary axis from below upwards the increase of the argument of P,.(z) equals kn, where k is an integer of the same parity as n, and !kl ~ n.
143
RESIDUES AND THEm APPLICATIONS
Prove that the polynomial P.(z) then has (n+k)/2 zeros in the right half-plane. lhNT. Represent Pn(Z) in the form Pn(Z)
= zn ( l +
: 1 + ..•
+ :: )
and apply the principle of the argument to the semicircle for a sufficiently large R.
lzl < R,
Res > O
96lJ. Find the number of zeros of the polynomial
z8 +z6 +6z4+1W+Sz1 +4z+1 in the right half-plane. 966. Find the number of roots of the equation
z'+2zs+sz•+z+2
= 0
in the right half-plane and in the first quadrant. 967. How many roots in each quadrant has the equation
2z4-3z3 +3z1 -z+ l = 0? 968. In which quadrants are the roots of the following equation to be found1 z4+zs+4z1 +2z+3 = 0. 969. Prove that the number of roots of the equation
z11n+°'z••-1+p• =
0
(°' and p are real numbers, °' ¥= 0, p ¥= O; n is a natural number), which have positive real part is equal to n if n is even. However, if n is odd, the number of them is n-1if°'>0 and n+l if°'< 0. lhNT. Consider the increase of arg (zll•+oczln-1+/JI) when the point describes the boundary of a right semicircle of large radius.
11.
If the coefficients of the polynomial Pn(Z)
= z•+0iz•-l+
... +an-1.::-i an
depend continuously on the real parameters a:, {J, then in order to find the relation between the number of zeros of Pn(z) situated in the right half-plane and the parameters it is possible to proceed as follows (commencing with the fact that each zero depends continuously on the coefficients of the polynomial):
144
PROBLEMS ON COMPLEX ANALYSIS
In the a.{J•plane, construct the lines Pn(i-r) = 0 (1' is a real parameter). that is, lines for the points of which among the roots of the polynomial there are purely imaginary (or zero) roots. These lines divide the a.fJ·plane into domains in each of which the number of zeros of Pn(Z) with positive real part is constant. This number can be found by taking an arbitrary point of the corresponding domain and applying to it, for example, the method of problem 96'.
In problems 970-972 determine the domains of the ix{J-plane in which the number of zeros of the corresponding polynomial Pn(z) with positive real part is constant; find this number m for each domain. 970. P(z) = z8 +az1 +cxz+f3. 971. P(z) = z8 -l-az1 +f3z+I. 972. P(z) = z8 +(ix+{J)z1 +(ix-{3)z+ix. 973. Let f(z) = Pn(z)+Qm(z)e-", where T > O; Pn(Z) and Qm(z) are polynomials prime to one another, where n > m and f(z) has no zeros on the imaginary axis. N is the number of zeros of the polynomial Pn(z) in the right half-plane. Prove that in order that the function f(z) should have no zeros in the right half-plane it is neces-
sary and sufficient that the point w = -
~:::: e-"•
should go
round the point w = 1 in the positive direction N times while the point z traverses the whole imaginary axis from below upwards (if Pn(z) has zeros on the imaginary axis it is necessary in the motion of the point z along this axis to avoid the zeros of P nCz> on the right by semicircles of sufficiently small radii). In problems 974-976 it is necessary to find the domains in the space of the coefficients a, b (that is in the ab-plane) for which all the zeros of the corresponding function lie in the left half-plane. a and b are real numbers, and T > 0. The use is recommended of the theorem of 973 and the method given on page 143. 974. z+a+be-'"• 976. z•+ (az+b)e-"•. 975. z2 +az+be-r: 977. Prove by means of Rouche's theorem that if the function f (z) has the expansion
w=
f(z)
=
w0 +crc(z-z0 )"+...
(ere ;i' 0, k ~ 1)
RESIDUES AND THEm Al'PLIOATION'S
145
in the neighbourhood of the point z0 then, for sufficiently small r > 0, there exists a (! > 0 such that any value w =F w0 of the circle lw-w0 1 zfl[1 + (a-b)•z•J. 4n1 n 1 n=l
n(+ 00
1010. cosh z-cos z = z1
1
4:.·n') •
n-1
1011. Let f(z) be a meromorphic function with a finite number of poles: ~.a., ... , a,,., not coincident with even one of the points z = 0, ± 1, ± 2, .... Prove that if there exists a sequence of contours {On}, tending to the point at infinity, and lim
Jf(z) cot nzdz = 0
(1)
n-+00cn
then m
co
}; f(n) = -n}; res [f(z) cotnz]z-arc. n=-OO
k=l
1012. Prove that if in the conditions of the preceding problem the requirement (1) is replaced by the condition lim n-+00
Jf ~z)
en
dz = 0 ' smnz
then
In problems 1018-1019 find the sum of the series, assuming the number a to be such that not one of the denominators becomes zero. 00
1013.
~ .L.J
n==-co
00
1 . (a+n)1
1016.
00
~
1014. ,L.,;
00
(-l)n
(a+n)• ·
1017. ~i (-l)n •
.L.J n1+a1
n-o
n--oo 00
1015.
2 (2n~l)ll.
n=O
2 n•!a• ·
n-o
00
~
1018 •
.L.J
n-o
(-l)n
(2n+l)B •
RESIDUES AND THEm APPLIOATIONS
151
(-:n - J'(C.l+ : • .F-(Co)
I
(4)
= .F(Co)-21/l(Co)
or .F"o)
l
=2
[.F+(Co>+.F-(Co)], .F+(Co)-.F-(Co) = l/l(Co) ·
(5)
If the contour 0 is closed and is traversed in the usual direction, then .F+(C) is the limiting value of the function .F+(z), determined within the contour (the domain D+), and .F-(C) is the function determined outside the contour (the domain D-)t. (See, for example, [l, Chapter III,§ 3] or [3, Chapter m, § 3]).
1021. Prove the formula (1) for n = 1, find the limit of the difference
Jc/J(C)df_
F(z+k)-F(z) __1_ k 2ni c
(C-z) 2
as k-+ 0. 1022. Prove formula (1) for any n by the method of mathematical induction. 1023. Prove that if 0 is a closed contour and the density of the Cauchy type integral +(C) and cf>-(C) are the boundary values offunctions analytic respectively inside and outside the contour 0, then
REMARK. If in the conditions of the problem one of the functions If>- or tf>+ is identically equal to zero the Cauchy type integral becomes the Cauchy
integral of the interior or exterior dome.in respectively.
1024. Let 0 be a closed contour. Find F+(z) and F-(z), if the density of the Cauchy type integral is the function (n is a natural number): (1) c/>(C) = (C-a)n;
1
(2) c/>(C) = (C-a)n (a is inside 0); 1 (3) c/>(C) = (C-a)n (a is outside 0). 1025. Find F+(z) and F-(z), if: (1) The function c/>(C) is the boundary value of a function analytic in D+, with the exception of a finite number of points ak, where it has poles; (2) The function cf> (C) is the boundary value of a function analytic in D-, with the exception of a finite number of points ak, where it has poles (among the points ak there may also occur the point z = oo ). 1026. Find F+(z) and F-(z), if
= CS+iC2-C+4i C"-3'2-4 circle !Cl = 3/2.
c/>(C)
+ log[(C-2)/(C-3n '2-4
and 0 is the 1027. Find F+(z) and F-(z), if c/>(C) = cot C and 0 is the circle ICl=5. 1028. Find F+(z) and F-(z), if c/>(C) = C2/(C2+1) and 0 is the real axis traversed from left to right. REMABK.
By an integral of Cauchy type taken along the real axis
F(z)
=
J--dT,
l oo l/>(T) -:n:i 2 • T-Z -00
is to be understood its principal value if it diverges in the usual sense.
INTEGRALS OF CAUCHY TYPE
155
1029. Find F+(z) and F-(z), and also the limiting values of F::1: on the contour of integration 0, if 0 is the circle !Cl= R, and 00
ef>(C) =a;+
.2
(ancosnO+bnsinnO)
n=l
is the uniformly convergent Fourier series of the real function tp(O) = ef>(Re18 ). 1030. (1) Let 0 be the circle !Cl = n/2 and/(C) a function analytic in the circle IC I (C) 1. Find F(z) outside 0, the limiting values F::l:(C) and the principal value of F(C) on 0. Evaluate in particular F(±i),F::1:(0) and F(O). 1032. Let 0 be the semicircle ICI = R, 0 < arg C< n (commencing at the point R) and ef>(C) 1. Find F(z) outside 0, the limiting values F::l:(C) and the principal value of F(C) on 0. Evaluate, in particular, F(O), F=(iR) and F(iR). Find also F'(O). 1033. Let 0 be the semicircle IC! = R, -n < arg C< 0 (commencing at the point R) and ef>(C) = 1. Find F(z) outside 0, the limiting values of F1=(C) on 0, F(O) and F'(O). 1034. Let the density of a Cauchy type integral ef>(C) = 1gn. Find F(z) outside 0, if the contour 0 is given by: (1) The boundary of the ring r < !z! < R; (2) The straight line Im C= n, traversed from left to right; (3) The boundary of the strip [Imz[ < n; (4) The semicircle !Cl = R, 0 < arg C< n (commencing at the point R); (5) The semicircle !Cl = R, -n < arg C< 0 (commencing at the point R).
==
=
156
PROBLEMS ON OOMPLEX ANALYSIS
In parts (4) and (5) find the limiting values of F*(C) on 0 and evaluate F(O). In problems 1035-1040 find F(z) outside 0, assuming that the contour 0 is an arc connecting the points a and b, and c/>(C) is the given function. 1035. efJ(C) = I. 1036. t/J(C) = c. 1037. = n(z-C)'. (2) K(z, f> is a function of 21, [ analytic everywhere in D and L(z, C) is a function of z, Canalytic everywhere in D except at the point z = C, at which it has a pole of the second order,
1
L(z, C) = n(z-C)• +l(z, C) , where l(z, C) is a function of z, C which is analytic everywhere. (3) K(z, C) = K(C, z), L(z, C) = L(C,z), l(z, C) = Z(C, z) (the symmetry of Green's function g(z, C) = g(C, z) is considered known). (4) Ifj(z) is an arbitrary analytic function in G, continuous in the closed region G, then
(J(z), K(z, C))=
Jf K(C, z)f(z)dxdy =f(C) G
(the "multiplicative property" of the kernel K(z, E}). (5) With the same conditions (f(z), L(z, C>) = - 2
n
I Ja_"g~z,_C> G
ozoC
f(z) dxdy = 0
(the "orthogonality property" of the kernel L(z, C>)· RBMAB:S:.
See the literature indicated in the remark to problem 1054.
167
THE INTEGRAL FORMULAE OF :POISSON AND SCHWARZ
§ 4. Poisson's integral, Schwarz's formula, harmonic measure If the real function u(C) = u(R, 0) is defined and piecewise continuous on the circumference C= RellJ (0 ~ 0 < 2n), then the PotaBOfl inlegral.
Ju(R, 2n
I
u(z) = u(r, rf>) = .2n
R•-r' 0) R•-.2Rr cos (0-4>)+r8 dO
(1)
0
defines in the disk lzl < R (z = rel) a harmonic function having at the points of continuity of u(Cl boundary values equal to u(C): limu(z) = u(C) z-+C
(z-+ Calong any non•tangentia.I. path). The corresponding function/(z) analytic in the disk lzl < R, is defined by Bchwarz's formula;
=u+w,
2n
/(z)
= .2~
J
u(C)
~: :
dO+iv(O)
0
(v(O) is an arbitrary real number).
1072. Prove the following assertions :
J In
I
(I) 2n
R"-r8 Jll-2Rrcos(8-4>)+r8dr=l;
0
(2) u(r, cf>)-u(R, 00)
f
In
I
= 2n
R1 -r8 [u(R,O)-u(R,Oo)] Jll-2Rrcos (0-cf>)+rl d(J;
0
(3) If lu(R, 0)-u(R, 00 )1 9
I
J
.-n 111-0,1 -0 0 1 < 9 , then
-
B1 -2Rroos (0-cf>)+rl > 4Rrsin2 : (5) If lcf>-00 1 <
;
;
and the condition of pa.rt (3) is satisfied,
168
PROBLEMS ON COMPLEX ANALYSIS
then iu(r,,P)-u(R,0 0)1
< e+
M(Jll-r8) 23tA -
where
f
2•
A= 43tRrsin2 : ,
M =
lu(R, 0 )-u(R, 00)id0.
0
1073. Prove that if
C+z C-z
C=Re'', z = re1•, then
=
(R2-rl)+i2Rrsin (t/>-0) R 8 -2Rrcos (O-,P)+r8
Using this identity obtain the following expansions:
r 2 (;r co
u(z) = u(O)+
2 (;
(a. cos n,P+bn sin n,P),
n=l
co
v(z) = v(O)+
(-b,. cos ncf>+an sin ncf>),
n=l
co
f (z) = f(O)+
2 o.z•,
f (0) = u(O)+iv(O),
n=I
where I•
a.=
!J
u(R, 0) cosnOdO,
bn =
! J••
u(R, 8) sin nOdO,
0
0 In
- - a.-ibn a) -'"'dav. OnR-- _I_f 3tRu (R ,ve 0
1074. Prove that for a harmonic function u(z) the Dirichlet integral is given by
D(u) =
JJ(u~+u~)dxdy =
~ + 2( :)" (a,,cosn+ n=I 00
v1 (z) = -v(O)+
2( !)"
(-b,,cosnef>+a.sinncf>),
n-1
where a,,, b,,, c,. are defined just as in problem 1073. Here Re/1 (C) = - Ref(C) = -u(C), Imf1 (C) = Im/(C).
170
PROBLEMS ON' COMPLEX .ANALYSIS
lzl > R make the substitution z = ~
HINT. In Schwarz's formula for
/f on the
and make use of the fact that C = R 8
Z1
ICI = R. In problems 1078-1083 find the functions f(z) (!zl < R) and / 1 (z) (lzl > R) defined by Schwarz's formula if u(C) is the given function. 1078. (I) u(C) = Re [cp(C)+vi(C)], (2) u(C) = Im [cp(C)+vi(C)], where cp(z) is a function analytic for !zl :( R, and VJ(z) is a function analytic for !zl ;;;:i: R. 1079. u(C) =Re 1080. u(C) = Re I/en. circumference
en.
1081. u(C)
=
Re log CC 1 (R
>
1).
1082. u(C)=Re-i/(c C1 ){forC>l,-V(c C1 )>0, R>l)· 1083. u(~) = Re log C. 1084. Prove that Schwarz's formula can be written in the form f(z) = _.;. ni
J ICl=R
u(C)dC -f(O).
C-z
HINT. Use the equality
C+z 2 C 0, that is, express the harmonic function u(z) and the analytic function f(z) = u(z)+iv(z) in terms of u(t) (-oo < t < oo). HINT. Make use of the conformal mapping of the half-plane onto the circle.
1086. Derive Schwarz's formula for the strip 0 HINT.
< Im z <
1.
Use the conformal mapping of the strip onto the half-plane.
The harmonic measurs w(z, oi:, G) of the boundary arc oi: at the point z with respect to the domain G is defined as a bounded function, harmonic in G, which is equal to 1 at the interior points of the a.re oi: and to 0 at the interior points of the remaining pa.rt of the boundary. The harmonic measure w(z, oi:, G) is an invariant in conformal mappings.
TH'E INTEGRAL FORMULAE OF POISSON AND SCHWARZ
171
In problems 1087-1090 the domain G is the circle lzl < 1 and ro(z, (Ji, 02 ) is the harmonic measure of the arc IX= (0 1 , 02 ), w = e19, 81::::;;; (J::::;;; 82. 1087. Using Poisson's integral prove that o, 1 l-r2 ru(z,0 1• 82)= 2n l-2rcos(O-)+r2d0, o, and in particular, that c.o(O, 01 , 02 ) = (08 -01)/2n. 1088. Find the level lines of the function dc.o (z, 01 , 0) /d(J for a fixed value of 6 (z = re'• is a variable point).
J
HINT. Prove that
~: = 2~ I::=: I·
where w' is the end of the chord from w which passes through z.
1089. Let us denote by w' the end of the chord from w which passes through the point z. Let IX be the arc (0 1 , 61), and IX'(z) the arc described by the point w', when w traverses the arc IX. Prove that the length of the arc IX'(z) equals 2nc.o(z, Ov 02). 1090. Find the level lines of the harmonic measure c.o(z, 01 , 01 ) of the arc (01 , 02). Using this prove that the integral defining the harmonic measure (see problem 1087), actually has the limiting values 1 on (61 , 02 ) and 0 on the complement (interior points of the arc are considered). 1091. For the half-plane Im z > 0 determine the harmonic measure c.o(z, a, b) of the segment (a, b), of the ray ( -oo, b) and of the ray (a, oo). What is the geometrical meaning of these harmonic measures 1 1092. Find the harmonic measures of the boundary rays of the angular sector. O R, Im z > 0. 1095. Find the harmonic measure of the boundary circular arc I' of the domain lzl > R, 0 < arg z < 2n. 1096. Find the harmonic measure of the boundary circumferences of the annulus r < lzl < R.
172
PROBLEMS ON COMPLEX ANALYSIS
In problems 109'7-1101 the domain G is bounded by a compound contour I', consisting of n simple, smooth contours I',.(v = 1, 2, ... , n). The contour is traversed in the positive direction with respect to the domain G; the norm.al n is inward with respect to the domain G. The period with respect to I',. of a function analytic in G is defined as the integral
Jdf(z)
r,, (see page 16).
109'7t. Prove that if the harmonic function u(z) is single valued in G, then the period of the analytic function /(z) = u(z) +w(z) along is equal to
r.
-f~:da. r,. 1098. Prove that for the complex Green function g + ik of the domain G (g(z, C) is the Green function of the domain G, k(z, C) its harmonic conjugate) the period along equals 2mo,.(z), where m,,(z) is the harmonic measure of I',. with respect to the domain G. Prove that
r.
2" w,.(C) = 1. HINT. The function u(C) harmonic in G can be represented in the form u(C) = - 1-
Ju (z) (Jg(z, t') da
2nr
an
(llff problem 1070).
l•.
Express in terms of w,,(z) the function u(z) which is bounded, harmonic in G and has constant values c,, on the I',,('1' = 1, 2, ... , n). 1100. Prove that for a function v(z) which is conjugate to a function u(z), single valued and harmonic in G, the periods p .. along can be represented in the form
r.
p,, = -
f ()aw..
r
(z)
..3-
u z - 0- w s . n
t For problems 1097-1104 see the references indicated in the remarks to problem 1014.
THE INTEGRAL FORMULAE OF POISSON AND SOHWARZ
173
1101. Let w,.(z) be the harmonic function conjugate to ro.(z) and p"., the periods of the function w,.(z) = ro,.(z)+iw.,(z) along I'". (1) Prove that
(µ, 71 = 1, 2, ••• 1 11.) •
'P/JY ='PY/I HINT. Use the representation
p"., =
__1_
J
wp
2n r
o(J)y da. 0n
(2) Prove that
" 2 •-I
tppy
(µ = 1, 2, ... , n) •
= 0
1102. Let C.(71 = 1, 2, ... , n) be arbitrary real numbers. Prove that if p." are the numbers defined in problem 1101, then the quadratic form
2"
... "
p.,pC.Op
~0
1
the case of equality occurring only when allthe "• are equal. llDrT. Apply the formula D(w) = -
f
OJ :
da
r to the harmonic function w(z) ...
I" C,,OJy (1)
(see problem 1084; the afgn is
..-1
ravened because n is now the inward normal).
1103. Prove that the quadratic form
is positive definite, that is, positive for all sets of values {c.}, excluding 0i = 0 11 = ••• = o...1 = 0. 116'. Prove that the system of equations 11-I
2 p.,".A" = B"
.. -1
(µ = 1, 2, ... , n - 1)
174
PROBLEMS ON COMPLEX ANALYt!!IS
(the A,, a.re unknowns) has a. unique solution for any B,.. Using this, prove that for any function u(z) harmonic in G, generally speaking not single-valued, it is possible to select constants A 1 , A 2 , ... , A 11_ 1 in such a. way that the harmonic function 11-l
u 1 (z) = u(z)+}; A.w.(z) v-1
is single-valued in G.
§ 5. Some singular integralst Let h(of>) be an integrable function with period 231; I.Jo, IJn, b11 (n = 1, 2, ... ) its Fourier coefficients; J(z) = u(z)+it1(z) and J 1 (z) = u 1 (z)+it11 (z) functions analytic respectively in the disk lzl < 1 and in the domain lzl > 1, defined by means of Bchwa.rz's integral
an J ..n
elt+z
1
c;-
h(t)-1-,- d t
e -z
(z
= rellf>).
(1)
0
The function h(•), defined by means of a singular integral with the Hilbert kernel - 1- cot !-t · 2n 2 ·
~
h(of>) = T[h(of>)]
J...
1
~ !n
J + JIn) ,
cf>-t l (.__. h(t) cot2 -dt = !n ~
0
0
(2)
·~
is said to be conjugate to the function h(cf>). The series 00
}; (-b11COSf'lcf>-l-a11 sinnof>)
(3)
11=1
is said to be conjugate to the Fourier series
-i +}; 00
(a11cosnof>+b11 sinncf>)
(4)
n-1
of the function h(cf>).
1105. Prove the following assertions: (1) The principal value of Schwa.rz's integral (1) at the point e1•, if it exists, is equal to ih(
0
~
~
k1 () cot 2 t d}k(t)dl
0
=n
0
2 (anb~1)_bna~1)), n=l
where (an, bn) and (a~1>, b~1>) are the Fourier coefficients of the functions k() and k1 (); (3) 21n
I
1 "{
1 2n
0
fl>I k(t)cotT-t -T -dl} cot-dT 2
2
0
= -k()+ or, if
k=
2~
111:
f
k(tf>) def>
0
T(k), In
T[T(k)] = T(i) = -k(ef>)+
2~ f k(t/>)dtf>. 0
1119. Applying the rectangle formula for the approximate evaluation of the singular integral
obtain the formula (5)
where
k1:,n = k
nk) • -k1:,n = k-(2nk) (2n 7
(k=O,
±1, ±2, ... ).
In particular, if n = 5 and k1;
=k(7~).
- -(nk) k1;=k10,
178
PROBLEMS ON' OOMPLEX ANALYSIS
then
-
I[
hk = -- 10 (h11:+i-hk
1)
3n
n cot 20 +(krc+s-hk-a) cot 00
5n
7n
+(hrc+6 -k1 _ 6) cot 00 +(h1:+7-h1:_7 ) cot 00 + (h1:+u-hrc_ 9) cot
:~ J·
(6)
1 t 1120. Let (l>(t) = - 231: [h(cf>+t)-h(cf>-t)]cot 2 • Assuming that h() can be continuously differentiated three times, prove that
where
1}Irc = max jh,(cf>)I
(k = 1, 3).
o.;;q,.;;;n
Using this and the estimate of error Rn of the rectangle formula for the calculation of k(c/>):
Rn= (b-ar (l>"(E),
24n
a,.,_;; E ~ b
(a=
o, b = n),
obtain the estimate
IRnl
~ 2~:1 [Mi+(;+ :)Ma]·
(7)
In particular, if n = 5
IR!< l ·65X l0- 2{M1 +7 ·6U·f3 ).
(8)
HINT. Expand 4'(t) by Taylor's formula with remainder term in Lagrange'11 form (with h'") and use the inequalities
a: cot a: .:;;; I, and a:Bcosa: ~ 1 fi sins a: """ or
1-a: cot a: . sm1 a:
~
1
179
SINGULAR INTEGRALS
1121. Prove that if k(+n) = k(), then formula. (6) (see problem 1119) acquires the form
~ + (k1c+ -k1c-a) cot ~~].
k1c = - 110 [ (k1c+ 1 -k1c- 1 )2 cot and if k(+n) =
(9)
8
-k(), then
In problems 1122-1128 calculate the values of h1c to five significant figures by one of the formulae (6), (9) or (10) and compare with the exact values. Also estimate the errors by formula. (8) and compare with the actual errors. 1125. k() = sin . 1122. 'k() = cos 2. 1126. k() = sin 3. 1123. k() = cos 3. 1124. k() = cos 4cp. 1127. k() = ( 1,
1128. k()
= { _,/,.
n '""
~ ).
0
if 0 ~ if /2
n
~ ~ n
~
and k(-)
n/2,
,/,. ./'
=
-k().
k(-) = -k().
~"'""'n,
REMARK. In problems 1H'7, 1128 compare the result with calculations, based on the expansion of h( 0 and Im z < 0 respectively. Prove that if k (x)e LiplX (0 a,
k-l
if
~-1- ~), na« .L.i k+cc a co
1--
(
k
x< -a,
if lxl 0, cc> 0 are not integers). Both the latter expansions a.re valid also for x = -a, where the function k(x) defined by the integral standing on the left remains continuous. HINT. For la:!> a consider the integral of l/[Co'(C-111)) along the bound· ary of the ring a < IC! < R with a out along the re.dial segment a < Re C = t < R, Im C= 0 (and deleting the circle IC-a:! < (/, if a: > a), and pass to the limit as R-+ oo (and (!-+ 0, if a:> a). For la:I (x-t-iy)dy = cf>±a>, X-+:1::00 1/: O
Jim ..!..J'l'(x-iy)dy = 1/'±rr,,, n
X-+:1::00
O
then T[cf>(t)]
= -{c/>(t)-
c/>-a>].
T[1J1(t)] =
{1/'~)- 1/'oo~'l'-- 00 }
(2) Find T[c/>(t)] and T[1J1(~] (Im t = 0, Im t = n), if the functions cf>(z) and 1/'(z) have poles at the point a (0 < Im a < n) and at the point b (-n < Im b < 0) respectively and elsewhere satisfy the conditions of part (1). 1147. Find functions defined by Schwarz's integral (15) in the strip 0 < Im z < n for the following functions k(t): (1) k(t) = Re cf>(t), cf>(z) is from problem 1146 (1); (2) k(t) = Re 1/'(t), 1J1(z) is from problem 1146 (1); (3) k(t) = Re cf>(t), cf>(z) is from problem 1146 (2); (4) k(t) =Re 1f'(t), 1/'(z) is from problem 1146 (2); (5) k(t) =Re 1/(t-a), 0 -roll
~
0 as
e ~ 0.
REMARK. The relation (19) also holds almost everywhere for roeL 2 • This enables us to define for co e L 2 the transformation Bro as the limit in mean of the transformation Be co.
1170. Let H« = L 2 n Lip IX, that is, H« is the space of functions which belong simultaneously to L 2 and Lip IX, Prove that all functions g(C) e H« satisfy the inequality
ISg(l)-Sg(-1)1 ::;:;; A«lul", where
HINT. Subtract from the difference which is being estinia.ted multiplied by -1/n the quantity
g(l)
ff
ff
4(;Md11
((;1-1)2 +g(-1)
4(;dEd11 ((;•-1)1
ReC>O
ReC>O
(prove that both sides a.re equal to zero!).
1171. Let w{C) e H« and w(C) =
w( u;v C+ u~v).
Prove that
-- 2-"lu-v'«lwl 'rol oc-
I
I
Gt
and using this, that
!Sw(u)-Sw(v)I ::::;; 2 "Aat'w'.oc:u-v1°'.
191
SINGULAR INTEGRALS
REJirAB:s:, From this result it follows that 8 is a bounded linear operator in H •.
1172. Let g(C) e H 11 and
I
1- _ _!.)g(C)dEd11 _!_ff (-C-z n C
T 0 g(z) = _
B
(in comparison with Tg the passage to T 0 g ensures integrability at infinity). Prove that 1--_!.)dEd11 = _..!...ff (-c-1 n C
l
(20)
B
and using this fact, that ITog(l)-g(O)-Sg(O)I ~
where 1
B" =-;;
Bcx!Y!cx,
ff IC-II 1ci11-1 dEd'fJ. B
HINT. Multiply g(O) by the left; side of (20).
1178. Let w(C) e H 11• Putting ro(C) = w[u+C(u-v)],
prove that IToro(u)-Toro(v)-w(v)(ii-ii)-Sw(v)(u-v)I ~ B 11 \wl 11 \u-1J[H 11,
and deduce from this the relation:
(T0 co(z)k= w(z), (T0 w(z))s = Sw(z). 1174. Show that the search for the quasi-conformal mapping w(z) of the whole w-plane onto itself, which satisfies the equation (q e Ha.)
WW= q(z)w:
by means of the substitution w(z) = z+T0w(z)
leads to the singular integral equation
w-qSw (see the remark on problem 1167).
=
q,
OHAPTEB IX
ANALYTIC CONTINUATION. SINGULARITIES OF MANY-VALUED CHARACTER. RIEMANN SURFACES § I. Analytic continuation 00
1175. The function /(z) =
2 t' is expanded into a Taylor
series n-o in the neighbourhood of the point z =a (lal < 1). For what values of a does this expansion permit the function f (z) to be continued analytically ~ oo
n
2 !.__ is n=l n
1176. The sum of tlie power series /(z) =
expanded
into a Taylor series in the neighbourhood of the point z = -1/2. What is the domain into which the function/(z) is continued by this! oo
1177. Prove that the function /(z) =
n
2 (- l)n+i!._n n=I
can be
continued into a larger domain by means of the series 1-z (1-z)2 log 2--2-- 2.22
(1-z)B
3.28
1178. The power series fi(z) =
l
00
n=l
l
zn
-
n
.2 00
and
/ 2 (z) = i:n:+
(-l)R
(z~ 2 )n
n-1
have no common domain of convergence. Prove that the functions / 1 (z) and f 2(z) are nevertheless analytic continuations of one another. 1179. Prove that the functions defined by the series 1 (1-a)z (1-a)2z2 1+az+a1z2+ ... and - - ... ' 1-z (l-z)2 (1-z) 3 are analytic continuations of one another.
+
192
i93
ANA.LYTIO OONTINUATION
1180. Let the power series
f(z) = a0 +a1z+ ... +anzn+ ... have the radius of convergence R = 1. By carrying out the change of variable z = Z/(i+Z) let us transform it into the form
f(z) =J[Z/(i+zn = F(Z)
= ea+eiz+ ... +c.Z"+ ...
Denoting bye. the radius of convergence of the series obtained prove the following assertions: (i)
e~
!'
function /(z), then (2) If
and if the point z = - i is a singularity of the
!.
e=
! < e<
i, then the equality
/(z) =
F(Z) = F( i z
z)
permits the analytic continuation of the function /(z) into a domain exterior to the disk lzl < i and interior to the circle of Apollonius
lz zii= e·
(3) If e = i, the equality given in part 2 analytically continues the function /(z) into the half-plane Re z < i/2. (4) If e > i, then the function f(z) is analytically conti-
I I
nuable into a domain exterior to the circle of Apollonius z z i =
e·
00
1181. Prove that the power series f(z) =
2 z n represents a func2
•-O
tion, analytic in the disk lzl < i and having the circle lzl = i as its natural boundary (that is, f(z) is a function which is not continuable beyond the unit circle). HINT. Using the identity
/(z)
= z1 +z'+
... +zaA:+J(z•t),
prove that for any point of the form < t < I).
J(tC> -+ oo as t -+ 1
co
C= ·vt1
(le is a natural number)
In problems 1182-1184 prove that the functions represented by the given power series are not continuable beyond the unit circle.
194
PROBLEMS ON COM"PLEX .ANALYSIS 00
~-, ~2n
1182. f(z) =- L.J --~,,-
·
n=O HINT. Use Pringsheim's theorem (see problem 642). 00
1183. /(z) = ,}; znl . n=O HINT. If p and q are coprime integers and
n;;;;;, q, then
(
2pnl
re q
)nl
= ,.n !.
00
~1
1184. /(z) =
znl
L_; rii"' n=O
REMARK. In problems
1182-1184
particular cases are
considered of
Hadamard's general gap theorem: 00
If the indices of non-zero coefficients of the power series /(z)
= I a,,zn
n=O form a sequence nu n 2 , •• ., in which n1:+ 1 > (I+ix)n11; where IX> 0, then the boundary of the circle of convergence of the series is a natural boundary of the function /(z).
1185. Prove that the series
.i:(l -zl n=l
l n)
n 1 1-
+
-z
in the domains
[z[ < 1 and [z[ > 1 represents two analytic functions which are not analytic continuations of one another (see also problems 696-699). 1186. Let /(z) and (z) be arbitrary integral functions, and S(z)
oo (--------1-zn l-zn-1) . Prove =,}; n=l I+zn I-j-zn-1
that the expression
1 1 1j!(z) = 2 [f(z)+(z)]+2S(z)[f(z)-(z)]
represents in the domain JzJ < 1 the function /(z) and in the domain [z[ > 1 the function (z). 1187. (1) Prove that if oi is a real irrational number the series
represents in the domains JzJ < 1 and Jzj > 1 analytic functions for each of which the circle [z! = 1 is a natural boundary.
195
ANALYTIC CONTINUATION
HINT. Prove that the sum of the series increasel'I without limit as z __,. e•imtzn along a. radill8 vector. REMARK. This problem is a. parlicula.r case of the following general theorem: Let L be a curve, closed or open, which has at every point a definite radius 00
of curvature. If the series
l;
converges absolutel;v· and the
Cn
points
n-1
a1' a 1 , ... , an, ... all lie on the curve L and are distributed on it BO that. any finite aro of the curve L always contains an infinite 110t of them, then the series
2 ---"-· an-z 00
.F(z)
c
=
n-1
represents a. function analytic in any domain not containing points of the curve L, and for which this curve is a. singular line (see GoUltSAT, A Oourae in Mathematical Analysis, Vol. 2, Chapter XVI).
(2) Prove that if oi is a real rational number, then the series of part (1) represents a rational function. 00
1188. Prove that the series
J;-,,-1n=l
n +n
(11"
=- e" 101 ") converges for
Re z > 1 and its sum has the straight line Re z boundary.
=
1 as its natural
00
1189. Prove that the functionf(z) = J; e-mlz is analytic for Re z>O m=o and has the straight line Re z = 0 as its natural boundary. 1190. Prove that the functionf(2!) defined in the half-plane Re z > 0 by the Dirichlet series 00
f(z) =
L
ane-Anz
n=l
where an= (-l)n+i, A.21c_ 1 = 2k, A.21c = 2lc+e- 2" (le= 1, 2, ... ), can be continued analytically into the half-plane Re z > - l. HINT. Write /(z) in the form 00
/(z)
=
J;(1-e-ze- 2")e--llcz k-1
and prove that in any finite domain belonging to the half-plane Re z ··. --1, 11-e-ze-llcl 0 by
*.
means of the Laplace integral 00
/(a)
=
Je-•e' sine' dt, 0
is to be continued analytically into the half-plane Re 8 > -1. 1192. Euler's Gamma function is defined in the half-plane Re z by means of the integral
>0
00
I'(z)
=
Je-'F-ldt
(F-1 = e(!)·It must be remembered that on the Riemann surface above one and the same point z, in addition to various logarithmic branch points there may also be found other points of single-valued and many-valued character.
1203. Explain for which values of z the values of w(z) a.re identical on all the sheets of its Riemann surface above the z-pla.ne, if: (1) w= (z2-9)y(z); (2) w =sin z + (z2 + 4) Log z; (3) w = sin z + (z• + 4)• Log z. Are the values of w' (z) identical at the same points 1 _.1204. For ea.oh of the functions: (1) w = zyz. (2) w = z1 Log z, w(O) = 0, verify that at the point z = 0 there exists a first derivative which is the same for all the branches and that a finite second derivative does not exist. In problems 1205-1212 expand ea.oh of the given functions w(z) into a series of powers of the local para.meter t in the neighbourhood of all the points of its Riemann surface situated above the given z points; indicate the domains of convergence of the resulting series. 1 1205. w= l+y( 2 -z), z = 1, z = 2.
1206. w
=
y[y(z-1)-2], z
=
1, z = 5, z = oo.
SINGULARITIES OF MANY-VALUED CHARACTER
201
1207. W= y[l+J/(z -1)], z = 1. 1208. w = y[(-i/z-a)(yz-b)], z = oo (a#= b). 1
= 0.
1209. w = eTz, z
sinJl'z 1210. w = --2-, z = 0.
z
1211. w= cotVz, z = 0. 1212. w = JI'sin z, z = o. If the function w = /(z) is single valued and the function z(w) inverse it is many valued, for the determination of the algebraic branch points of the function z(w) it is neceBBary to find the zeros of /'(z), the multiple poles of /(z) and investigate the behaviour of /(z) at infinity. In this, a point z0 #: oo corresponds oo an algebraic branch point of order k-1 of the function z(w), if in the neighbourhood of Zo the Laurent expansion of the function /(z) has the form
oo
00
= w0 + 2
/(z)
o.(z-z0)n
(01c
#:
0)
n=k
or 00
/(z)
If Zo
=
=
2
en(z-z.)n n=-k
(o-1c
#: 0).
oo, then the given expansions must be of the form
(ore #: 0)
or -00
/(z)
= Wo+
2
enzn n=-k
(ll-Tc
#: 0).
In problems 1213-1219 determine the singularities of z(w) if
w is the given function. 1213. w = z(l-z). 1214. w = z3-3z.
ums.
w=
1216. w =
1217.
(l~z)2
(-z ) 1-z 2
•
z-a
W=~l
z-
(0 0 onto the interior of the polygon P, is determined by the Schwarz-Christoffel formula: (1)
where - oo < IJi < a 1 < .. . < an < oo a.re points on the m-e.xis which correspond to the vertices Air A 1 , ... , An of the polygon P; 0 and 0 1 a.re complex constants. The mapping of the upper half-plane onto the exterior of the same polygon ia effected by the function
f n(z-ai/kz
F(z) = 0
n
1
o k-1
dz
_ +Ou
(2)
(z-b)•(z-b)•
where b is· the point of the upper half-plane corresponding to the point at infinity of the w-pla.ne; ale a.re points which correspond to the vertices Ak of the polygon (now oo > a~> ~ > ... > ak > - oo); the fJkn a.re the n
external angles of the polygon (/Jk
= 2-or:t, 2
k=I
/Jk
= n+2). If it is
a ques-
tion of the mapping of the exterior of a polygon some of the points Ak may be geometrically coincident.
1321. Prove that in formula (1), of then points aa:, three can be assigned arbitrarily, and the remaining n-3 points and the cont For thia section see [I, Chapter VIII, § 7], [3, Chapter II, § 3]; B. A. FucHs and B. V. SHA.BAT, l!'imotWna of a Oomple:a Variable, (l!'unlatsii laomplelamogo peremennogo), hd ed., Chap. VIII, Fizmatgiz, 1959. English translation published by Pergamon Press (1962). 211
212
PROBLEMS ON COMPLEX ANALYSIS
stants 0 and 0 1 (in all n+I real parameters) are determined from the relations 0 1+1
J lf'(x)ldx =
~
°1+1
101
J
~
n n
lx-atj«t-ldx = l1
t-1
a1
(i=I,2, .. .,n;an+l=a1, andj an
oo
a1
=J + J), an
-oo
where the l 1 are the lengths of the sides A 1A 1+1 (An+i = A1), 0 1 = f(a 1) = w 0 (w0 is the affix of the vertex A 1) and arg 0 = () (() is the angle of inclination of the side AnA1 of the polygon to the real axis), The given relations exceed the unknown parameters in number, however: they determine these parameters uniquely. Explain this and show which of the relations can be discarded. 1322. Prove that if one of the vertices of the polygon is the image of the point at infinity, for example an= oo, then the mapping function is of the form z n-1
f(z) = 0
Jn (z-at) t-ldz+01. 11
0
t=I
HINT. Make the transformation C= -1/z, if none of the at equals 0 and C= - - 1 - , where a at, if one of the points at= 0 (k ... 1, 2, ... , n-1).
z-a
+
1323. Prove that if one or several of the vertices A1; is at oo, then formula (I) remains in force, if by 1Xt31: is understood the angle between the corresponding rays at the finite point of their intersection, taken with a negative sign.
A.y
2) FIG.
32
213
CONFORMAL MAPPINGS (CONTINUATION)
HINT. Let At= oo. If OCA: < 1, then consider the polygon P', cut off from P by the segment Ai,At' where Ai, and At' lie sufficiently far from the sides At-iAk and AtHAt (Fig. 32 (1)) and in formula (1) for P' pass to the limit as
Al: -+ oo, A/: -+ oo. If, however, oc1c ;;;i:: 1, then join A_tA/i' in P by a polygonal l.irie (see Fig. 32(2)) and by enlarging it similarly extend it to infinity.
1324. Determine the quantities °'1:, which enter into formula (I) for the infinitely distant vertex of the "polygon" formed from parallel rays represented in Fig. 33.
A,~~Az
A A,=======
A 1 ~Aa
1)
2)
A,-A3 A
z
AJ
3} Az
A,=====~
A* A+
----
A,=======;:
Az
6)
5)
A, A,c::=======! A1 A,
As======;; A• 11.
A3=======?Az
~ ,_A• '\
tZ/(
Aa
--
A
A~'
8)
7) FIG.
33
HINT. Pass to the limit as recommended in the hint to problem 1323.
1325. (I) Prove that in the mapping of the unit circle !z! (z)
=
f'(z) f(z)
is single valued in the z-ple.ne. REMARK. This formula immediately gives the solution of a number of problems of Chapter II, for example, problems 281, (2), 282(1), 284, 290. It is recommended that these problems should be solved again e.nd the constants entering into the general formula determined.
1343. Map the exterior of the double angle (Fig. 41) onto the exterior of the circle lzl >I subject to the conditions: w(oo) = oo, w'(oo) > 0. 1344. Prove that the function w = f(z), which maps the upper half-plane Im z > 0 onto the exterior of the star of problem 1342 is of the form n
f(z) = 0
n
(z-aa:)11"
k-1
-
(z-z0 ) (z-z0 )
'
where z0 is the point of the upper half-plane which is transformed
into oo.
0
FIG.
41
1345. Let the domain P in the w-plane be bounded by the ray [O, oo), by n - I segments, going from w = 0 to the points B 1
CONFORMAL MAPPINGS (CONTINUATION)
221
(i = 1, 2, ... , n - 1), and m - 1 rays going from the points D. (8 = 1, 2, ... , m -1) to oo, such that their continuations in the opposite direction pass through the coordinate origin (Fig. 42). Let At (le = 1, 2, ... , n) be the vertices of P at the origin, rtt'lt the cor-
42
FIG.
responding angles, 0 1(j = 1, 2, ... , m) the vertices of P at oo, i'J"" the corresponding angles, Ai Bi A 2 ••• An Oi Di ... Om a positive traversal of the boundary of P. Prove that the function w = f(z), which maps the upper half-plane Im z > 0 onto P, is of the form
n 0 n n
f (z) =
(z-ak)«a,
(4)
k=l m
(z-e1)'JIJ
j=l
where ak, e1 are the points on the x-axis which correspond to the vertices At, 01. The points b,, a. on the x-axis which correspond to the vertices B,, D. are the roots of the equation
2,
2
- 'JIJ -=0
k=l
j=l
J
n
m
-rtk- Z-!Zt
(5)
Z-C·
a.
How are the parameters 0, ak' bi. Cj, determined 1 What is changed if one of the parameters ak, b,, Cj, a8 is equal to 001 HINT. Prove that
n = n n-1
n
/'(z)
/(z)
m
'\'"1 IXk '\'"1 "/J = L.J z-ak - L.J z-e1 k-1
J=l
i-1
(z-ds)
s=l
n
k=l
n n
m-1
(z-bi)
m
(z-ak)
1-1
(Z-Cj)
222
PROBLEMS ON COMPLEX ANAJ,YSIS
1346. Prove that formula. (4) of problem 1840 still applies to the case where the domain P, contains two rays [O, oo) (Fig. 43). Here the vertex at the coordinate origin is taken into account also in the case when there issue from it only two rays forming a single straight line.
JJ,
/' 81
c,
In problems 134'7-1351 map onto the upper half-plane Im z > 0 the domains of the w-plane given in the corresponding diagrams subject to the given conditions, and determine the parameters a and b (a > 0, b > 0).
c,[
~
c
c,
FIG.44
A,1~ FIG.
c,
46
134'7. The domain indicated in Fig. 44,
w (.A 1 = 0, B = ke'-., .A 1 , 0 = oo) -+ z (-1, b, 1, oo). 1348. The domain indicated in Fig. 45,
w (.A 1
= 0,
B
= ik,
.A 1 , 0 1
= oo, D = iH,
01
= oo)
-.z(-a,O,a,
!
,oo,-!)·
223
CONFORMAL MAPPINGS (CONTINUATION)
1849. The dome.in indicated in Fig. 46,
w (..4.1 = 0, B1 =
-
he,_, ..4. 8, B 8 =he*, ..4.3 , 0 = oo) -+ z (-1, -b, 0, b, 1, oo). 1350. The dome.in indicated in Fig. 47,
w (A1 = 0, B =he'••, A11 , 0 1 = oo, D =He,.•, 0 1 = oo) -+
8,~Bz
C
z(-a
-1
I
0, b,
.!.., b
.!..).
oo ' a
c
FIG. 46
I I
'
FIG. 47
'\\
/
\ I
1)
2) FIG. 48
1351. (1) The dome.in indicated in Fig. 48, (1), w (oo) = oo, w (± 1) =
± 1.
(2) The dome.in indicated in Fig. 48, (2) (the angles between the cuts are TC/n; the extreme segments form with the corresponding rays of the real axis angles n/2n),
w(oo) = oo,
w(A 1 , An) -+
z (-
1, 1).
1352. Let the domain P in the w-plane be the horizontal strip of width H with cuts going left to oo from the points B 1 (i = 1,
224
PROBLEMS ON' OOMPLEX ANALYSIS
2, ... , n - 1) and right from the points Ds (8 = 1, 2, ... , m - 1) {Fig. 49). Prove that the function w = f(z), which maps the upper half-plane Im z > 0 onto the domain P is of the form n
f(z) =
2~
m
log (z-a1c)-
k=l
2 ~log
(z-c1)+0,
(6)
J=l
where a1c, c1 are the points on the x-axis corresponding to the vertices A1c and 0 1 of the domain P; the h1c are the distances between the cuts on the left, and the l1 are the corresponding distances on the right. The points b1(i = 1, 2, ... , n - 1) and d8 (8 = 1, 2, ... , m - 1) on the x-axis corresponding to the vertices B 1 and D,, are the roots of the equation n
'1 h1c .L.J z-a1c
k-1
m
-
'1 .L.J J=l
l1 - 0 z-c1 - •
(7)
How are the parameters 0, a1c, b1o c1, ds determined 1 What is changed if one of the parameters a1c, b1o c1, becomes equal to 001
a.
FIG. 49 HINT. The problem reduces to problem 13'6. It is also possible to find the functionf'(z) which is single valued or to derive directly from the SchwarzChristoff'el formula.
1353. Let the domain P be the upper half-plane Im w > 0 with horizontal cuts going leftwards to oo from the points B 1 (i = 1, 2, ... , n) and to oo on the right from the points D. (8 = 1, 2, ... , m) (Fig. 50).
CONFORMAL MAl'PINGS (CONTINUATION)
225
Prove that the function w = /(z) which maps the upper ha.If-plane Im z > 0 onto the domain P is of the form n
m
f (z) = L.J ~!!I!.. log (z-a1;)- ~!:!..log (z-c·)+Az+B, 1' L.J 1' J k=l
(8)
J=l
where ~. cJ are points on the x-axis corresponding to the vertices A1: and OJ of the strips of the domain P, and k", Z1 a.re the distances
FIG. 50
between the cuts going in one direction. The point A 0 is transformed m
>0
into oo, A
and Im B =
2
l1 • The points b1(i = 1, 2, ... , n)
J=l
and d. (8 = 1, 2, ... , m) on the x-axis corresponding to the vertices B1 and D. a.re roots of the equation n
m
~~- ~-l1 -+An=0. L.J ·z-a1; L.J z-c1 k=l
How a.re the para.meters A, B, a", b1, cj, if A 0 passes into a. point a0 #= oo, then /(z) =
(9)
J=l
n
m
1:-1
J=l
a. determined 1 Show that
2 ~log (z-~)-2 ~log (z-c
1)
L-H
A
n
z-a0
+--log (z-a0)+--+B,
(10)
226
PROBLEMS ON OOMPLEX ANALYSIS
where n
H= };h1" L= k=l
m
};zi, A 0 onto the domain P, is of the form n-1
f(z) = -Az2 +Bz+O+
,2 ~ log (z-a1c), k=l
where the a1c are points on the x-axis corresponding to the vertices A1c of the strips of the domain P, and k1c are the distances between the cuts. The point A 0 maps into oo, A> 0 and Bis a real number. The points b1 (i = 1, 2, .. ., n) corresponding to the vertices B 1 are zeros of the derivative J'(z). How are the parameters A, B, 0, ~. b1 determined 1
OONFOBMAL MAl'l'INGS (OONTINUATION)
227
HINT. See the hint to problem 1353.
1355. Let the domain P be the w-plane with horizontal outs going to oo on the left and on the right (Fig. 52). Prove that the function
FIG. 52
w = /(z), which maps the upper half-plane Im z > 0 onto the domain P, is of the form n-1
/(z) =
2 k-0
2l1 m-1
·k" -log(z-arc):n;
J-0
A 0 )+--+--+B, z-a z-eo (ll)
-log(z-c1 :n;
0
where A > 0, 0 > 0, k 0 = Im (Drn - B1), l0 = Re (D1 - Bn), and the remaining parameters (including b1 and d11 for the vertices B 1 and D.) have the same values as in problem 1353. How are the parameters A, B, 0, arc. b,, CJ, a. determined 1 Show that if ao = oo, then n-1
/(z) =
2n k-1
2l1n m-1
k" -log (z-arc)-
J-0
0 -log (z-c1 )---+Az+B, z-c0
(12)
228
PROBLEMS ON OOMPLEX ANALYSIS
where A > 0 and 0 > 0. If the parameter ak(Tc ¥= 0) or c1 (j ¥= 0) is equal to oo, then in formula (12) the corresponding term is absent. HINT.
See the hint to problem 1353.
1356. Map the upper half-plane Im z > 0 onto the domains in the w-plane indicated in Fig. 53 (all the dimensions are given on the corresponding diagrams) for the given conditions; find a and b (a> 0, b > 0).
FIG. 53
(1) (A 1 , B, A 2 , 0)-+ (-1, b, 1, oo); (2) (A 1 , B 1 , A 2, B 2 , A 3 , 0)-+ (-1, -b, 0, b, 1, oo); (3) (E, 0, D,)-+ (-1, 1, oo); (4) (E, 0, D)-+ (-1, 1, oo); (5) (E, 0, 0)-+ (-1, 1, oo); (6) (A 0 ,B,A 1 )-+ (oo,O, l); (7) (A 1 , 0, 0)-+ (-1, 0, l); (8) (A 0 , B 1 , Av B 2)-+ (oo, -(1 +a), -1, O); d = Re(B2 -B1 ); (9) (A,B,O,D)-+ (oo,-l,a,l);d=Re(D-B);h=Im(B-D).
229
OONFORMAL MAPPINGS (OONTINUATION}
§ 2. Conformal mappings involving the use of elliptic functionst The integral z
u
=
J
dz y[(l-z1) (l-lc1 :z:2)]
0
.p
=
d.P y(l-lc•sin•.p)
J
=
{l)
F(,P, le)
0
is known as the normal elliptic integral of the first kind in Legendre'a form. The inverse function z = sn (u, le) (2) is one of the fundamental elliptic functiona of Jacobi, known as the sn function of Jacobi. Two other elliptic functions of Jacobi a.re connected with it:
en (u, le)= y[l-sn1 (u, le)],
dn (u, le) = y[l-lc1 sn1 (u, le)],
(3)
called respectively the en- and dn-functiona of Jacobi. If there is no necessity to indicate the modulus le, then they a.re simply written sn u, en u _and dn u. It follows from (1)-(3), that dsnu du
- - - = cnudnu,
dcnu du
(4)
- - - = -snudnu,
ddnu du
=
-lc1 snucnu.
In what follows, if nothing is said to the contrary, it is assumed that 0 The integrals
<
le
< I.
(5) 1
K' = Jy[(I-:z:•)dtl-Tc'•zs)] = F( ~,le') 0
are known as complete elliptic intll{l1"ala of the jirat lcind, and le' as the complementary parameter.
= y (l -
lc2 )
t For this section see: N. I. AxHIYEZER, Elementa of the Theory of Elliptic Functiona (Elementy teorii elliptichealcilch funlcta-ii), Gostekhizda.t, 1948; Yu. S. Sm:oBSXII, Elementa of the Theory of Elliptic Functiona with Applicationa to Mechanica (Elementy teorii elliptichealcilch funlctBii B prilozheniyami le melchanilce), Gostekhizdat, 1936; A. M. ZHURAvsxrr, Textbook on Elliptic Functiona, (Spraooohnik po elliptichealcim funlctaiyam), Gostekhizdat, 1941; A. BETZ, Konforme Abbildung, Berlin, 1948.
230
PROBLEMS ON COMPLEX ANALYSIS
1357. Investigate the mapping of the z-pla.ne and the ,P-pla.ne by means of the normal elliptic integral of the first kind (1). Continuing this mapping by the symmetry principle, prove that sn u is a doubly periodic function with periods 4K and 2iK'. HINT. Consider the mapping of the upper half-plane Im 111 > 0 by means of the principle of corresponding boundaries. Trace the variation of the argument of du/dz when 111 is varied along the :i:-u:is.
1358. Map the upper half-plane Im~> 0 onto the recta.ngle in the w-plane with vertices ±a, ±a±ib (a > 0, b > 0). 1359. Investigate the mapping of the z-plane by means of the elliptic integral I
w=
f J"[(l-z•)~'•+klz•)]' z
Find the inverse function and prove that it is doubly periodic with periods 4K, 2K + 2iK'. HINT.
Make the substitution
111
== y(l-P).
1380. Investigate the mapping of the z-pla.ne by means of the elliptic integral
Find the inverse function and prove that it is doubly periodic with periods 2K, 4iK'. HINT.
Make the substitution
111
== y'(l-k't1).
1361. Verify Table 1 and Table 2 (apart from the la.st column). The dash (') denotes passage to the complementary parameter k'. HINT. Table 2 is obtained from Table 1. Let B be the rectangle with vertices (0, K, K+iK', tK') in the u-plane and B 1 its image in the u,,-plane. To establish the relationship between 1111 = sn (a,,, Ji) and 111 - sn (u, Ji) map the image B onto the image B 1, preserving the correspondence of vertices. This applies to all the rows of Table I except the last where as R there is taken the rectangle with vertices (0, K', K'+tK, tK) and a correspondence is esta. blished between 1111 and 111' = sn' u = sn (u, k').
1362. Consider the mapping of the z1-plane onto the Ui·pla.ne by means of the elliptic integral of the first kind Ui = F(,P1, ii)
((z1 =sin 0. Consider the cases g3 > 0, g3 < 0, g3 = 0. Find the periods of p(w). HINT. Consider the mapping of the upper half-plane Im z > 0 by the principle of corresponding bowidaries.
1368*. Investigate the mapping of the z-plane by means of the normal elliptic integral of the first kind (17) for real g2 , g3 and LI < 0. In particular, consider the case g2 = 0. Find the periods of p(w).
238
PROBLEMS ON OOMPLEX ANALYSIS
HINT. Since .d < 0, two of the quantities 8 1, 8 8, 8 8, are complex conjugates, and one is real. Let 8 8 be the real quantity, 8 1 = a.+ip, 8a = a.-ip 0). Consider the mapping of the semicircle lz-e.I = le1 - 81 1, Im z > 0 by means of the principle of corresponding boundaries and continue this mapping by the symmetry principle.
1369. Find the mapping onto the upper half-plane Im w of the triangle ABO with the given conditions: (1) (A= 0, B = ro > 0, 0 = ro(l+i))-+ (oo, -1, O);
>0
in)
ay3 (2) ( A=O, B=a>O, 0 = 2 -es -+(oo,-1,1);
!'!!)
ay3-es -+(oo,-1,0). (3) ( A=O,B=a>0,0=2 HINT, Use the solutions of problem 136'7 (the case g8 = 0) and 1368 (the case 91 = 0). REMARK. For this problem see also problems 1330-1332.
1370. Map the doubly connected domains 1-15 of the z-plane shown in Fig. 55 onto the circular ring (h < lwl < es and determine the modulus µ = e.le1 (see page 29). In problems 1371-1373 map the given domains onto the unit circle ltl < 1. 1371. The rectangle Re luj < K, Im jitl < K' (0 < Ii < 1). Find the positions of the vertices in the mapping. 1372. The interior of the ellipse jz - lj + jz + 1J = 2a (a> 1) with the cuts [-a, - 1], [1, a]. 1373. The interior of the ellipse jz - lj + jz + lj = 2a (a> 1). Find the positions of the foci in the mapping. 1374. Map the exterior of the unit circle jtj > 1 onto the domains 1 - 3, of the z-plane indicated in Fig. 56. 1375f. Let R be the rectangle IEI < K, l'l'/I < K' in the plane of u = E+in. Using Table 2 (see problem 1361) and the principle of corresponding boundaries, prove that the functions z dnu -l = sn u-J-ig en u ' .:_ = Z(u)+ dn u (en u+~ sn u+ik) +mu
l snu-f-igcnu (all the para.meters are real, / 1 g• < 1, m > 0, 0 0).
a
1385. w =!?.__log 2:n;
(z- _!_)z ·
1386. w=log(1+ : 2 ) 1387. w =log (z2
•
+ :2 )
Q
1388. w = az+ 2:n; log z
•
(a > 0, Q > 0).
I'
(a >0, I'>O). 1389. w = az+2:n;i. log z 1390. Investigate the character of the flow in the region lzl w
= a(z+
~2 ) + 2~i log z
~
R, if
(a> O, I'> 0).
Consider the cases: I'< 4naR, I'= 4naR, I'> 4naR.
1391. Find the complex potential w(z) of the flow in the whole plane formed by the vortex sources {(ak;Qk,I'k)} (k=l,2, ... ,n) and having at infinity the given velocity V00 = Ve1«. 1392. Is it possible for stream lines to emerge from a point hav:ing: (1) a vortex, (2) a doublet, (3) a vortex and a doublet together~ 1393. Find the law of variation of a vortex source, doublet and multiplet, situated at a point a or a.t oo, for the following schlicht conformal mappings of the neighbourhood of these points (Ci =fi 0, C-1 =fi 0):
(1) C= a+Ci.(z-a)+ ... ;
246
PBOBLEMS ON COMFLEX .ANALYSIS
(2)
Ca: X + c_l
(3)
+co+ ... ; C= eiz+eo+ ...
(4)
z
C=
+ ... ;
C-1
z-a
1394. Find the law of variation of the vortex source for the n-sheeted mappings:
C= ix+c
11
(z-a) 11 +
... ,
c_ .. .,,. =oc+-+ ... , z" In problem 1395 there is established the 81Jmmett'y principle, which together with conformal mapping is widely used for the construction of flows (see problems 139'7-1405).
1395. Prove that the fl.ow can be continued by the symmetry principle a.cross a rectilinear or circular segment of a stream line or equipotential line, a vortex source passing into a vortex source, a doublet into a doublet, a multiplet, generally speaking, into a set of multiplets of the same order. Find the stl't"ngth and intensity of the reflected vortex source and the moment of the reflected doublet. Rmt:ABK. It follows from the symmetry principle that if there are recti• linear or circular segments on the stream line or equipotential line the fl.ow muat be symmetrical with respect to this curve. This imposes a definite restriction not only on the singularities of the fl.ow outside the given lines but also on them or at their end points (if any).
1396. A fl.ow in the z. plane is formed of a finite number of sources, vortices and doublets. (1) Find the necessary and sufficient condition for the circle lzl = R to be a stream line, if the sources, vortices, doublets: (a.) a.re not situated on this circle; (b) a.re all situated on it; (c) some a.re on it and some not. (2) With the same assumptions find the conditions for the circle lzl = R to be an equipotential line. 1397. Find the complex potentials of the flows in the upper half. plane Im z > 0 with the given singularities and the velocity Voci· (1) The velocity V00 = V.
247
APPLICATIONS TO MECHANICS AND PHYSICS
(2) The vortex (a; I') and velocity V00 = O. (3) The source (a; Q) and velocity Vco = 0. (4) The doublet (a;p) and velocity Vco = 0. (5) The vortex sources {(a1c;Q1c,I'1c)} (k=l,2, ... , n), the doublet (a; p) and velocity Vco = V. What can be said of the behaviour of the fl.ow at 001 (6) The vortex source (O; Q, I') and doublet (O; p); Vco =0. What values can the moment .P of the doublet assume 1 Is the :ftow always possible if r #= 01 1398. In the circle izl < R construct flows, having respectively: (1) a vortex (a; I'); (2) a doublet (a; p). 1399. Find conditions for the possibility of constructing flows in the circle lzl < R, if: (1) There exist only the sources {(a1c; Q1c)} (k = 1, 2, ... , n,) situated inside the circle; (2) In addition to the sources of part (1) there are sources {(a~; Q~)} (k = 1, 2, ... , m), situated on the circumference izl = R. In both cases find the complex potentials of the flows. 1400. In the domain Jzl > R construct flows having respectively: (1) The vortex (a; I'), the velocity Vco = 0 and circulation at infinity co = 0; (2) The doublet (a; p), the velocity Vco= 0 and circulation
r
l"'co=O; (3) The velocity Vco = Ve111 and circulation I'co = 0; (4) The velocity Vco = Ve'11 and circulation I' round the
circle
izl = R.
REMAIUC. The last two examples of problem 1400 give the streamlining
ofa circle with given velocity at infinity, without circulation and with circulation (see, for example, [3, Chapter III, § 49]).
In examples 1401-1405, using the symmetry principle construct flows with given singularities (at infinity and at angular points the velocity is equal to zero). 1401. In the domain Jzl > 1, Im z > 0, with the vortex (ia; I'), a>O. , !!.
1402. In the angle O 0, Im z > 0, with the source (l;Q). 1404. In the first quadrant Re z > 0, Im z > 0, with the source (1 ;Q) and the sink (i; -Q).
248
PROBLEMS ON COMPLEX ANALYSIS
1405. In the first quadrant Re z > 0, Im z > 0, with the source (l+i; Q) and the sink (O; -Q). 1406. Construct a fl.ow in the whole z-plane, if it is known that in the upper half-plane Im z > 0 there are vortex sources { (ak; Qk, I'k)} (k = 1, 2, .. .,n) and a doublet (a; p), the x-axis is an equipotential line and the velocity V00 = Vei". Is such a flow always possible? 1407. Construct a fl.ow in the whole z-plane if it is known that in the circle [z[ < R there are vortex sources {(ak;Qk,I'k)}(k=l, 2, ... , n) and a doublet (a; p), the circle [z[ = R is an equipotential line and the velocity V00 = Ve1". Is such a flow always possible? 1408. In the simply connected domain D, bounded by the contour 0, construct a fl.ow with streamline 0, having the vortex sources {(ak; Qk, I'k)} (k = 1, 2, ... , n). Is such a fl.ow always possible~ 1409. In the domain D, bounded by the contour 0 and containing the point at infinity, construct a flow with stream line 0, which has the vortex sources {(ak; Qk, I'k)} (k = 1, 2, .. ., n) and the given · velocity V00 = Ve1". Is such a flow always possible?
In problems 1410-1417 the streamlining of bounded and unbounded contours (they must be stream lines) is considered. The problems are solved by means of conformal mapping onto the exterior of a circle, the upper half-plane and a rectilinear strip. 1410. Construct the streamlining of a bounded contour 0 with given circulation I' and velocity V00 = Ve1". What is the mapping effected by the complex potential w(z) in the case I'= 01 1411. Construct the streamlining of the ellipse x 2 /a2 y2 /b 2 = 1: (1) With given velocity V00 , without circulation; (2) With given velocity V00 and circulation I'. 1412. Construct the streamlining of a flat plate [xi ~ 0, y = 0: (I) With given velocity V00 , without circulation; (2) With given velocity V00 and circulation I', defined by the condition that one of the ends of the plate is a point of departure of the flow (the Zhukovskii-Chaplygin postulate). 1413. Construct the streamlining of a Zhukovskiit profile with given velocity V00 and circulation I', defined by the Zhukovskii-
+
t See problem 274.
.Al'PLICATIONS TO MECHANICS AND PHYSICS
249
Chaplygin postulate (the sharp end of the profile must be a point of departure). In problems 1414-1417 construct the streamlining of the given contours. 1414. The parabolas y2 = 2px (from the inside and from the outside). 1415. The right branch of the hyperbola x1 /a8 -y1/b• = 1 (from the inside, and from the outside with the velocity V00 = 0). 1416. The half lines:- oo < x < - 1, y = ± n. 1417. The half lines: 1 < [x[ < oo, y = 0. In problems 1418-1422 periodic flows {V(z+co) = V(z)} and flows in curvilinear strips (channels) are considered. For the construction of these flows the curvilinear strips must be conformally mapped onto a rectilinear strip, and then the fl.ow continued by the symmetry principle and use made of the expansion of a meromorphic function into a series of partial fractions. In problems 1418, 1419 investigate the singularities, construct schematically stream lines and equipotential lines and determine the velocity at infinity in the strip of periods for periodic flows with the given complex potentials. '· 1418. (1) w 1419. w =
=
2Qn log sin z; (2) w
=
r.
2ni log sin z.
in cot z (0 < arg p < ; )·
1420. In the rectilinear strip 8: 0 < x < co of the z-plane construct the fl.ow formed by the vortex source (a; Q, I'), a e 8, having the given velocities V (x+ioo) = V1 , V(x-i oo) = V1 • Is such a fl.ow always possible 1 Construct schematically stream lines and equipotential lines if I' = 0 or Q = 0. HINT. Continue the fl.ow by the symmetry principle and use the result of problem 1418.
1421. In a rectilinear strip of the z-plane 8: 0 < x < co construct the flow formed by the doublet (a;p), a es, having the given velocity V(x ± i oo) = iV. Construct schematically the stream lines and equipotential lines. 1422. In a curvilinear strip of the z-plane 8, bounded by contours 0 1 , 0 1 , construct the fl.ow streamlining 0 1 , 0 1 , having
250
PROBLEMS ON OOMPLEX ANALYSIS
given vortex sources, and doublets in S and given velocities Vv V2 at the points at infinity !J1 , !J 2 of the strip S. Indicate sufficient conditions for the existence of such a flow.
A fl.ow is said to be doubly pM'iodic if its velocity w'(z) is an elliptic function. Elliptic functions are doubly periodic meromorphic functions, possessing periods 2 - 41> =
!+
ik the chessboard arrange-
ment (k is an arbitrary real number); the speeds of motion in . . 1y b y 2f I' cot h -l:nk an d 2f I' tan h -z:nk t h ese cases are given respective
(I' is the magnitude of the intensity of the vortex). REMABX. It is assumed that as y-+ ± oo the velocity V(m+iy)
remains
bounded.
1448. The fl.ow of a fluid in a rectilinear channel with impenetrable walls a:= ±k is caused by a vortex chain consisting of vortices of intensity I', situated at a certain instant at the points a+ikl (k = 0, ± 1, ± 2, .. ., ; -k O).
1457. w = piz+2qi log..!:.. (p
> 0, q > 0).
z
n
1458. w = piz+2i
.2 k=l
qk log-1-
z-a1c
(p
> 0,
q,.
> 0,
al O; (2) Onto the exterior of the circle !ti > R, if a= oo, where t(oo) = oo and t'(oo) = 1. 1474. Construct the electrostatic field formed by the point charges {(a1:; 2q1:)} (k = 1, 2, ... , n) and the dipoles (a; p) in the simply connected domain D.
Let g(C,z) be the Green's function of the domain D (see page 165), the boundary of which consists of the piecewise smooth simple contours I'u ... ' I'n; also let ti be the inward normal to and let be traversed in the positive direction with respect to D. If u(z) is a function harmonic in the domain D and continuous on I', then it follows from Green's formula that
r
r
r,,
r
J
u(z) = - 1u(C) 8g(C, z) 'da 2nr an. or u(z)
=
J[u ana log IC-zl l 1 aum] -log IC-zl --a,;-
l 2nr
da.
If the domain D contains the point at infinity and the function v(z) is harmonic there, u (oo) must be added to the right hand sides of the given for· mulae. Then in the neighbourhood of the point at infinity the Green's func· tion g (z,oo) can be represented in the form g(z, oo) =log lzl+Y+o(
i}.
The quantity
/' = lim [g(z, oo)--log lzl] is known as the Roben's constant of the closed set representing the com· plement of D on the z-plane; the quantity e-7 is known as the capacity of this set.
1475. Prove the following assertions (n is the inward normal): . 1 1 (1) g(z, a)= log-1 - -1 - 2-
z-a..
n
f ag(C,
a) 1 . . --!l-log-1, . - -1 ds if a#: oo,
F
MZ.
.,-z '
263
Al'PLICATIONS TO MECHANICS AND PHYSICS
(2)
g(z,
oo) =
I
r- 2n
J
ag(C,oo)
an
r
log
I
IC-zl
ds,
if zeD and the domain D contains the point at infinity; 1 ag(C,oo) 1 . (3) 2n an log IC-zl ds = ,... If z e D and the domain r D contains the point at infinity;
J
(4) ;nf
ag~~,a)
ds =I, if a#: ooor ifa = ooeD.
r HINT. In part (1) use the symmetry property of the Green function g(C, z) = g(z, C) and the integral representation of the harmonic function in terms of its boundary values. In part (2) use the integral representation of the function log!C-zf+g(C,z)-g(C,oo), which is harmonic in D, pass to the limit and use the symmetry property g{oo,z) ... g(z,oo). In part (3) proceed in the same way but start with the function log IC-zf-g(C,oo)t. The function l
tJ0 (z) -2qlog-1- 1
z-a
is known as the logarithmic potential of the point charge (a; 2q). In the extended z-pla.ne tJ0 (z) represents the logarithmic potential of the two point charges: (a, 2q) and (oo; -2q). Let the contour I' satisfy the conditions indicated on page 262, and let and • be real and continuous on The integral
am
r.
"(z)
=
J
(!{C) log
r
IC~zl
da
am
is known as the logarithmic potential of the simple layer wUh density (in three dimensions there corresponds to it the potential of a charged cylindri·
cal surface with base r and surface density of charge : • that is bearing the charge
~ da8 on the element of area cfal).
The function tJ(z) is continuous in the finite z-plane and is harmonic everywhere outside I', except at the point z = oo, where it has a logarithmic singularity "(z)
=
-2q log l•l+o{
i),
2q
=
f
e(C)da
r
t See R. N:mvANLINNA, Eindeutige analytiache Funkf.ionen, Chapter V, § 2, Springer, Berlin, 1936,
264
PROBLEMS ON COMPLEX ANALYSIS
(this means that the potential "(z) corresponds to a charge (oo; -2q)). The integral tJi.{Z)=
J
r
8
11m
1 log IC-zi ds
an
r
is known as the logarithmic potential of the double layer with density 11((:) (if is the boundary of a domain there is on I' a distribution of dipoles with axes directed along the inward normal to I'; 11((:) is the density distribution of dipole moments). If 8((:,z) is the angle between n and a vector going from Cto z, and d ((:, z) is the angle subtended by the element of arc ds at z, then 111(21) =
J r
11((:)
cos 8((:, z)
IC-zl
J
ds =
11((:)d((:, z).
r
In particular, for a closed contour I' and 11 ((:) = 1
f :n l'
I
2n, if z is inside
log
IC~zl
ds
=
z
r,
r,
n,
if
0,
if z is outside
is on
r
(see also problem 1066). The Green function g(z, a) of the domain D can be considered as the po· tential of the electrostatic field formed by the point charge (a; 1) ifthe boundary I' of the domain D is earthed. Problem 1475, (1) shows that in the case a 00 the earthing of r is e~uivalent to the placing on r of a charge
*
of linear densityg((:)
= - 2~ ag~~ a)
• In this case, by part (4) of problem
1475, the total quantity of charge is equal to -1. In the case a = oo the point charge (oo; 1) and the earthing of rare together equivalent to the placing on
r
of a charge with density given by
em= - 2~ au(~~oo)
of total value -1 and to the addition of a field with constant potential y (see 1475, (2)). In these oases the given distributions on r are said to be induced by the charge (a; 1).
In problems 1476-1479 find the density e(C, a) of the distribution induced by the charge (a; 1) on the contour and the corresponding potential v(z,a) of this simple layer for the domains bounded by the contour 1476. I' is the real axis, Im a > 0. 1477. (1) I' is the circle lzl = R, lal < R; (2) I' is the circle lzl = R, lal > R (consider, in particular, the case a = oo). 1478. I' is the real-axis segment lxl R, y = 0, a= oo. 1479. r is the ellipse x"/a.2+y2/(32 = 1, a= 00. 1480. Considering the Green function of the domain D containing the point z = oo to be known, solve Boben' s problem:
r
r.
<
APPLICATIONS TO MECHANICS AND PHYSICS
265
Find the density of the distribution e(C) on the boundary I' of the domain D due to unit charge creating outside D and on I' a constant potentialt. HINT. See problem 1475, (3) and (4).
In problems 1481-1483 solve Roben's problem for the given domains D. 1481. D is the exterior of the circle jzj > R. 1482. D is the exterior of the segment jxl < R, y = O. 1483. D is the exterior of the ellipse x2/as+y2/b2 = I. In problems 1484-1487 find the capacity (see page 262) of the closed sets. 1484. jzj < R. 1485. jxj R, y = 0. 1486. x2/a2+y2/b2 I. 1487. jz2-a2[ a2 (a > 0). 1488. On the simple closed contour I' let there be given a real function c/>(C) continuous and differentiable along the contour. Prove that the real part of the Cauchy type integral -1 c/>(C) dC is the logarithmic potential of a double layer 23ii r C-z with density c/>(C) and its imaginary part is the logarithmic
<
<
<
-J
:!: ·
1 2 31: 1489. Prove that the function v(z), bounded and harmonic in the upper half-plane Im z > 0, can be represented as the logarithmic potential of a double layer: potential of a simple layer with density -
00
v(z) =
~
f
v(t)
a~ log
It I zj dt.
-oo
If however v(z) is regular at infinity, then it can also be represented as the logarithmic potential of a simple layer:
f
00
1 v(z) = v(oo)-31;
-00
ov(t) I dt. - 0- l o g 1 1 n 1t-z
t For the general Roben's problem which requires a non-negative distribution of unit charge on a given set E to be found in order that the corresponding logarithmic potential should assume one and the same value at every point of the set E, see the book by R. Nevanlinna mentioned on page 263.
266
PROBLEMS ON COMPLEX ANALYSIS
1490. In the upper half-plane Im z > 0 find the complex potential of an electrostatic field if its potential v(z) assumes on the real axis given piecewise continuous values. Write the potential function in terms of the harmonic measures of the corresponding segments of the real axis (see page 170): (1) cf> in the interval (-oo, a), 0 in the interval (a, oo); (2) cf> in the interval (a, b), 0 in the intervals (-oo, a), (b, oo); (3) c/>1 , c/> 2 , ••• , cf>n respectively in the intervals (-oo, a 1 ), (a1 , a 2), ••• , (an-l• an) and 0 in the interval (an, oo) (here a1 1 , c/> 2 , ••• , cf>n respectively in the intervals (-oo, a 1 ), (ai, a 2) and so on. HINT. In part (1) use conformal mapping onto a strip; in the remaining parts use the method of superposition (it is possible also to use Schwarz's integral formula for the half-plane, see page 170).
1491. Find the complex potentials w(z) and the potentials v(z) in the following doubly connected domains with a given difference d = v2 -v1 of potentials v1 , v2 on the boundary contours: (1) In the circular ring r 1 < lzl < r2; (2) In an arbitrary doubly connected domain D. 1492. Prove that if D is an arbitrary doubly connected domain and on each of the contours bounding this domain the potential function assumes constant values (v1 and v2 ), then
w(z)
=
i(;2-v1 ) logt(z)+e+iv1 , v(z) = v12-v1 log lt(z)!+v1 • ogµ ogµ
where t(z) conformally maps D onto the ring 1 < !ti < µ (µ is the modulus of D) and the boundary contour with potential v1 passes into the circle ltl = 1; e is a real number. 1493. Find the complex potentials in the given doubly connected domains (the potentials v1 and v2 on the boundary contours are constants). (1) In the exterior of the circles lz± al = R (a> R) (v1 is the potential on the circumference of the circle on the left). (2) In the exterior of the circles lzl = r 1 (potential v1 ) and lz - al = r 2 (a> r 1 +r2 ). (3) In the non-concentric circular ring, bounded by the circles lzl = R (the potential v1) and lz-aj = r (0 < a < R - r). (4) In the ellipse x'l-/a2+y2/b" < 1 with a cut along the segment joining the foci (the potential on the ellipse is v1 ).
267
APPLIOATIONS TO MEOHANIOS AND PHYSIOS
(5) In the exterior of the segments 1
<
)x)
<
!,
JI= 0
(0 < le< 1). On the segment on the left the potential is v1 • (6) In the exterior of the segments !xi < 1, y = ± n. On the upper segment the potential is vp
1494. Let D be a multiply connected domain with boundary I', consisting of n piecewise smooth contours I',, (le= 1, 2, ... , n), and let ro,,(z) be the harmonic measure of I'" (see page 170). If the domain D is bounded we shall consider I'n to be the exterior boundary contour. Prove the following assertions: (1) If the domain D is bounded, then
1 O; the source (a; q). 1506. In the circle lzl < R; the source (a; q). 1507. In the half-strip lxl O with the source (ih; q) (h >0). 1508. In the rectangle lxl O, ta.n-1-
a
<
(1-i); (4) ±
(1-1-i), ±
'
+n
for a
<
0 and b ~ 0,
0.
1 iy'3 y'3 i 3. (1) 1,-2±-2-; (2) ±2+2·
(3) ±
= .:!..sgnbt·
2 b
-i;
~2 (y(3)+i). ± ~2
(J-'(3)-i), ±J-'2i;
~2-(1-i);
(y'[y'(2)+1]-iJ"[y'(2)-l);
(7) ±(2+i);
(2k+: )n (2k+{)n] +isin 3 3
(k
=
0, 1, 2);
[ (2k+l)n-tan-1 3/4 . . (2k+l)n-tan-13/4] (9) H JI 5 cos 5 - - - - - +isin 5 (k = O, 1,2, 3, 4).
(k
=
O, 1, 2, ... , n-1).
t sgn b denotes Kronecker's symbol : sgnb sgn b
= =
1
for b> 0,
-1
for b
274
<
O.
275
ANSWERS AND SOLUTIONS
12.
z8
2n . • 2n) = z 1 + (Z1-Z1) ( COSn±iBID n •
13. z, = z1 +z 3 -z 2 • 14. The ratio (z8 -z1 )/(z1 -z1 ) must be a real number (the condition is necessary and sufficient). 15. The anha.rmonio ratio (z10 z1 , z1 , z4 ) =
z,-za: Z1-Z4
Zz-Za Z9-Z4
must be a real
number (the condition is necell8a.ry and sufficient). 16. SOLUTION. In the proof it can be assumed (without loss of generality) that the straight line in question is the imaginary axis and that all the points considered are on the right of it (if this is not the case it is necessary to mul· tiply all the z1: by some number of the form cos a:+isin a:). It is then obvious 1 that Re z1: > 0 and Re - > 0 for any k. Zk
19. The interior of the circle of radius R with centre at the point z = 21o; the exterior of the same circle; the circumference of the same circle. 20. An ellipse with foci at the points z = ±2 and major semi-axis 5/2. 21. The interior of the left hand branch of the hyperbola with foci at the points z = ±2 and real semi-axis 3/2. 22. The straight line perpendicular to the segment connecting the points z1 and z1, and passing through the middle of this segment. 23. (1) The straight line Ill = a and the half-plane on the right of it; (2) the half-plane below the straight line y =a. 24. The strip -1 < y < o. 25. The interior of the angle with vertex at the coordinate origin and sides which form with the real axis angles equal respectively to a: and fJ; the interior of the same angle with vertex at the point z0 • 26. The parabola y1 = 2a:+ 1. 27. The half-plane bounded by the straight line a:+y = 1 and containing the coordinate origin. 28. The straight line passing through the points z1 and z1 (the point z1 being omitted) ; the circle with the segment connecting the points z1 and z1 as diameter (the point z8 being omitted). 29. (1) The interior of the domain bounded by the segment 0.;;;:;; a:.;;;; 2n of the real axis and one turn of the Arohimidean spiral r = ~ ; (2) The set of points determined in part (1) and the complement to the interval (0, 2n) of the real axis. 80. (1) A family of circles touching the imaginary axis a1> the coordinate origin, and the imaginary axis itself (the equation of the family is; O(z1 +y1 ) = 111); (2) The family of circles touching the real axis at the coordinate origin and the real axis itself. 31. (1) The family of hyperbolas 1118-y• = O; (2) The family of hyperbolas lllY = 0/2. 32. Each curve is a circle, being the geometrical locus of points the ratio of distances from which to the points z1 and z1 is constant (the circle of Apollonius with respect to the points z1 and z1 ). 33. The family of circular arcs with ends at the points z1 and z1 (this family also contains the two rectilinear segments with ends at the points z1 and z1 ; one of these segments contains the point at infinity).
276
PROBLEMS ON COl'rlPLEX ANALYSIS
34. (1) Each curve is the geometrical locus of points, the product of the distances of which from the points z =--L.ancLz.-= l is constant (lemniscates with foci z = ± 1). For ..1. > 1 the curves of the family are simple closed curves, for ..1. < 1 they separate into two simple closed curves which for ..1. -+ 0 shrink to the points ± 1. For ..1. = 1 we have the lemniscate of Bernoulli; its equation in polar coordinates is rl = 2 cos 2•. (2) Lemniscates with foci at the points z1 and z20 where z10 z1 are the of the equation z1+az+b = 0. The lemniscates consist of a single curve if ..1. >
..1.
=
V(.
lza;zil),
v(
and of two curves if ..1.
V(
lza;zil).
For
lzi;Zil) we have a lemniscate of Bernoulli with the double point
(z1 +z1 )/2. 35. The spiral of Archimides r = •· 36. The logarithmic spiral r = e•. 37. (1) :ii; (2) 2:n/3; (3) 2:n; (4) :ii; (5)
a:
38.
<
y
o. a:•+y•
.
~+i11
E = :z:1+y1+1 ' 11 = a:1+y1+1 , C = a:1+y1+1 ; z = a:+i11 = 1-C •
at.(! ,o, !)· (-! ,o, !). (o. !,!). (v; ,- ~2 , !). All the four points are situated on the equator, their longitudes being respec· tively equal to 0, :ii, ; , - : (the longitudes are calculated from the initial meridian, situated in the ~. (:-plane). 40. The circle of radius tan CP/2+:ii/4) with centre at the point z = O. The "South" pole corresponds to the coordinate origin, the "North" pole to the point at infinity. 41. ( 1) Half meridians of longitude cx; (2) Parallels of latitude p = 2 tMC 1 r -:n/2. 42. (1) Diametrically opposite points on the Sl\llle parallel; (2) Points mutually symmetrical with respect to the initial meridian (that is, differing in the sign of the longitudes); (3) Points mutually symmetrical with respect to the equatorial plane (that is, with the same longitude and with latitudes differing in sign only). 43,. Z1.Za = -1.
44. Points obtainable one from the other by a rotation of the sphere through 180° about a dil\llleter parallel to the real axis of the z-plane. 45. (1) The eastern hemisphere; (2) The western hemisphere; (3) The hemisphere - ~ < ex < ;
(cx is the longitude); (4) The hemisphere ; < lcxl 2;
(2) A compression for lz+ll
I
< 2'
a stretch for
I
lz +II> 2;
> I, a stretch for lzl < I; (4) A compreBBion for Re z < 0, a stretch for Re z > O; (5) A compression for lz - II > 1, a stretch for lz - II < I. (3) A compression for lzl
152. S
= Jf1f'(z)l 8 da:dy, G
ua. y2(e8"- I). 154. 2e 2 (e1 - I).
L
=f
1/'(z)lds.
I
155. The domain D is the ring e < lwl < e•. It is not possible to apply the formula of problem 152 since the mapping is not one-one.
CHAPTER II 1156. w = (l+i)(l-z). 157. w = (2+i) z+l-3i. 158. (1) z0 = -1+3i, 8 = 0, k = 2, w+l-3i = 2(z+l-3i); (2) z0 = 2+2i, 8 = n/2, k = 1, w-2-2i = i(z-2-2i); (3) there is no finite fixed point; (4) if a = 1 there is no finite fixed point; if a+. 1, then ( w1 -az1 w1 -az1 w1 -az1 ) Zo = - - - , 8 = arga,k = lal, =a z - -- , 1-a l-a l-a
w----
(5) Ifa = 1 there is no finite fixed point; if a .p 1, then z0 = _b_,
l-a
8=arga,k=lal,w- 1 ~a =a(z- 1 ~a)· U9. (1) w = az+b; (2) w = -az+b; (3) w = -i(az+b); (4) w == az+bi. Everywhere a and b are real numbers and a > 0. 160. (1) w = z+bi or w = -z+l+bi; (2) w = z+b or w = -z-i+b; (3) w = z+b(l+i) or w = -z+l+b(l+i). Everywhere b is a real number. It is possible for correspondences to exist between points lying either on a straight line parallel to the boundaries of the strip, or on parallel straight lines symmetrical with respect to the centre line of the strip. The mapping is not uniquely determined if the points lie on the centre line of the strip.
161. (1) w = (z-a)/h; (3) w = (4) w =
(2) w = -z+,,,a+h +i;
y(l+k1 ) -i(-j-+tan-ik) b e z, y(l+k1 ) -1(.!!.+tan-1k) . e 2 (z-ib1 ). b1-b1
162. w = el 11 Rz+w0 • 163. (1) The family of straight lines u = l/a parallel to the imaginary axis (not including the imaginary axis itself); (2) The family of straight lines 11 = - l /b, parallel to the real axis (not including the real a.xis itself); (3) The family of circles b(u•+v•)+u+v = 0, touching the straight line tJ = -u at the origin (also including this straight line); (4) The pencil of straight lines v = -ku; (5) The family of circles passing through the coordinate origin and through the point Wo = l/Zo (the straight line which passes through the points w = 0 and w = w0 also belongs to this pencil); (6) The cissoid u 1 = -vl/(v+ 1). 285
286
PROBLEMS ON COMPLEX ANALYSIS
164. (1) Into a family of circles touching at the point w = h the straight lines, respectively parallel to the imaginary and real axes (including also these straight lines); the equations of these families a.re;
+ (v-hg)8]-(u-h1) = O; (0-y0 )[(u-h1 ) 1 + (v-h8 ) 1l+ (v-h1 ) = 0, (O-:i:o)[(u-h1 ) 1
where z0 = :i:o+iyo, h
= h1+ih8 ;
(2) Into the family of circles with centre at the point w = h ( lw-hl =
!) andafamilyofraysissuing from the point w ...
h (arg (w-h)
=-cc).
165. (1) The equation of a family of circles of Apollonius with respect
I I=
to the points z1 and z1: z-zi z-z1
.A. The ends .A. and B of the diameter lying
on the straight line passing through the points z1 and z1 (Fig. 60), divide the J.. 0, then the area a.re situated on the right of the direction z~~ if8 < 0, they are on the left of it);
B FIG. 61 (4) to the upper semicircle there corresponds the right angle indicated in the figure. 166. The semicircle lwl < 1, Im w < O. 167. The domain containing the point w = 0 and bounded by arcs of the
lw+ !i I= 3/4. removal of the points of the disk Iw -
circles
lwl
=I
and
168. The domain obtained from the lower half-plane (Im w
! + ~ I<
<
0) by the
~2 which are in this half.
plane. 169. (1) The domain bounded by the straight line Rew= 1 and the
. 1 circe
Iw- 211 =2'l tangential to 1t;.
I ~I
(2) The domain bounded by the tangential circles w -
and
lw- ! I= ! .
=
~
288
PROBLEMS ON COMPLEX ANALYSIS
170. The doubly connected domain the boundary of which coneists of the
=
straight line Re 10
!
and the circle / 10 -
171. (1)
10
=- :
(2)
10
= ~(~-1)+hi da-~ z
(3 )
10
=
172.
(1)
=_
(3)
10
175.
10
(2 )
2i(z+l) • 4z-l-5i '
(l+i)z+l+3i (l+ilz+ 3 +i ;
173. (1) w = 10
= : or
/
= : •
+hi; 10
= ~(~-1)+1+hi; ~-da z
d1(z-da) . z(d1+d2 )
10
174. (l)
+l+hi or w
:
(2)
= (-1+3i)z+l-i.
(l+i)z-l+i
=
'
(1+2i)z+6-3i 5(z-i) •
10
10 =
(2) 10
iz+2+i z+l ;
1-i (3) w = - 2 -(z+l).
(l-4i)-2(1-i) = z2z(l-i)-(4-i) . ;
z (3-i)- (1 +i) • (l+i)(l-z)
=
= IZ-1 .z-i ;
the upper half.plane maps onto the unit disk.
176. (1) 10 = (az+b)/(oz+d), where a,b,o, dare real numbers and ad-bo> O; (2) 10 = (az+b)/(cz+d), where a,b,o, d are real numbers and ad-bo < 0; (3) 10 = i(az+b)/(cz+d), where a,b,o, d are real numbers and ad-bo < O. 2 2z+l) (2)10= -2 ( - • 177. (1) 10 = 2 _ 21 ; z-2 178. w = (B-z)f(B+z); the image of the upper semicircle is the quadrant u> O,v < O. 179. (1) 2 ~i ; 180. (1)
lzl =
(2) :
+i.
2; (2) the straight line a:
=
! ; Iz - ~ I= ~ ; (3)
.!. ;
(5) lz-21ol = y(lz0 18 -l) (that is, this circle is 2 symmetrical to itself with respect to the unit circle); (6) (a:Z+y•)•- (a:2-y•) - O (a lemniscate); (7) a curvilinear triangle (4) a:a:o+YYo =
'h vertices ' ' l wit at t h e pomts =-, Z1
l l =-, =-, Zs
and wi'th
Bl'des
. which are arCB o f
Zs
circles passing through the point z = 0 (one of the arcs may be a segment of a straight line). 182. (1) O(a:)
=
cx+2 arg (a:-/3); (2) 10'({1) =
e(•-f) 1 / 2b
289
ANSWERS AND SOLUTIONS
(3) If b ;;;;;;,. 2, then the whole half-plane is compressed, if b < 2, then the domain within the circle lz-pj < y(2b) is stretched. (The circle lz-pj = y(2b) is termed iaometrio.) z-2i t(.!.+e)z-(a+bi) 188. w = (z-i)/(z+i); (2) w = i ( b" • 2 .; (3) w = e 2
z+
= Bi--. +w0 • z+1 w = -(z-2i)/(z+2i)
184.. w 185.
z- a- i)
i
z-i
zi+2
186. w = - 4 21 _ 2 _ 41 • 187. w =Tee
-~++ars~) z-zi, where k is a real number the sign of which is z-z.
identical with the sign of arg z1 /z1• Rays issuing from the point w = 0 in the half-plane Re w > 0 correspond to circular arcs, situated within the disk lzl < l and passing through the points z1 and z1 in the z-plane; the semicircles with centre at the point w = 0 in the half·plane Re w > 0 correspond to arcs of the circles of Apollonius with respect to the points z1 and z1 situated within the disk lzl < l. w-b z-a = el11.-= 188. ---w-b z-a·
z-a z-a
w-ii
189. - - =i--_-.
w-a
190. (1) O ()
=
oi:-+2arg(el-a)
=
oi:-+2tan-1
sin-Asin8 A , where COS- COS 8
a= Aelll.
=
(2) w'(O) (3)
(l-lal 1 )el•,
w'(a)
=
el• l-ja!•"
Ifa#= O, then the domain lying inside the disk lz -
!I< V(,:i.
-1}
is stretched, and the domain outside this disk is compressed. (The circle
lz-
~
1-v(
1; 11 -1) is isometric.) If
(4) max
191. (1) w = (4)
IdZI= dw
2z-l 2_
w-a
1-i.iw
21
=
;
= 2iz+_!_ ·
el•~. 1-'iiz
w-b .ffl-bw
z-a
JU. R1----=el«R1--- .
193. w =- (l-z)/(z+2l
R~-ih
o,
Iaz I=
. dw mm
l+lal 1-lal' (2) w
a=
2+iz '
then lw'(z)I 1-lal l+lal.
(3) w
=
-iz·,
== 1.
290
PROBLEMS ON COMPLEX ANALYSIS
19'. w
=
2 .
z-2+i 2·
u+ 2 -
i
z-a R 8e•11 - - · (2) R•-az'
195 (1) w =
.
z-a 11 -· RB-az'
w-b . - - = e'
R•-bio
z-a (3) w = R 1 - -
RB-az'
where a is a real number and Jal < I. (l+k)z-(za+kzi) a, where k =-. f(l-lzsl 1)elar1(1-i.z,), (1-k)z- (z1 -kz1) l-Jz1J1
196. w =
JI
Jal=
lz1-z1l · - - - Jl-z1z1J+y[(l-Jz1J2)(1-lz111)] · az-l+Jl'(l-a2) l-y'(l-a2) 197 • w = ± [l-y(l-a1 )]z-a' !I= 2 a
200. (1) w = z; (2) w
=
z+2-y3
.
l+( 2 -y 3 )z; (3) w =
t
z-2+)1'3 l-( 2 -y'3 )z"
toa. Circles passing through the point Zo and having at this point a tan· gent determined by the vector h. (R-lci)z-R1 204. w = z-(R+lci) , where k is a real number, k O (fork= oo it
+
is necessary to put w = 1). 209. If Jal sin ; , then the transformation is hyperbolic. If we use the notation:
• J. I I . fJ 1 =e mn2=asm,z
1(11+f-P) • zI = e 1(11+-i-,.+.8) , then the transformation is written in the
W-Zi
form - - -
w-z1
= KZ-Z1 --z-z · 1
291
ANSWERS AND SOLUTIONS
210.
I'=
r
= 2 tan-1:;
~ + 0 [(:
1'-
r
r = ~ +o [(
n sin,,
:n
for sma.11
~;
for small : .
z..);
Zo = 2 tan-1 (I +a:. tan l-a:0 cosy l-a:0 2 I' < y, if Zo < 0, and I'> y, if Zo > 0. 215. (l)w= -20/z; (2)w= -(2iz+I+2i).
211.
= Y+2 tan-1
216. w
==:~~~::==~el«,
l! =
~ + v[(~r-ll
217 h = 5/4· w = 2el« 4z- 3 or w 4z+3 · '
= el« 4z+ 3
4z-3 ·
z+24 2 2z 218. w =---el« orw = - - - els (! = - 3 . z+24 3z 219. w
z-z' = .il--! or z-z
z-~
w = .il--, where .il is an arbitrary complex number, z-z1
1
I
Z1 = z1+ u1=
~~-~ d
I
• Zs= z1+
~~-~ d
,
d
= lz1-z1I•
:d ('1+d1-r:- y ([d -(r1+r )1) [dl-(r1-r1) J), 1
1
1
I
~= 2ffM+d1-f1-f-y([d2-(r1+r1)I] [dl-(r1-r2)1]);
I) µ ( orµ
=
I
(d+r1-u1) (r1-u1) (d+r8 -u1 ) (r1-u1)
--------
I
I"
.HO. (1) µ = 2; (2) µ = 5+2 y6. 223. The group will be finite if a/n is rational. 225. The fundamental domains (some of their possible forms) are shown shaded in Fig. 62. Equivalent boundary sides are connected by arrows. Points with figures are fixed points of the rotations leading to the groups (the figures indicate the number of rotations). For the last five examples the parallelogram of the doubly periodic semigroup is indicated; in example (7) this is a square, in examples (8) and (9) it is a rhombus with angles 120° and 60°. RElllARX. It can be shown that apart from a linear transformation groups (3)-(9) exhaust the groups of linear transformations with one limit point (known as the limit point of a set of points equiva.lent to one another)t. w-l z-1 w-i z-i 226. (1) w = elotz; (2) - - = el«--; (3) - - = el«--. w+l z+l w+i z+i '
t See L. R. FoRD, (1929), Amomorphic junctionB, §§ 59,60, McGrawHill (reprint Chelsea, 1951).
292
PROBLEMS ON OOMl'LEX ANALYSIS
w-a =e"'~
4
( ) l+aw
l+az ·
232. (1) and (2). The construction is obvious; (3) The equidista.nts of the "straight line" a.P (a. and p a.re the "infinitely distant" points of this straight line) are circular arcs with ends at a., 8 (known as hypercycles); (4) The li· miting lines of a pencil of "parallel straight lines" with common point "at infinity" a. are circles touching (internally) the unit circle at the point a. (they are known as oricyclea). 288. (1) For the construction of a "rectilinear" triangle with angles tf> 1, r1> 1 , r1> 8 , construct a circular sector OAB with central angle LI = n- (r1> 1 +r1> 1 +r1> 8 ) draw the "straight line" AB, and through the points A and B "straight lines" at angles t/> 1 and t/> 3 to AB, intersecting at the point O. The triangle ABO is the one required. 234. (1) w
= i(z1 -a1 );
235. (1) w
= z1/a1 ;
(2)w
=
y(z- ~ }-iy'(p/2).
(2) w = y(z/a); (3) w
I
I)
I
2
2 4)
J)
l
7)
)'• = 2A (z+l 21 _ 2
8) FIG. 62
ANSWERS AND SOLUTIONS
293
286. The domain is bounded by Pascal's lim~on u = B(ooslfi+moos 21/1), "= R(sinlfi+m sin 21/1). If the coordinate origin in the w-plane is transferred to the point w =-Rm, then we obtain the equation of the ~n in the usual form (in polar coordinates): (! = B(l+2m cos 0). For m = 0 Pa.sea.l's lim~on becomes a circle, for m == 1/2, a cardioid with cusp w = -B/2. The images of the circles lzl = r < 1 are also lim~ns of Pa.seal the polar equations of which are easily obtained by transferring the coordinate origin to the point u• = -Rmr2 : (! = Br{l+2mr cos 0). The images of the radii of the circle arg z = a: are parabolas passing through the coordinate origin: m(u sin 2a:-tl cos 2a:) 2 +R sin a:(u sina:-t1 cos a:) = 0. To the radii a= 0 and a = 1r correspond the segments of the real axis O~ u~ B(l +m) and R(m-1) ~ u ~ O. 237. The domain bounded by the parabola u = -t18 and the curve Q = 2 cos (0/3), 101 < Sn/4 (Fig. 63).
Fm. 63 238. (1) The domain bounded by the epicycloid u = "= R z=
B( ooslfi+ oo:nl/I ) .
(sin lfi+ sinnnl/I ) , having (n-1) cusps which are the images of the points
n-V(-1); (2) the exterior of the hypooyoloid: u = B (cos lfi+ oos7~1/1 ) , V
. ) •+l t1=B(sinl/l-smnnl/I , having(n+l)ousps(imagesofthepoints z= 1). 289. (1) lml ~ l/n. The domain is bounded by an "elongated" epicycloid (epitroohoid), that is, the trajectory of a point at a distance mB from the centre of a circle of radius B/n, rolling outside a circle of radius B(n-1)/n. (2) lml ~ l/n. In the first case the exterior of the unit circle, and in the second case its interior is mapped onto the exterior of a "shortened" hypooyoloid (hypotroohoid).
294
PROBLEMS ON COMPLEX ANALYSIS
a
~J
(V
l-J-_ 241. (1) w = ( _ z )' ; 1-z 242. (1) w
=
ici '
2 (y(4)+l)ea za (2) w = ---""---:-""""'---1 +3 j/4 c4)-2) e
l
240. (1) w =z•;
(2) w
=-
z•+ l ; (3) w 2z 2z1 -J-3iz-J-2 w = 2z1 -3iz+2 •
z1 -J-2iz-J-l iz•+2z +i;
=
-2zl+3z-2 . 2zl-J-3z+2
243. w = [(z-1)/(z-J-1)]8. l
244. (1) w =
1
z• +R• .!. ( .!.
):
(2) w
=(
z•-R•
z•+R•
a
245 • w = -(2z-J-y(3)-i)I. 2z-y(3)-i 246. w = - ( 2z-J-y(3)-i 2z-y(3)-i
(
250.
)a.
151. 152. 153. 254.
2z+ y(3)-i)•
247 " w
= 2z-y(3)-i •
.. ~ 8
=
249. w
= [ z-y(2)(l-J-i)
Q• • w
z~ -R~). w=e~'(z+l )f. z-1 w = y[(z+ 1)/(1-z)]. w = y[(z-J-i)/(i-z)]. w = y[(z-z 1 )/(z1 -z)]. w = y[(z-J-R)/(z-R)]. tel
a
255. w = e - ... y(z-i) • 256. w = y(z1 -J-h1 ).
·(2z-J-y(3)-i)I. i 2z-y(3)-i z-y(2)(1-i) ]'
158. w = (
~:::~:
r.
157. "' =
•
y(zl-J-hl)
-----"-'Z
For one choice of the branch of yz this function
gives the solution of problem 158, and for the other the solution of problem 159. 159. See the answer to problem 158. 260. w = z/(1-z)• 261. To the circles 4u• ---,1
+
""
(R+ ~) (R- ~)
lzl = R (Fig. 1
=
64) there correspond the confocal ellipses
l (to the circle lzl = 1 corresponds the segment
11 = 0, -1 ~ u ~ l); to the rays arg a = a: there correspond branches of the confocal hyperbolas u 1 /cos•a:-'11 /sin1 a:= l (to the ray arg z = 0 corresponds the ray"= 0, u;;;;;.. 1, to the ray arg z = n, the ray "= O, u~-1; to the rays
argz =±.;!..corresponds the axis u 2
=
0).
ANSWERS AND SOLUTIONS
v
FIG. 64 162. (1) and (2). The exterior of the ellipse
4t1• 1
+
4111 1
(R+ !) (R- ! )
=
1
(Fig. 65, (1), (2)) ;. (3) and (4). The whole plane with a cut along the segment [-1, l] (Fig. 65, (3), (4)); (5) and (6). The wholepla.newithcutsalong the rays (- oo, -1] and [1, oo], on the real axis; (7) The lower half-plane; (8) The upper half-plane; (9) The upper half·plane; (10) The upper half of the interior of the ellipse 4'11 4111
(R+Rl).+( B-R1) =l; 1
(11) The lower half of the interior of the ellipse 4u1 4v1
(R+ ! ).+(R- ! )1=1;
(12) The right half of the interior of the ellipse 4u1 4v1
(R+
!
r
+ (R- ! )' =
with a cut along the segment [ 1,
1
!(R+ ! )];
296
PROBLEMS ON OOMPLEX ANALYSIS
(13) the domain between the branches of the hyperbola u 8 /sin1 r1.-v 8 /cos8 r1.=l
FIG. 65 4ul
263. (1) To the circles lzl =R correspond the ellipses
(R-
~
4tJ! + ----,-,-
r
(R+
~r
= l (to the circle lzl = l corresponds the segment u = 0, -1 =::;;;; v =::;;;1), to the rays arg z = r1. correspond the branches of the hyperbola v1 /sin1 r1.-u,lfcos•r1. = l (to the rays • = 0 and • = :n; corresponds the axis v = 0, to the ray • = :n;/2 corresponds the ray u = 0, v ~ 1, to the ray • = -:n;/2 corresponds the ray u = 0, v=::;;;; -1); 4ul 4tJI (2) Tothecircleslzl=Rcorrespondtheellipses ( 1
a•) +( as)•
R+R
R-R
= l (to the circle lzl = a corresponds the segment v = O, -a=::;;;; u =::;;;;a), to the rays arg z = r1. correspond the branches of the hyperbola
a 1 cos1 r1. = l (to the ray arg z = 0 corresponds the ray v - 0, u ~ a, to the ray arg z = :n; corresponds the ray v = 0, u =::;;;; -a, to the rays arg z =
±;
corresponds the axis u
=
0);
(3) The families of confocal ellipses and hyperbolas obtained from the corresponding families of problem 261 by a rotation through an angle y and a magnification with coefficient lol (the centre of the magnification is the coordinate origin).
.297
ANSWERS AND SOLUTIONS
264. (1) w (2) w
ef• = -(z+ y(z1 -c1)); c
= _2_b (z+ y[z1 a+
(a1 -b1 )]); in both cases for one choice of the
branch of the root we obtain a mapping onto the exterior of the unit circle, and for the other onto the interior; (3) w =
265. w
=
az-by[z1 -(a1-b•)] a•-b• .
A(z+ y[zl- (a1 -b1 )]), where A is an arbitrary comple:ir num•
a-b ber; µ = y(al+b1 )-y(bl+kl)
! (a+ !)] a>O; rays [-oo, ~ (a+ !} ] [-1,+oo)
266. The whole plane with a cut along the segment [-1, the whole plane with cuts along the
if
if
and
a< O.
270. w
-v( ! [(~1 +
271. [1+ 272.
;. )+ (z•+ :.)]).
4(l~h)](w+ ~) = 1
w+;; =
4a
(l+a)•
(
1)
z+z-
(e;• e:•)- 2(1h~h). +
(1-a)1 -2 (l+a)•;
,
{l+a)1
w (0) = ~;
the length of the arc corresponding to the cut equals 2 cos-1
6a-l-a1 (l+a)I ; it is
equal to n for a= 3-yS. 1
l
a+-+b+2(z+-H-(a+-l-)+(b+-H a b 1 ; w'(O) = - - 4- - ; 273. w+ w = 2 ( l ) ( l) a+a + b+b
298
PROBLEMS ON COMPLEX ANALYSIS
the lengths of the arcs corresponding to the cuts are equal respectively to:
274. (1) The image of the circle 0 is a circular arc with ends at the points ± 1, inclined at the point l at an angle 2ct to the real axis; the exterior of the given circle is mapped onto the whole plane with a cut along the given arc;
!I
u
Fm. 66 (2) The image of the circle 0 is (see Fig. 66) a closed curve (Zhu· kOtJBkii profile) with cusp at w = 1, the tangent at this point ma.king an angle 2ct with the real axis; the circular arc with ends± l discussed in part (1), is contained in the domain bounded by the Zhukovekii profile; the exterior of the circle 0 is mapped onto the exterior of the Zhukovekii profile. 276. (1) The image of the circle 0 is a closed curve consisting of two circular arcs with common ends at the points ± 1, the tangents to these arcs at the point l ms.king with the real axis angles respectively equal to 2ct-M and 2ct+ (n-cx)c5; the exterior of the circle is mapped onto the exterior of the domain bounded by the given arcs; the image of the circle 0' (see Fig. 67) is a closed curve with an angular point at w = 1, the tangents at this point being inclined to the real axis at angles respectively equal to 2ct-M and 2oi:+ (n-cx)c5; the image of the circle 0 is contained in the domain bounded by the image of the circle O'; the exterior of the circle O' is mapped onto the exterior of the image of this circle. (2) The interior of the circle 0 is mapped onto the exterior of the domain bounded by a.res of circles passing through the points -1,1, the tangents to these circles s.t the point l forming with the real a.xis angles respectively equal to: (a) 2oi:+ (n-cx)c5, 2oi:+ (2n-oi:)c5, if the function w(z) is defined in the z-pl.a.ne with a cut a.long an arc of the circle 0 lying in the
299
ANSWERS AND SOLUTIONS
lower half-plane, (b) 2oc-d, 2oc- (n+oc)d, if the cut defining the function w(z) is made sloug an arc of the circle 0, which lies in the upper hslf. plane. lJ
FIG. 67
w-l
276. w+l
=
(
zelY+ie'~ ) _,1.. ,wherey= oc, if P> O,andy-oc+nif P 1/2 with a cut along the 1/2~ a:~ I. 279. The whole plane with cuts slong the rays y = O, y = 0, -oo < a:~ -1. 280. The angle -n/n < arg z < n/n with a cut a.long
a: ~ -1 /2 and segment y l
~a:<
0,
oo and
the ray y = O,
ii!~ a:< oo. 281. (1) The whole plane with cuts along the rays
=
lwl ~
V
4,
I
arg w = 2kn (le= O, 1, ... , n-1); (2) w= (l+z•)11 .
f/C4)z
n 282. (1) w = y[z•+v+vcz•-1)]; y'2
f[l+y'(l-z')] = -1-[y(l+z1 )l+Jl'(-z1 )]. z zy2 1
281. (1) w
= (oc•+oc-•)-n f/{zn +r•+ y[(zn+z-•)1 -
Solution. The function C=
! (z• + 21~)
(ocl'+oc-•) 8]}.
maps the sector onto the lower hslf.
300
PROBLEMS ON OOMPLEX ANALYSIS
plane, the points oi: a.ndoi:e :'passing into the points ±2-(oi:n+_.!._), Next, ix"
2
it is necessary to compress the half-plane 1'
=
1
2 the unit semicircle required.
1'+ y(pl-1).
184. w = ..
and map it onto
(ix"+ ix-")
The function w
=
y
T
is the one
.!.( zl +z-I + V![(zR +z-S)a- (oi:I +oi:-8)']).!n
n n )= ( oi:I +oi:-s n
(2) w
n
11
..
n
n
n
11
I[ y(b Ji'(z'+c')+ y(a•+o•)] and +o y(z +c')
JI
w =-
T =
C
1
1) -
1
I
p {y(z•+o•)+oi:+y[y(z1 +c'+oi:)•-tJ8]}. I
I
where oi: = 2" [t'(a1 +o')-y(b'+cl)], tJ = 2" [y(a1 +o1)+ y(b1 +o1)]. 185. w
= y[y(z1 +c')+ y(a1 +o1)].
188. W=-*(v[(:::r+1J+v[(:::r-1J)· 187. w =
y'[ y~oi::{:;•::.;~~(:•-I) l
. . M I, Im T> O,which
~
~«,
where
C=
z+y(z•- I) is the
the upper half-plane is mapped onto the
(T+
~ ), mapsontotheupperba.lf-plane.
Applying the symmetry principle we obtain the mapping of the interior of the right hand branch of the hyperbola onto the whole plane with a cut along the ray ( - oo, -1]; this latter domain is easily mapped onto the upper half. plane. I
REMARK.
The factor y 2 does not play any pa.rt, since the transformation
w' = kw (k > 0) maps the half-plane onto itself. rs
""''tr
292. w = [e::~ 1, 0 < tJ < IX (for IX = 2n onto the exterior of the unit circle with a cut along the ray"= o. 1 ~ u < oo); (7) The domain e« < (! < efl, y < tJ < d (for d-y = 2n this domain is a concentric ring with a cut along the segment tJ = y, e« ~ (! ~ efl). 100. The angle 0 < arg (z+n) < n/n; the strip 0 < '!/ < n. 301. (1) The rectangular Cartesian net u = 0, O; (2) A straight line; (3) The strip 0 < " < IX; (4) The half-strip u < O, 0 0, - n/2 329. w = em-osh . The fiunction u-Btrip c a: a; < n/2 onto the upper half-plane: the points ±n/2+ai then become the
ANSWERS AND SOLUTIONS
305
points ± oosh a. Hence it is easily obtained that the function w = sin-1 sin z oosha will map the given half-strip onto itself so that the rays a: = ± n/2, a.::;;; y < oo will correspond to the rays u = ± n/2, 0.::;;; " < oo. On applying the symmetry principle an infinite number of times we see that the function found is the one required. b sin-l i sinh z
330. w
=
cosha ' sin-1 (l/oosh a)
. 1 1 331. w = sm- "'CciSiil.I
.. / [( .
+V
1
sm-
1
)1
"'CciSiilJ -
(
1 • 1 sin z ) ] sm- "'CciSiil.I
. 1 sinz sm- oosha·
V!['"'-:~ -~-:~].
332. w = ..
sm- cosh a - sm- cosh a
333. w == sin-1 e21s . Solution. The function C= ells maps the strip 0 < a: < n/2 onto the upper half-plane and the function w = sin-1 Cmaps the upper half-plane onto the half-strip - n/2 < u I; 2, if n = I. 424. ni/2. 429, (1) 2/3; (2) 1 - 2i/3. 425. (sin a)/a. 445. Convergent. 446. Convergent. a} ( 426. e" 1 + 2 · 447. Divergent. 427. (1) I; (2) -le; (3) I-le. 448. Non-absolutely convergent, 449. Non-absolutely convergent for ~ :;(; 2lm (k = O, ±1, ±2, ... ), divergent for ~ = 2kn. 450, Converges absolutely. 459. e. 451. Diverges. 460. I. 452. Absolutely convergent. 461. I. 462. I. 458. Divergent. 454. Absolutely convergent. 468. 1/4. 455. B =I. 464. l/e. 456. 00, 465. I, if !al ::s;;;; I; Iflal, if !al> 1. 457. o. 466. I. 458. 2. 467. (1) B; (2) B/2; (3) oo; (4) 0; (5) B"; (6) B, if !Zol ::s;;;; I, and B/llol• if lz.I > I. . 310
311
ANSWERS AND SOLUTIONS
<
r1/r2• 468. (1) R ;;;:i:: min (r1, r 1); (2) R ;;;:i:: r 1r 1; (3) R 1 1 469. (1) z/(l-z) ; (2) -log (1 - z); (3) 2 log [(l+z)/(1-z)]; (4)log (l+z).
470. 471. 472. 473.
Divergent at every point. Convergent (non-absolutely) at every point except z =l. Abeolutely convergent. Convergent (non-absolutely) at every point except z
=-
I. lkirl
474. Converges (non-absolutely) at every point except the points z = eP (k = 0, 1, ... ,p -1). 475. Converges (non-absolutely) at every point except the points z = (1 ± iy3)/2 and z = -I. 476. Converges absolutely. 477. For example, Cn = (-l)n. 479. SoLtrTION. Let us first investigate the convergence at the point z = 1. ~ (-l)b'n) with denominators k 1, k•+ 1, ... , The terms of the series ,,::,, n n-1 (k+l) 2 -1 have the sign (-l)k; let us denote the sum of these terms by (-l)kO'k and prove that O'k-+ 0 monotonically. We have 2k+l (k+l)1 -k1 =--;ca--+Oask-+oo, k• O
I.
00
(see problem 615), for 0
<
<
lzl
,J; nan 3 --+2n; - +2 zan 1 z n=l
+
+-2-]zan-1 ~[(-1)" 22nR1n n•n (2n) !
.L.J
for n
<
lzl
<
2n .
n=l 00
687. \-, b" - all for fzf .L.J nzn
>
max (la!, !bl) .
n=l
,2: ;.." 00
588.
0
where
o_,,. =
21k-a
(
- i 221:- 1- - 3-
22k-s
+-5-
- ...
n=2
... +(-l)1'+l
C-11i+il
=
20-11c (k = 1, 2, ••. ), for fzf
>
2; i tan-1
!·
00
n=O
where o- 1 = 2i tan-1 1/2, 0-1Tc
=
21lci ( tan-1 1/2-
00
,J; ;: + ,J;
k-1
,2:
m=O
2k~l ).
0 ;_,." ,
n=l
( 2m1;l~):m+t),
= 20- 11: (k = 1, 2, ..• ), for 1 < lzl < 2. 689. (1) Yes; (2) Yes; (3) No (the point z = 1 is not an isolated singularity); (4) No; (S) No; (6) No; (7) No (in any ring round the point z = 0, the function is not continuous); (8) No; (9) Yes; (10) Yes, if ex is an integer or zero; no in all the remaining cases. 590. (1) No; (2) Yes, both branches can be expanded; (3) No; (4) Yes, all three branches have expansions; (5) No; (6) Two branches of the four determined by the conditions (l+yl) = ± y'2 have expansions; (7) No; (8) No; (9) No; (10) Yes, all six branches have expansions; (11) No; (12), (13) and (14) Yes, all the branches have expansions; (15) No; (16) No; (17) All i;ne branches have expansions except for two determined by the values sin-1 y'2/2 = n/4. •'-c1t+11
y
594. Onto the whole plane with a cut along the ray e-loit, :
~t <
oo.
+ ... , then denoting by w 0 the value not 595. SOLUTION. If f(z) = assumed by the function f(z) in the unit circle, we consider the funct.ion z+c 2z2
320
PROBLEMS ON COMPLEX ANALYSIS
f(z) / ( 1- f :: } • It is single valued in the circle lzl
expansion
1
~~
= z + ( o1 +
~o } z1+ .. .
m and a regular point if n ~ m; if n < m, then z = oo is a zero of order m-n; (3) A pole of order n+m. I 633. Examples& (I) z1 ; (2) --;- +z; (3) l/(1n - I).
z
834. (I)
a Z-CI:
(a #- 0) or az+b (a#- O);
of the numbers am different from zero) or
ao+a1z+ ••• +anzn (Z-C1:1) (z-ix1 )
...
(Z-IXn-1)
(an #- 0, IXk #- a1 for k #- l). 636. (1) z0 is a removable singularity; (2) z0 is a pole of order n, if cf>(z)
is single-valued in the neighbourhood of the point z0 , and a pole of order nm, if cf>(z) is m-valued in the neighbourhood of this point; (3) z0 is an essential singularity. 637. (1) The point z: is a pole of order n if y' is a rectilinear segment, and a regular point of multiplicity n, if y' is a circular arc, that is, J(z) - f(zt)
322
PROBJ,EMS ON COMPLEX ANALYSIS
= (z-zcT)" t/>(z) where t/>(z) is analytic in the neighbourhood of the point zf and t/>(zf) ¥< o. If z: = oo, then this condition is written in the form /(z) - /(oo) = z-nt/>(z), where t/>(z) is analytic at infinity and t/>(oo) .fi O; (2) Z: is an essential singularity. 689. (1) -1; (2) O; (3) O; (4) O. Ha. (1) An essential singularity at z = oo; the exceptional value is 0 (and ro!); ez-+ 0 if a:-+ - oo (art:-+ oo, if a:-++ oo); (2) An essential singularity at z = O; the exceptional value 0 (and oo);
-1
-1
e 11 -+ 0, if z -+ 0, for example, along the path y = 0, a: < o (e 11 -+ oo as z -+ 0, along the path y = 0, a: > 0); (3) An essential singularity at z = 0; there are no exceptional values (not counting oo); cos .!._ -+ oo for a:
z
== 0,
y -+ 0;
(4) An essential singularity at z == oo; the exceptional values are i and -i; tan z-+ i, if y-+ + oo, and tan z-+ -i, if y-+ - oo; (5) An essential singularity at z == oo; the exceptional value is -1; tan• z-+ - 1, if '!J-+ ± oo. 1 6'8. For example, /(z) = (l+z)I = 1-2z+3zl- ... ; 11111 = -n, 'an+i
= n+l.
= o.
650. 61H. 652. 658. 654.
(! (! = (! = (! (!
660.
(!
655. (! = 656. (! = 657. (! = 658. (! = 659. (! =
n, a= a.
1, '1 = 3. = 1, '1 = 3. = 3, '1 = 2. 1 = 2m, '1 =
SOLUTION.
Similarly
oo ztn I 2 --=n=O (2 .n)I 2
(cos yz +
8
where ot
"\"'1 zn nL::.o (2•.n)! = 21
J; (2~n
=
1
.n).
n=O
V-1,
1/2,
'1 =
1.
l .
00
whence
== Jl'5. = 1. = 1. 1, '1 = y2.
2, '1 1, '1 1, '1
=
(cos
COB
iyz),
'Vz)+cos i j/z).
1/4 (cos otVz+cos ot1 Yz+cosot3 V'z+cos ot•Vz), zn 2~ = -4l~ ,,c.; cos afc 'l /z and k=l 00
whence
n=O (.....-.n).
so on.
661. (! = 00. 662. SOLUTION. It is easily seen that it is sufficient to consider values of z> O. I
I
f e.z:t'dt Then _o_ _
art:
· 665. (1) fl*= fl• a• .::;;;a1+a1; (2) fl*= fl, a•= max (a1, d~ .,-z)
r radius r+~ (~ >
where as
r
e.::;;;e1 •
Since
can be taken the circle with centre at z and
0 is arbitrary), it follows that M 1 (r).::;;;
M(2r+:: (r+~l
,
that is fll.::;;; fl• and hence, fl1 = Cl• Hence also from the inequalities given above we conclude that a 1 = a. Another possible method of proof is based on the theorem given on page 99. 669. fl = 1, a = l/e. 675. fl = 1, a = 2. 670. fl - a, a"" oo. 676. h( = 678. fl = o. 679. h( O. 740 and 741. The integral converges uniformly in any closed interval of the real axis not containing the coordinate origin. 742. The integral converges uniformly in the half-plane Im z ~ 0 with the semicircle lzl < r deleted, where r is any arbitrarily small positive number.
<
326
PROBLEMS ON COMPLEX .ANALYSIS
°'
'748. The integral converges uniformly in the intervals 0 < :s;;; z :s;;; {J < 1 and 1 < y :s;;; z < oo. '744 • .An example is:/(t) =- e11 for n < t < n+e-n' (n =I, 2, ••. ) and/(t) = 0 for all other values of t. '74li. :I:c = a:a = 0. '749. flJc = - oo; a:a =-+ oo. '746. :l:c = :I:a = - 00. '750. :l:c = - l; a:a =+ oo. '751. :l:c = 0; flJa.,. I. '74'7. :I:c = a:a = + oo. '748. a:c = - oo; a:a =I. '752. Divergent at a.11 the points of the boundary. '753. Absolutely convergent. '714. Convergent at the point z = 0, convergent (non.absolutely) at the remaining points of the boundary. '755. Non-absolutely convergent at all the points of the boundary. '765. SOLUTION. On evaluating the integrals present in the obvious inequalities :!.. ~ n I
f
sm•n+l:i;da;
<
0
I
f 0
sin•na;da;
<
I
f
sin•n-la;da;,
0
. [ (2n)!! ] 1 l n [ (2n)ll ] 1 1 we obta.m (2n-l)ll 2n+l < 2 < (2n-l)ll .2n. In order to prove Wallis's formula it remains to establish that the difference between the extreme members of this set of inequalities tends to zero as n .,.. oo. '76'7. (1) No; (2) Yes. '769. (1) Divergent; (2) Diverges (to zero); (3) Convergent; (4) Convergent. '7'70. Converges non-absolutely. '7'71. Divergent. '7'72. Convergent if p
>
!,
the convergence being absolute if p
> l;
divergent if p .s;;; 1/2. '7'78. Absolutely convergent if p '7'74. Absolutely convergent. '7'76. lzl < I. '7'7'7. lzl < 2. 778. lzl < oc. '7'79. lzl > 1.
> 1; divergent if p .s;;; 1. '780. lzl < l/e. '781, lzl < oo. '781. lzl < oo. '781. 1•1 < oo. '784. lzl < oo. '791. SOLUTION. (1) From the series C(a) =- I+ ~ + ~ + ... by subtract·
ing the series for 2-•C(a),weobtain (1-2-•)C(a) = l+
the terms
!, ~s ,
:. + :. + •.. ;
for which n is divisible by 2 a.re absent from the right-hand
side of this equation. Similarly (I-2-•)x (I-3-•) C(a) the terms
~+ =
l+ :.
+:.
+ •.. ,
for which n is divisible by 2 or by 3 being absent from the
right-hand side. Generally (I-Pi"') (1-p;"') ... (1-p;;;')C(a)
=
l+
J; ! ,
(1)
327
ANSWERS AND SOLUTIONS
the summation on the right of (1) extending to those indices n (greater than unity), which are not divisible by any of the numbers p 1 , p 8 , .. ., Pm· It is easy to prove that for Re a ;;ii. 1 + c5 (c5 > 0) the sum of the series on the right hand 00
side of (1) tends to zero as m-+ oo, and, consequently, C(a)
fl (1-p;•) = 1. m-1
(2) Since it follows from the test of absolute convergence (see problem '768)
n 00
that the product
(1-p;;;•) converges for Re a;;;;i. l+c5, the function C(a)
m=l
has no zeros for Re a> 1. '792. SOiiO'TION. It follows from the proof of the preceding problem that
n 00
for any c5 > 0 we have
(1-p;Ci+"»
= C(l~c5)
• From this it is easily
n-1 00
concluded that lim
fl (1-p;),
1-oO n=l 00
it is obvious that the product
fl (I-p;;1) n=l
00
quently, the series
I
n=l
p;;-1 also diverges.
diverges (to zero), and conse·
CHAPTER VII 798. res [/(z)]:r= ± 1 79'. res [/(z)]:r=I
=
795. res [/(z)]:r= -
1
=
-1/2; res [/(z)]:r=o = l; res [/(zll:r=co
i
= 4;
-1/4; res [/(z)]:r= -I
=
res [/(z)]z=co
O.
= O.
(2n)I (-l)•+i (n-l)I (n+l)I; res [/(z)l:r=co
=
(2n)!
= (-l)• (n-l)!(n+l)I · 796. res [/(z)]:r=o = l; res [/(z)]:r=±l = -1/2; res [/(z)]:r=co = 0. 797. res [/(z)]:r=o = O; res [/(z)]z=l = l; res [/(z}]z=co = -1. 798. res [/(z)]:r= - 1 = 2 sin 2; res [/(z)]:r=co = -2 sin 2. 1 5 4 (sin 3-i cos 3);
799. res [/(z)]:r=o - 1/9; res [/(z)]:r=sl = -
res [/(z)]:r=-al = - 514 (sin 3+i cos 3); res [/(z]:r=co = 217 (sin 3~3). 800. res [/(z)]:r= d:+i,.
=
-1 (le= 0,
±
1,
±
2, ... ).
I
801. 802. 808. 88'.
res [/(z)]:r=lcn = (- l)k (le = 0, ± 1, ± 2, •.. ). res [/(z)]:r=kn = 0 (le = 0, ± 1, ± 2, ... ). res [/(z)]z•lcn = -1 (le = 0, ± 1, ± 2, ... ). (1) res [/(z)]z=1 =res [/(z)]:r=co = O; (2) res [/(z)]:r=1 = -res [/(z)]z=CO = -143/24. co
805. res [/(z)]z•o
==
-res [/(z}]z=co
=
2
nl (nl+ l)I •
n-o
806. res [/(z)]z=o = res [/(z)]:r=co = O. 807. res [/(z)]z= - 1 = -res [/(z)]:r=co = -cos l
808. res [/(z)]:r= -a = -res [/(z)]:r=co
=-sin 2
41n [2co (2n-l)! (2n)! n-1
809. res [/(z)]:r=o = 1/2; res [/(z)]
810. res [/(z)]z=o ..!!.
=
(-1)• . (n+l)I' if n
=
0, if n
= 0 or
n
l
d:ni = -2le z--.-
< 0, and also .
if n
co
")1
41n+1 ] + -.J (2n)I (2n+l)I • n-0
• (le=
:n:i
1,
±
2, ... ).
> 0 is odd; res [/(z)]z=o
> 0 is even; res [/(z)]:r=co = 328
±
-res [/(z)]:r=o·
329
ANSWERS AND SOLUTIONS
811. res [J(z)] 1 z-bi"
=
l (-l)k+1-k 1 1 (k :rr;
=
± 1, ±2, ... );
00
2 ~ (-l)k 1 :rr;• ,L.J ---,CS- = - 6. k=l 812. res [f(z)]z=fcl"' = (-l)k2k8:rr;1 (k = 1, 2, ... ). . . 2tk(211k-l) (2k)! B 1t, 813. res [f(z)],.=o = 0, if n is odd, res [J(z)],.=o = (-l)k+l if n = 2k (k = 0, ± 1, ± 2, ... ), where B1 t are the Bernoulli numbers (see 1 problem IH5); res [f(z)] (k+ 1) = (k = 0, ± 1, ± 2 ... ).
res [J{z)]:r=oo
=
I"
z=
( k+2 l )":rr;n
814. 815. 816. by the
l; - 1. O; 2. -2e2knai, if yl = 1 and Logl = 2km; 0 for the branch defined value yl = -1. (a-b)2 s17. ±-8- ·
818. (1) a.-{J (for all the branches); (2) e'% - efl (for all the branches). 819. (1) 2b' + 2 .131 - - /51 + ... , if Log l = 2km; l (2) -21+
820. res [f(z)],.=o if Tan-1 oo =
1
sTI·-
l
5 • 61 + ... (for all the branches).
. = k:rr;, if Tan-1 0 = k:rr;;
res [J(z)]s=oo = -
(2k+l):rr; • 2
( 2 k+l):rr;.
2 821. res [f(z)],.=o = 0, if n;;;.. O; res [J(z)]z=o =Log
P'
if n = -1, and
res [f(z)]z=o = - 1 - (a."+i-f1n+1), if n~ -2; res [J(z)],.=oo = _l_ (a,11+1 n+l n+l -[Jn+ 1), if n;;;.. O; res[f(z)],.=oo = -2k:rr:iifn = -1 and Logl = 2km,and res [J(z)],.=oo = 0, if n ~ -2. 822. -2 0001· o_11f>' {a) O-t(k-i)(a) 828. (1) .Alf>(a); (2) 0_11f>(al+-1- 1- + ... 1) ! - ·
+-vc=
82'. (1) n; (2)-n. 825. (1) n(a); -nlf>(a).
826.
~.
'(a) 827 • .AB. k
_
828. ~(-1)" no-n(/JR*.~_:i)rt+i.
,L.J
n-1
-·
m 829. - y 2 • 880. 831. 832. 883. SH. 885.
-2:rri. -m/121. ni. -2m/9. l.
o.
330
PROBLEMS ON COMPLEX ANALYSIS
2n+1 • 836. - - - • i f n;;;:i: -1, (n+I)I
and 0 if n < -1.
838. 0, if,. < I; ± I if,. > l (the sign depends on the choice of the branch of the integrand). I 841. entl/(l+e"). 839. 7v(I+y'2). 842. 2n/v(a1 -I). 843. 2na/(a1 -b1 ) 111 • 840. n/(l+a). 844. (2a+b)n/[a(a+b)]111 • 845. 231/(1-a1 ) if lal < l; 231/(a8 -l) if !al> l; 0 (the principal value) if !al = 1, a ± l (for a = ± I the principal value does not exist). . n(a1 -f-l) . n l-a11 846. n(a'+l)/(1-a1 ), if lal 1 ,. n /2 , if n -_I . 2.4.6 ... (2n-2) 2 855. n/y2. n • 854. 857. ( - l)k-1/(2ih-z)k. ab(a+b)
+
+
858. (1) ~(cos 1-3 sin I); 3e' ne-lallJ 860. • 2 lbl
(2) ~ (3 cos I+sin 1), 3e1 861.
..:!.. e-Jalllsign a. 2
859. ~ (2cos2+sin2) • 2e' 863. ni, if t> O; 0, if t = O; -ni, if t < 0. 864. n (2sin2 - 3sin3). 865 • ..:!.. (cos l - l/e8). 5 866 • ..:!.. [sina+e-Cav'a)/I (sin (a/2)+y3 cos (a/2))]. 3 867. n(e-labl-1/2) sign a. 869. ~ [2- (2+ labl)e-fallf] sign a. labl 868. ~ (1-e- ) sign a. 4b' 2b8 870. n (lbl-lal)· 871. n/2. 872. 3n/8. I'(p) cos 873. (1)
aP
n:
n:
I'(p) sin ; (2) - - - - ; i n order to verify the correctness aP
331
ANSWERS AND SOLUTIONS
the answer for -1 < p < 0, it is sufficient to note that for these values of p the integral converges and the function in the answer is analytic. 87'. - 1 p
n r (1) -p cos -2p
876. _l_ r p-l
875. -1 p
•
(2-) cos n/2p (for p = l the integral is equal to n/2). p
878. n/sin p:;i. 881.
882.
n • r (1). -p ~m -2p
880. n/2 cos np • 2
n(l-p) - - - (for p = 1 the integral equals 1/2). 4cos np 2
~ ~pl smpn sm l
:::: :
, if l #:-
o, and~
~~: :::
smpn 888.
if l =
o.
-.-:;g[211 ( 1smpn
.!.)-1] . 2
n ( . pn pn ) 889. - . - - sm-+cos--1 , smpn 2 2
886. np(l-p) • 28-11sinpn pn ) , 887. - .n- - ( 2L2 COS--1 smpn 4
~/4/y3, 890. 7'J'
891. If a does not belong to the interval (-1,1), then 1 = -- y(a:- l), where y(a1 - l) > 0 for a> 1 (in· the plane with a cut along the seg· ment [ - 1, l] the quantity v(a 8 -1) is single valued) for a = ± el•, 1
n
= ± y( 2 sinor;)
1(7-.!.)
e c
1
ni
;
for a= iy, I= y'(l+ y•) sign y; for a< -1,
I= y(a~-l); for -1< a< 1, I= 0 (the principal value). 892. If b does not belong to the interval (0,1), then I = ,..:!-bll- 1 (b- l)-ll • smp:;g where (b-1)-11 > 0, bll- 1 > O for b > l; if 0 < b < 1, then I = -nb11-1 X(l-b)-czcotp:;g (the principal value).
898. Io= 2 ; 2 , Iu = ; ( ; 2 -Sit). Iat+1 = - ; ( : 2 -Sat+1) where . h esumof "terms of t h e series . 1- l 1.3 ( 1) 1.3 ... (2S1-l) 8 v 1Bt 2 2 . 4 - ... + - v 2 .4 ... 2,, • n n 1.4 ... (3n-2) 8 4 I 9 . o = y3, In= y3 3.6 ... 3n (n = 1,2, ... ).
+-
SH. n/(n sin : ) •
898. 2 !aJ•;( 2!al) ( : log !a!-1- 3: } •
896 ~l I I • 2!al og a•
899 ' -:n. . n
900. 897. 8nlal (n8+4 log• !al).
2
1og 2.
901. :;il/4.
332
PROBLEMS ON COMPLEX ANALYSIS
-+-1 -1 -(for
903. (1) - 1 I-a
:ri;
a= 1, I= 1/2); (2)
2a(log2a+~2)
oga
- _I_. I+a2
T' ,r -,:..J.
2n
.... ..:i_, {:: L 2a. 927. n+l, if na
along
~1ogJC+B1 +.!., 2ni C-B 4
F+(C) =
~logJC+BJ _ _!_, 2m ~-B 4
F-(C) =
1 1C+B1 1 1 F(C)= 2mlog C-B +7; F(0)=2· F+(iB) = 3/4, F-(iB) = -1/4, F(iB)
=
1/4; F'(O) = -i/nB.
1033. F(z) =~Log z+BB (a single-valued branch in the z-plane with ...ni :za ou.t along 0 is defined by the value Log 1 = O; for lzl > B it is identical with the similar branch of problem 1032)1 F+(C) =
2~ log1~=:1+1/4,
F(O) = -
!"' , F'(O) =
_
F-(C)
=
2~ log1~:;1-3/4;
_i_, nB
t At first glance the solution is obviously:
~ 2ni
1
J~=~[log C-z
-1
2m
(1-z)- log (-1-z)]
=~log 2ni
z-l • z+l
However, it is necessary to verify that the final transformation really leads to the given branch of the logarithm, since the equation log z1 - log z1 = log z1 /z1, is not generally true. This remark must be home in mind in what follows also.
341
ANSWERS AND SOLUTIONS
1034. (1) 0, if lzl < r or izl > R, and l/z", if r < lzl < R; (2) l/zn, if Im z > n, and 0, if Im z < n; (3) 0, if !Im zi < n, and -1/zn,
2~zn
> n; (4)
ifllmzl
[Log
!:=: - k~l ( 2k~;)-~•k-i] •
if m = [;] •
2
Log It+z =log/:!l_+z/+iLfc{arg (C-z)-argC} is the single-valued branch R-z R-z in the z-plane with a cut along 0, defined by the value Log 1 = 0. The limiting value of this branch on 0 has imaginary part n/2 on the left and (- 3; ) on the right. This determines the limiting values F±(C) on 0; F(O) =
~ l+(-l)n-l ; (5) the function F(z) is the same as in the prenRn
2ni
ceding section, only Log RR+z is the single-valued branch in the z-plane
-z
with a cut along the semicircle 0, with the same value Log 1 = 0. The limiting value of this branch on 0 has imaginary part 3n/2 on the left and ( - ; ) on the right. This defines the limiting values of F± (C) on 0; F (O)
=
_1_ 1+ (- l)n-1. nRn 2ni
identical in R-z the disk izl < R, generally they are different; thus, for example, at oo in the first case the branch has the value -ni and in the second +ni. However, In parts (4) and (5) the branches of Log R+z are
REMARK.
F (oo) = O.
1035. Log b-z = log / b-z1 + iLlc arg (C- z) is the single-valued branch a-z a-z in the z-plane with a cut along 0, defined by the value Log I = 0 (the branches are defined in the same way in problems 1036-1040). b-z 1036. b-a+z Log a-z. 1037.
n '\1 L.J Ankzk-1 +zn
b-z
Log
a=z ,
Ank =
bn-k+l-an-k+l n _ k -i- 1
k=l
1038.
b-z oo ~n cnAnkzk-1 + q, (z) Log - - . ~ a-z n=lk=l
b-z..) b-z - . - - Log 1039. -l - ( Log a-z0 a-z z-z., n
1040.
b-z + ~ Ak (z-z0 ) kb-z - - Log--" n [ Log a-z., a-z ~-Zo) l
k=2
l
At= k-1
[
] l l (b-z0 )k-1 - (a-z 0 )k-f '
F (zo)
= -An+i·
1]
,
342
PROBLEMS ON OOMPLEX .ANALYSIS
1041. (1) F+ (z) =log (z+R), F- (z) =log( l+
. ,,,
,,,
!) ;
,,,
,,,
F+ (C) =log !R~+•-p F- (CJ =log 2 oos2-l2; (2) F+ (z) =log (R-z)+ni"t, F-(z) =log F+
(1- :);
m- log 2R lsin : I+' n~t/I , F- (CJ= log 2 lsin ~I+ i n~t/I
of the angle 4> =
11org
C corresponds to the conditions of the problem).
1042, (1) F+ (z) == O, F- (z) = log ( 1F- (C) =log (1-
n;
(2) F+ (z) =log
!);
F+ (C) = O,
z~l'
F- (z)
= 1
(the sense
~
~
R+z
1 [
= O;
F+ (C)
c logC-I' F-(C) =
zlll:-1
1043, F (z) = 2ni ,.:::- ";;i' Log R-z - 2 ,.:::- (2k- l) Rd-l n=l
]
O.
where m
k=l
=[;], and Log [(R+z)/(R-z)] has the same value as in problem 1034, (4). This determines the limiting values .ll':i:( 00
If
~
C) on O.
If lzl
<
R, then
C C-+1 ~ .
1l' (z) -= ,.:CJ onzn, where On = 2 ni log C- l n-o c 1044 F+ (z) = -
:i [
2+y'z log
~~~: ] ,
:~:: ] - rz.
F- (z) .... - :, [ 2+y'z log
1045. F+ (z)seO, F-(z) =Log z-b (the branch of the logarithm is de-
z-a
termined by the conditions of the problem). 1, F- (z) == 1- y'[(z-a)/(z-b)]. 1046. Jl'+(z) 1047. F+(z) = z-la- (1-A)b, .ll'-(z) == z-Aa- (l-A)b-(z-a)A(z-b)l-.1.,
==
1048. F+ (z) =Log z-Cbo, F- (z) =Log z-Co,
z-
z-a
1050. -:i:l+z-3·5. 1011. (1) .!.. (tog - 21 - ) 2 z-1
1
;
(2) .!.. (tog _-r_ ) 2 1-T
1
-
n• • 2
t If Log z and log z are the branches indicated in the conditions of the problem, then Log z =log (-z)+m.
343
ANSWERS AND SOLUTIONS
1058. If
l•I> 1 and
~ 1 11 R+z zeO, then F' (•) =2-, log--Log-R +F'1 m z-1 -z
where F'1 (z) is an analytic function for
(•),
l•I > 1, and the branch of Log BB+z -z
is chosen just as in problem 1084, (4). From this the behaviour of F(z) at the points ± B at the ends of the arc 0 is obvious.
BF Ba: =
1059. F'(z) = - 211 log log 1/r+2• log log l/B,
~~·
- 2 log log l/r
~··
- 2 1 +m.oglogl/R, u .ll.. = -2'loglogl/r-2 1 + 2'loglogl/R, ogr • ogr BF' elll BF' 1 1 1 ;;z= of>(Z) =logl/r' az=-2loglog-;--logr+2 loglog1f, 1068. D (/)is the area of the domain (}', onto which/ (•) maps the domain (Jo
I
I
Rl-C• ; (2) log 1069. (1) log B(z-C)
J "(C)
1·-c1 •-.C ;
,e.. -e"'I
(3) log e•• _ e•C.•
2K
1 1976" "(•) = 2n
" (oo) =
rl-B• ~ (0- 4>)+rl dO'
0
2n
2~
.Bl-2Br
J "m d8.
0
1078.
(1) /(•) - 4>(•)+'1'
('!'), /
1 (•)
= -4> (':)- 'P(z),
where
v(O) - Im/ (0) = Im [4> (O)+'P (oo)]; (2) t1
/(z)
=
-iof>(z)+if/l(~B),
/1(11)
""'iof>(~I) +i'P (z),
(0) =Im/ (0) ... Im [-iof> (O)+i'P (oo)].
1079. (1)/(z) = z•, / 1 (z) = -B••/•" (here and in the answers to problems 1080-1088 the value of v(O) is taken as equal to zero). 1080. /(•)
z" = B••, / 1 (z) =
-l/z11 •
1081. /(•) = - log ( 1- ;. ) (log 1 =- 0), / 1 (z) =log (1-1/z). 1082. /(•)
= y(R~-•) (/(0) =
! ), /1(11) = -
y[z/(z-1)].
1083. /(z) ===log B, / 1 (z) === - log R.
J u(t) 1085. u(z) = n 00
1
-oo
J u(t)c:U + iO. 00
yc:U
1
(t-a:)•+y• , /(•) = ni
~
-oo
(for the existence of the first integral the piecewise continuity and boundedness of u(t) in the whole interval (- oo, oo) is sufficient, for the second, in
344
:PROBLEMS ON COMPLEX ANALYSIS
addition, for example, the function u(t) must be of order ·It~« (a.> 0) at infi. nity). 1086. /(z)
== u(z)-f-it1(z)
J
J
~
=
1 2f
~
n(t-z)
u(t) coth - 2-
1 dt -2f
-~
n(t-z) U1(t) te.nh-2dt,
-~
where u 1(t) =- u(t+i) (for the existence of the integrals it is sufficient, for example that u(t) and u 1(t) should decrease at infinity like
lti~+« , a.>
0) •
1088. Circles inside the disk lzl < 1, tangential to the circle lzl == 1 at the point el9. 1090. Arcs of circles, connecting the points el9,, el91 inside the circle lzl < 1. z-b 1 1 1091. ) = - cos 34>.
k
h11; .Approx.
0 1 2 3
-1·00002 -0·58780 0°30902 0°95108 0°80903 0.00000
4
5
346
PROBLEMS ON COMPLEX ANALYSIS
k
hk
0 1 2 3 4 5
-0·84925 -0·78768 -0·64004 -0·44958 -0·23107 -0·00000
hk
Approx.
Actual
-0·85247 -0·78556 -0·64039 -0·44940 -0·23120 -0·00000
4 ( cos3tf> cos51/> ) 1128. h(tf>) = --;;- COSl/>--3-1 -+-5-1--··· • k
hk
0 1 2 3 4 5
-1·1868 -1·1240 -1·0611 -0·8430 -0·6248 0·0000
hk
Approx.
Actual
-1·1662 -1.1345 -1·0815 -0·8585 -0·5694 0·0000
1131. (2) T[tJ>(x)] = -i[tJ>(x)-21/>a(x)-tf>(oo)], T[!p(X)] = i[tp(x)-2tpb(x)-tp(oo)], where lf>a(z) and '/lb(z) are the principal parts of the expansions of tf>(z) and 'Jl(Z) in the neighbourhoods of the points a and b. No.of problem
h(x)
1132
sin h
f(z)
f1(z)
eiM:
-e-IA: -ie-IAll 1 - z-bi
1133
-cos.Ax
-ieiAz
1134
b - z1+b2
1135
x(l+x2) (y2)(l+x')
1 z+bi y2-iz (y2)(1-z 2-i(y2)z)
y2+iz (y2)(1-z1+i (y2)z)
1136
1 Im (x-a+i'b )"
1 -(z-a+ib)n
1 (z-a-ib)"
1137
1 Im (x-a+i'b) n
1 (z-a+ib)"
1 (z-a-ib)"
347
ANSWERS AND SOLUTIONS
U.'1. If h (:i:)
== 1
in the interval (-1, 1), then h (:i:)
for l:i:I > l; if h(:i:) = 1-l:i:I, if h (:i:)
=
h (:i;) =
then
= - -1 n
:i:-1 log-:i:+l
.!.. [:i: logw-l -log:i:-l] · r
n
1 :i;I sign :i:, then h (:i:) = -; log w-1 •
:i:+l '
1141. By the rectangle formula: h(l) = 0·60716; h(2) = 0·34003; h(3) = 0·20116; h(4) = 0· 14595; h(5) = 0· 11524. By the asymptotic expansion: h(2) = 0·34; h(3) = 0·2015; h(4) = 0·14595; h(5) = 0·11524. 1143. By the rectangle formula: h(l) = 0·49985; h(5) = 0·19257. Exact values: h(l) = 0 · 5; h(5) = 5/26 ~ 0·19231. 1146. (2) T[l/I (t)]
T ['I' (i)]
= - i [• (t)-21/1,.(t)- l/loo~l/l-cc], = i ['JI (i}-2'1'b ('ij- '1'oo+2'1'-oo], n
1/111 (z) =
z-a) ,
'1 (- l)k- 1 At
L.J
dA:- 1 { (k-l) 1 dzA:- 1 coth2-
k=l
m
'Jib (z}
'1 (-l)k-1 B1: ·dk-1
= L.J
(k- l} 1
z-b) ,
(
dzk-l coth - 2-
k=l
n
where
m
,J; (z~:)A: ,
2
k=l
k=l
(z~~)A:
are the principal values of the expansions
of the functions If> (z) and 'I' (z) near the points a and b. 1147. (The notation is the same as in problem 1146) (1) l/l(z)-i Im lf>oo+2lf>-oo;
(2) 'I' (i)+i Im '1'oo+2'1'-oo;
(3) l/l(z)-1/111 (z)-i Im l/lcc~l/l-oo +4>11 (i); (4} 'l'(z)-'l'b(i)+i Im 'l'oo~'l'-oo +'l'b(z);
(5)
1 I z::a2
z-a
z-o
I coth2-+ 2 coth2- ;
(6) coth (z-a) -
1
2
z-a
coth2- +
I
2
z-a
coth2- .
1148. -:;ill"'. 1150. -n•i-n. 1149. -n1/i-•. 1101. nl/-r:•. 1151. =F n 1i-P+9 (the sign - if p and p+q have the same sign, the sign+ if they have different signs; 0 is counted as the sign +).
348
PROBLEMS 0111' COMPLEX ANALYSIS
:ii• ll5S. - T+bi • I
llH. -:n1 oosAToosµT,ifA > µ;:ii 1 sinATsinµT, if A O, plus, if b < 0).
ll56. log l-T + ..!. [(1og-T-) 1 -:ii•]. T 2 1-T
ns7.
~+Tlog l:T + ;•[(1og 1 ~T}9-:n1].
nss.
~[(1og1~Tf-:n•]+ ~n~k[~1og1-;T +(~ + k~1 + ... k=l
... + ~; 1 )] ll59.
-1!
T•-p-l
p=O
Lpl
! [1og :~: -:ii•], 1
T [ R+T ] R+T 1180, "j" log8 R-T-:n1 -2R log R-T·
ll61.
j[tog•:~;-:n•],
ll62. 2R [m- logR+T] + .!.[1og•R+T -:ii•]. T R-T 2 R-T ll63.
R+T ] 1 21 [ log•R-:n-:n
. R+T +2:n.Rt+TlogR-T'
llH. __!._log•R+T - 2m. 21'8 R-T T
1)1 -
i1 ~ (~~1~:.J.
k-0
CHAPTER IX 00
1175. /(z) =
~ .L.i
(z-a)"
n-o (1-a)"+i
;
this expansion continues /(z) analytically
inside the circle lz-al < 11-al which does not lie entirely within the circle lzl < 1, if a does not belong to the interval (0,1).
1176. /(z)
00 (~)"(z+~)" = log -32 + .L.i ~ ; the n
. . Iz+ 211 < 23..
sanes
circle of convergence of this
n-1
18
1191.
SOLUTION.
The substitution e' =a: reduces the integral to the form
00
/(a)
=
J
sin; da;
(11:8
=
eB 101 X),
Integrating by parts, we obtain /(a)
1 00
= cos 1-
J 1
Re
cos a: da:. The latter integral converges in the half-plane ::z:11+1
a> -1. 1194. 0 < Re z < 1, -1 < Re z < I. 1196. The point z = 1 is a simple pole with residue one. 00
1197.
SoLUTION.
Let us use the notation / 1 (z) =
J
e-tl/l(zt)dt. The ana-
o
lyticity of the function / 1 (z) in the circle lzl < 1 follows from the result of problem 535 and the genera.I properties of Laplace integrals (see page 110). Integrating by parts (n+l) times and using the inequalities of problem 535, we obtain for lzl < 1 n
/i(Z)
oo
= - 2,1 zk [e-tlfi(lc)(zt)r +zn+ 1 f k=O
O
e-tlfi(n+ll(zt)dt
O
n
=
oo
J; anz"+zn+i J e-tlfi(n+l(zt)dt.
k-o o From the estimate for llfi(n+l)I it follows that the second term on the right hand side of the last equality tends to zero as n-+ oo (lzl < r). In order to prove the second assertion let us take any point z e G. Then, as is not difficult to prove, inside and on the boundary of the circle on Oz as diameter there a.re no singularities of the function /{z). Hence for a sufficiently small d > 0 within and on the boundary 0 of a circle of radius (lzlf2)+d, concentric with the one already constructed, the function /(z) is also analytic. Thus 3-19
350
PROBLEMS ON COMPLEX ANALYSIS
the equality o,, =
2~ j ~:~ dC holds for the ooemcients On of the expansion
00
/(z)
=
2
o,,zn and consequently,
n-o
co
~ znt",!(C) converges uniformly on 0, it follows that lfl(d) £..I n!1o•+i n-o
Since the series
=~
J!(C)entcdf. The maximum of the quantity Re (z/C) on Oie equal to
me
..
jzi'~,,,.- q <
1 (for the proof of this assertion it is sumoient to consider the
case when z is real and positive as a rotation round the coordinate origin does not change the quantity Re (z/C)) and consequently, ll/l(d)I < .Aett (..4. 00
is a constant). It follows from this that the integral
J e-tl/l(zt)dt
converges.
0
1201. If Log 1, then z = 0 and z = oo a.re logarithmic branch points, the domain of indeterminateness of w (z) at these points coinciding with the whole extended w-plane. 1261 and 1262. z = 0 and z = oo are logarithmic branch points with the domain of indeterminateness for w(z) coinciding with the extended w-plane. 1263. z = 1 and z = oo are logarithmic branch points where lim w
==
=Z-+00 lim w = 1264. z
,, .... 1
oo and z ... 0 is a simple pole for all the branches of w(z) except one.
=
0 and z = oo are logarithmic branch points and lim w z....O
... umw = oo. Z-+00
1265. The branch points a.re the same as for sin-1z (that is, an infinite set of algebraic branch points of the first order above z = ± 1 and two logarithmic branch points above z = oo, lim w = O; z = O is a pole of the first Z-+00
order for all the branches of the function except one). 1266. z = ± i are logarithmic branch points, lim w = oo; z = 0 is a pole z .... :I: I of the second order for all the branches of the function. 1267. The surfaces w(z) and z(w) are the same as for the logarithmic function (logarithmic branch points above the points 0 and oo ). The mapping is easily obtained by means of the parametric representation z = eC. w = ~.
368
PROBLEMS ON COIVIPLEX ANALYSIS
1268. The Riemann surface for w(z) has an infinite number of sheets with one logarithmic branch point above each of the points z = a, z = b and two logarithmic bmnch points above z = oo. The surface is obtained by joining together an infinite number of sheets of the z.plane with cuts going from the points z = a, z = b to co. These sheets correspond to the single-valued branches of the function w+2:iin(n = 0, ± 1, ± 2, ... ). On going round the points z = a and z = b these branches pass successively into one another and this determines the character of the attachment of the sheets. 1269. The Riemann surface for w(z) has an infinite number of sheets with one logarithmic branch point above each of the points z = a, z = b, z = c and three logarithmic branch points above z = oo. The construction is the same as the preceding. 1270. The Riemann surface for w(z) has an infinite number of sheets with one logarithmic branch point above each of the points z = kn (k = 0, ± I, ± 2, ... ). As sheets, z-planes with cuts going from the points z = k"l to oo (for example, along vertical straight lines) can be taken. Two sheets are joined together simultaneously along all the cuts, on one side, as in the construction of the surface of the logarithmic function. At oo there is a transcendental singularity, the limit point of the logarithmic branch points. 1271. (1) In every connected part of the Riemann surface of the function !; = (z) above the z-plane to which corresponds a connected part of the Riemann surface of the inverse function z = 4>-1 (/;) above Gt, w(z) represents a single analytic function; (2) In every connected part of the Riemann surface of the function !; = lf>(z), located above Gz: (this is the domain Gt, transferred to the z-plane), w(z) represents a single analytic function; (3) The same as in part (2). The particular case of w(z) indicated in the conditions of the problem always represents a single analytic function. 1272. w(z) consists of the two analytic functions ±z. 1273. w(z) consists of the two analytic functions ±zt/s. 1274. w(z) consists of the p analytic functions wkzm1/111, (w
=
e21ti/P,
m n1 p = H.O.F of (m,n); k = 0, 1, ... ,p-1; m 1 = p'
n ) = -p·
1111
1275. w(z) consists of the n integral functions wkez:/11 (w = e " , k = 0, l, ... , n-1). 1276. w(z) is a single n-valued function with algebraic branch points of
the (n-l)th order above z = kn (k = 0, ± 1, ... ). At oo it has a non-isolated singularity, the limit of the algebraic branch points. 1277. w = nLogz+2nik(k = 0, l, ... ,n-1) is n distincti analytic func· tions. 1278. w(z) consists of the functions z+2nik (k = 0, ± l, ... ). 1279. A single infinitely many valued function with logarithmic branch points above z = 0, ± l, oo. 1280. w(z) is a single infinitely many valued function with one logarithmic branch point above each of Zk = 2nik (k = 0, ± 1, ... ). At co it has a nonisolated singularity, the limit of the logarithmic branch points. The Riemann surface of the function w(z) is simply connected and is obtained by joining together an infinite number of sheets of the w-plane with cuts (no two having
369
ANSWERS AND SOLUTIONS
points in common) going from the zk to oo simultaneously along all the cuts, but only 1!81. The Riemann surface is the same logarithmic branch points above z = nk (k 1282. The Riemann surface is the same logarithmic branch points above
z= n:
(two sheets are joined together along a definite side of them). as in problem 12110, only with = 0, ± 1, ... ). as in problem 1280, only witb
(k
=
0,
± 1, ..• ).
1283. w(z) consists of the functions ±z+2kn (k = 0, ±1, ••. ). 1284. w(z) consists of the functions z+kn (k = 0, ± 1, ••. ). 1285. (1) Let r 1 = mi , rz = ms, r = r 1r 8 = m be irreducible fractions n1 n8 n and p = (m10 n 1 ) (the greatest common divisor of m1 and n 1 ), q = ('m2 , n 1). Then (z'1)'2 consists of p distinct n-valued analytic functions, equal to
p
p
II
yl.z' = y'l.(y'z)m
=
p
q
II
yl.y'(zm), and (z•2)•1 consists of the q functions yl.z'. One of them, namely zr, always belongs to both cases. In particular, (z2/ 8 ) 8 / 1 3 = ± z, (z3/B)B/8 = yl.z. (2) Let r 1 = m 1/n1; r1 = m 1/n8, r = r 1+r2 = m/n be irreducible
,,
fractions and p = (n1, n 1 ). Then z•1z•2 consists of p distinct n-valued analytic functions equal to yl.z'.
n. ' n1 • n~ = ...!. and n 2 = 'P 'P 'P Then w(z) consists of p distinct analytic N-valued functions with the para. 1et N = ni.n. . t h e preceding notation (3) U smg --· ,
"
metric representation: z = tN, w = y'l (tm1~+tm211~). The Riemann surfaces of N
all these functions above the z-plane are the same as that of the function yz.
n,
1286. LetN=L.C.M.ofm and
= .!!... ,, • Then w(z) is a single
p= (m,n),q
=(m+n, mn)and~+.!. 'P p m n
N-valued analytic function having N /n algebraic
branch points of the (n-l)th order above z = 0, N/m algebraic branch points of the (m-l)th order above z = l and N/q algebraic branch pointl' of the (11-l)th order above z = oo. 1287. One nm-valued analytic function having one algebraic branch point of the (n-l)th order above z = 1, n algebraic branch points of the (m-l)th order above z = 0 and one algebraic branch point of the (nm-l)th order above z = oo. 1288. Two distinct four-valued functions differing in sign with the same
•
Riemann surface as the function yz. These functions have each one branch for which the point z = 1 is a pole of the first order. 1289. An infinitely many valued function with one algebraic branch point of the (n-l)th order above z = 1 and n logarithmic branch points above each of the points z = 0, z = oo. In order to construct the surface it is necessary to join n surfaces for Log z with the cut [l, oo) on one of the sheets of each. The curves ensinn8 = kn(k = 0, ± 1, ± 2, ••. ) divide the w-plane into domains corresponding to the half-planes y ;;e: 0 (Fig. 96, for n = 2. C= w8 is an auxiliary plane).
370
PROBLEMS ON COMPLEX ANALYSIS
FIG. 96 1290. An infinitely many valued function with one logarithmic branch point above z = 1 and an infinite number of only logarithmic branch points above z = 0 and z = oo. The Riemann surface is obt,ained by joining an infinite number of surfaces for Log z with the cut [1,oo) on one of the sheets. Above z = 0, z = oo the surface has only logarithmic branch points though they are infinitely many, and above z = 1 ordinary points and one logarithmic branch point. The curves eu sin ti= (21c+l):n and t1 = 2Tc:n (Tc= 0, ±1, ±2, ... ) divide the w-plane into domains, each corresponding to the z-plane with the cut -co I = It (k = 1, 2, ... , n) and the direction of one of the sides of the star. One value of «k ie chosen arbitrarily. 1348. w
=
O(z-l)ll«!z+l)ll«-1'
0
= :
111:-1•(1-111:)••-1,
1345. The parameters a!.'e determined from the values of lf(bi)I, lf(ota)I, known from P and one of the directions of the sides of P being given. Three of the parameters (ak, b,, "J• d8 ) are chosen arbitrarily. If one of the parameters «t or "J is equal to infinity, then (4) and (5) apply if the factor and term with this parameter are neglected. If one of the parameters b1 or d8 is equal to oo then (4) and (5) remain in force without alteration. 1847. w
= O(z+l)•1(z-1)•1,
1348. w
=
0
ay'{;~:~.). o =
= h(b+l)-•1(1-b)--.,
b
= «1-«1 , «1+«1
J"(hH), a= J"(h/H).
1849. w = oz1-1•cz•-1)•, 0 = hb••-1c1-b•)-•, b = y(l-2111:). 1350. w
=
J"(hH) (
:t:t( :=~r,
where a and bare determined from
379
ANSWERS AND SOLUTIONS
the system of equations
a•ztb«•
= i/(h/H),
IX1 (
1851. (1) w = [Tn(z)]l/n, where Tn
are Chebyshev polynomials;
T~(z)~z8 - l)
(2 ) w = [
~ -a) =
CX1 (
~ -b).
1
= "2 [(z+Ji(z1 -l)")+ (z-Ji(z1 -l)")]
]1'"
1852, The parameters lll'e determined from the valuesofRef(bl) and Re/(d8 ) known from P and the position of one of the sides of P being given, Three of the parameters (a,., b;, oJ• d8 ) are chosen arbitrarily. If one of the parameters or OJ equals oo, then (6) and (7) still apply if the corresponding terms are neglected. If one of the parameters b; or d 8 equals oo, then (6) and (7) apply without alteration (see the answer to problem 1845). 1853. The parameters are determined from the values of Ref(b;) and Re f(d8 ). Three of the parameters a,., bi, OJ• d 8, are chosen arbitrarily. In formula (10) two of the parameters ak, b;, "J• d8 are chosen arbitrarily. 1854. The parameters are determined from the values of Ref(b1) and the position of one of the sides of P. Two of the parameters are chosen arbitrarily. 1855, The parameters are determined from the values of Re f (bi) and Rej(d8 ) and the position of one of the sides of P. Three parameters are chosen arbitrarily. In form1ua (11) two parameters corresponding to the vertices are chosen arbitrarily.
a,.
h
h
:Tr
:Tr
1856. (1) w =....!log (z+l)+ ....!. Iog (z-1) 0 = -
hi log (b+l)- halog (1-b), b = ht-ha; :Tr
(2) w = ....! log :Tr
=
hi+~
:Tr
h
b
+ 0,
h1 (z1 -I)+:Tr
h1 hg log z+O, 0 = - - log (l-b2 ) - - log i, :Tr
:Tr
v(2hih:hs): z+a1 +h11 (3) w = -h11og--- og -z-aa - - . The parameters a 1, a 1 are n l+a1z n 1-a1z
determined from the h1(!1 -a1)
system of equations: a/11a.'•2 = e-,.,
=h1(~1 -a.);
!
+~)].where f(z) is the mapping of part
(4) w
=f [
(5) w
=~log Tn(z), :Tr
(z
where Tn(z) are Chebyshev polynomials (see
the answer to problem 1351); h (6) n =-[log(z-l)+z]; :Tr
(3);
380
PROBLEMS ON' COMPLEX ANALYSIS
z+I
2
.
.
(7) w =log I-z +Az,A- b'-I' where b is determined from the
. b+I 2b equation log b-l + b'-I =cl; (8) w =log (z+I)-Azl-z+const, where A=~ and a is determi·
ned from the equation log a+
~(a-~) +cl= O. In particular, if cl= 0,
then
a- I; (9) w =
h/n log (z-a)+O/(z-a)+Az+const, where A = h/2na,
0 = 27:m (l-a 8) and a is determined from the equations log I-a+.!. = ncl • In particular, if h = 0 then a = 0, A = 0 = d/4. I+a a h 1357. The upper half-plane Im z > 0 is mapped onto the rectangle with
± K, ± K+iK',
vertices
corresponding to the points
±
I,
± ~.
In the
•·plane the same half-plane corresponds to the vertical half-strip with the base (-n/2, n/2), the points given above corresponding to the points
±
n/2,
±
n/2+i log
l~k'
. The whole z-plane with the cuts (- oo,-1],
[I, oo) is mapped onto the rectangle with vertices K ± .X',-K ± iK', with the corresponding pairs of points I/k and -1/k at the sides of the outs. In the •-plane a vertical strip is obtained passing through the points ± n/2 (Fig. 102).
@
iK' Ill D~C ---JV A~1 8 B'
C'
I
D
C'
I
c"
'
B'
-f
/(
-if(' FIG. 102
z
1158.
10
= .ii
J
0
dz
y[(I-z1 )(1-klzl)]'
z = sn ( ': , k) , where .ii and k are "'
determined from the relations K'/K = b/a, a= .ilK.
381
ANSWERS AND SOLUTIONS
1859. See Fig. 103. The inverse function is z = en (w, k).
JJ" D'
@)
21<
c
-tiO
I
n -,/
< v<
0
!vi
n -,/
< v 4nq.R, then IZtrl > B (the second stagnation point is in this case inside the circle lzl = B). See, for example, [3, Chapter III, § 49].
1391. w(z) = Ve-l11z+
i: I't:~Qk
k=l
log (z-ak)· At infinity there is a doublet
m
n
with moment 2nVe-f11 and a vortex source of strength Qoo =
and
k=l
n
intensity I'oo = -
2 Qk
2 rk. k=l
1
r
189!. (1) No; (2) Yes; (3) Yes (for example, the floww =-+..--=-log z z ...m has a stream line originating from the coordinate origin). 1893. In a schlicht conformal mapping a vortex source is transformed into a vortex source of the same strength and intensity. A multiplet is transform!i!d into a set of multiplets of the same order inclusively. A doublet is
401
ANSWERS AND SOLUTIONS
transformed into a doublet with the following relations between the moments: (1) (a; p)-+ (a:; pc1 ), (2) (oo; p)-+ (a:; pc_ 1 ), (3) (a; p) -+{oo:
(4) (oo; p)-+ { oo;
L) C-1
:J,
1394. In a conformal mapping onto n sheets a vortex source is transformed into a vortex source with the strength and intensity decreased by a factor equal ton. 1895. The law of the change in the vortex source is (a* is the point symmetrical to the point a): (a; Q, I')-+ (a*; Q, - I') in the case of a stream line and (a; Q, I')-+ (a*; -Q, I') in the case of an equipotential line. The law giving the change in a doublet is more complicated. In the case of a rectilinear stream line: (a; p)-+ (a*; p'), where the vectors p, p', are drawn through a and a* respectively, symmetrical with respect to the stream line. In the case o! a circular stream line lzl
=
R: (a; p) -+ {a*; -
~ p ).
if a of: 0 and
(a; p) -+ ( oo; :. ). if a= O. In the case of rectilinear and circular equipotential lines using the same notation we have respectively: (a; p)-+ (a*; -p'); (a; p)-+ (a*; ;: p ). (O; p)-+ (oo; -
p/R1).
1896. (1) In all cases the singularities of the flow must be symmetrically situated with respect to the circle lzl = R (see problem 1895). In particular the axes of the doublets must be situated on this circle, they must be tangential to it. If vortices occur on lzl = R they must also have doublets. Moreover, the total strength and intensity must be equal to . ~ 1 ~ • zero, for which: ~ Qa:+ 2"~ Q1 = 0, where Qk are the strengths of the
sources within lzl = R and Qj are the strengths of the sources on lzl = R; = 0 where the are the intensities of the vortices on 1z1 = R. (2) The singularities of the flow must be symmetrically situated with respect to the circle lzl = R (see problem 1875). In particular, the axes of doublets situated on lzl = R must be orthogonal to it. If there are sources on lzl = R they must also have doublets. Moreover the sums of the strengths
Iri
rj
and of the intensities must be equal to zero, for which:
2 I't+ ! Ir';
= O, where the I't are the intensities of the vortices inside lzl = R and the rj are the intensities of the vortices ~n lzl = R; Qj = 0, where the Qj are the strengths of the sources on lzl = R.
2
1897. (1) w
(3) w
= Vz+c = ~
(c is a constant); (2) w
log [(z-a) (z-ii)]+c (at oo there is a source of
strength -2Q);
p (4) w= 2n
z-a = -2r:n:i.log--_-+c; z-a
1
p
1
z::a+ 2n z-a+o;
402
PROBLEMS ON COMPLEX .ANALYSIS
p 1 p 1 +--+--+liz+c 2:n z-a 2:n z-i 7
n
.J; Q1c);
(at oo there is a source of strength P
+ -r+•Q -.-logz+ c,
k=l
(6) w = -2where Imp= O. A flow with 2:ii• :nz a vortex without a doublet is impossible.
r
z-a
1398. (1) w = - 2 • log R• _ +c; :it' -az pl p*l . (2) w = - - - + - - - + c , if
2:n z-a
2~z +
and w =
2:n z-a*
a+ 0
: : z+c, if a = O { p* = :. ) , n
1899, (1) w
= .}; ~! log [(z-a1c) (R1 -a1cz)]+c, k-1 n
(2) w
n
if .}; Q1c
=
O;
k=l
=.}; ~!log [(z-a1c) (R -a1cz)] 1
k-1 m
+.}; ~~ log (z-a;)+c, 1-1
1400. (1)
r z-a w = -2 • log R• _ +c; :iii -az
(2) w =
L -1-+
p* - 1-+c ( a*
1 = Ra , p* = - iR•1 p ) ;
2:n z-a* R 1 Velac (3) w = Ve-lacz+ - - +c; z R•Veloc I' (4) w = ve-lacz+--+-. log z+c. z 2m 1401
' w
140Z. w
2:n z-a
= _.!.__ 1 2:iii
= 2~
(z-ia) (az+i) , og (z+ia) (az-i) '
log (z 8 +a 8 )+c.
0 •
1408. w=
2~ log(z -l)+o. 1
403
ANSWERS AND SOLUTIONS
1404.
Q
w= 2n
z -l log --+c. z1 +I 1
The flow is possible if
tX
1405, w
= ± ~ {if I 2
= 2~
log
(I+!) +c.
n
I't iF 0, then at oo there is
1'=1
n
,2; I't) .
a vortex of intensity -2
k-1 n
1407. w
~ [r"+'Q" = £.J 2ni
r"-'Q"
- ]
log (z-ak)+ ~-log (Rl-atz)
k ... 1
1' I p• I RI Vel« +-. --+-. --+ ve-l«z- -z- +a 2n z-a 2n z-a• ( a•
RI = ii'
n The flow is possible if I't = 0 • k-1 1408. Let t =- /(z) ma.p D conformally onto the unit disk to ""' fl)[J(z)], where
1
p•
= Ba• p ) •
2
!ti < I. Then
n
2
Q" = 0 • k-1 1409. Let I = /(z) map D conformally onto the domain
11: = /(at), with the indispensable condition
ltl > I with the
normalisation /(oo) = oo, /'(oo) > 0. Then with the condition we have w =- !J>[J(z)], where
n
2
k-1
Q" = 0
404
PROBLEMS ON COMPLEX ANALYSIS
1410. In the notation of problem 1409, w = !li[J(z)], where
Ve-lex Veiac I' !li(t) = /'(oo) t+ f'(oo).t + 2mlogt+c. For I'
=
O, w(z) maps the exterior of 0 onto the exterior of the segment
[- !'~:), /'~:)] w(oo)
=
of the real axis of the w-plane with the normalisation
oo, w'(oo) = Ve-lex.
1411. (I) w(z) =
_!_b [(az-by(z1 -c1)cosa:+i(bz-a y(z1 -c1))sina:]+const a-
(Voo = Velex, c = y(a1 -b1));
(2) w(z) =
...!'.'.._ [(az-b y(z1 -c1)) oos a:+i(bz-a y'(z1 -c1)) sin a:] a-b
rm log (z+ y(z -c )}+ const.
+ 2
1
1
1412. (I) w(z) = V(z cos GC-iy(z1 -c1 ) sin GC)+const (Voo = Vefex);
(2) w(z = V(zcosa:-iy(z1 -c1 ) sin GC)+
{m log (z+v (z -c ))+ const, 1
1
where I'= -2ncV sin GC (c is a point of departure). 1418. Let theZhukovskiiproflle be obtained by the mapping z = (C+C-1 )/2 of the circle IC-Col= I I-Col= R > I, Co= I-Be-IP (0 ~ fJ < n/2). Then for circulation I' and V oo = Veicx ; w(z) = VR ( z-Co+ y(z1 -I)
2
B
Relex ) e-lex+ z-Co+ y(z•-I)
r
+ 2 mlog [z-Co+Y (z1 -I)J+c, where I'= - 2n BV sin (or:+ {J)(I' is determined from the condition w'(I) = 0 corresponding to the Zhukovskii-Chaplygin postulate). 1414. w(z) = y[z - (p/2)]+o is the streamlining of the parabola from outside; w(z) = i cosh y(2z-p)/2yp is the streamlining of the parabola from inside. I 1!. _ mcx 1!. mcx 1415. w(z) = y.I [(z+ y(z1 -c1 ))1P e IP -(z-y(z•-08))1/1 e I/I ]+const
is the streamlining of the right hand branch of the hyperbola from outside b
(tan GC = - , fJ = n-a:, c = y(a1 -f-b1 ) and a i ..!!... 1!.. w(z) = y 2 [(z+y(z1 -c1 )).., + (z- y(z1 -c1 )).,. l+const is the streamlining of the right hand branch of the hyperbola from inside. 1416. w(z) is determined from the equation z = enwto+nw/v (the values of the stream function on the streamlined half-lines are taken as equal to± v).
405
.ANSWERS AND SOLUTIONS
1417. w = cosh-1z =Log [z+Jl'(z1 -l)] (the values of the stream func· tion on the streamlined half-lines are taken as equal to 0 and n). 1418. (1) A fl.ow with period n; at the points kn (k an integer) there
are sources of strength Q; the points ; +kn are stagnation points. The velocity at infinity in a strip of periods is V00 = V(:i: ± i oo) = =F
~ i.
See
Fig. 129 for the stream lines and equipotential lines. (2) The same, only instead of sources at the points kn there are vortices of strength I' and V(:i: ± i oo) = =F I'/2n. In order to construct the field the stream lines and equipotential lines in Fig. 129 must be interchanged.
FIG. 129 1419. A fl.ow with period n; at the points Tm there are doublets with moment p; the velocity is V(:i: ± i oo)""' O. For the stream lines see Fig. 130. 1420. If V1 = iV, then the solution is possible and unique for V1
=i{v- 2~):
·(v--;;-Q) +
_ I'+iQ 1
"' -
. n(z+ti) + • n(z-a) + -I'+iQ 1 ogsm 2w 2ni ogsm 2w '
2ni
1421. w
=
p n(z-a) 2n cot 2w
P
2n
1
0•
cot n(z+a) +iVz+o. 2w
1422. Let t = /(e) map S conformally onto the rectilinear strip St so that D 1, D 1 pass into points at infinity of St. If there exist non-zero derivatives /'(D1 ),J'(D8), the velocities V1, V1 touch the boundary of S at infinity and one of them is given arbitrarily, then for St the problem reduces to problems 1420, 1421; the solution exists and is unique.
406
PBOBLEMS ON COMPLEX il.ALYSIS
f>·O
arg p·f FIG.
130
407
ANSWERS AND SOLUTIONS
c
A. • ~8 .8A ~B Cc:e,
]J
A
~B Cce8
A
~ e1 cCc:e1
A
z)
1)
]) c = A
3)
FIG. 181
A
40S
PROBLEMS ON OOMPLEX ANALYSIS
1413. (1) It is necessary and sufficient that the numbers M and 0 should be real. In this case the curves Re u = ± Q).
r.
1 r." loge "'= 2nr log a+ 2 i
418
PROBLEMS ON COMPLEX .ANALYSIS
(2) /(z) = !l>[u(z)], where
u-a) u-a "" 1--
(J (
I'1-I'1 l 2W r. o +~ og-(--::.-) + 2 ,.. u+ , i
(J
2w
and u(z) - a+!:_+ ... maps the domain D onto a circular ring. z 14150. The trajectories of the vortex are closed curves within the rectangle with centre of symmetry at the point a+bi. For a vortex close to the centre of symmetry these trajectories are nearly ellipses. If the vortex is at the point a+bi itself these trajectories are stationary. Solution. Let Zo = m0 +iy0 be the position of the vortex in the rectangle considered. Continuing the flow by the symmetry principle we obtain a doubly periodic flow with periods 2w == 4a, 2w' = 4bi, having in the rectangle of periods two vortices of intensity r at the points ± Zo and two vortices of intensity -I' at the points± i 0 • Hence
w(z)
(J (z-zo)(J (z+zo)
r
= - - . log
2ni
1'° 1'°
91 (z~0 }(J1 (z!zi0 )
•
consequently•
w'(z)
==
f'ni
{[ccz-z0 ) -
ia (z-z.>]+(ccz+z.i>- i, (z+-zo>]
- [ C(z-i0 ) - 2: (z-z0 )[ -
[
C(z+zo)- 2: (z+io)]}.
• to t h e lim'it as z -+ z.,, we · tirom t his -2r . -1- and passmg Sb u tractmg m z-zo obtain:
dmo
. dyo
I'
,.
2a;
•
,.
2i
'"(Om
I'
p'(2mo)-p'(2iYo)
""dt-'dT = 2ni [,( o+ 2•Yo)-,( Yo)-, -o>l = 4m p(2lllo)-p(2iyo) ' whence
419
ANSWERS AND SOLUTIONS
(since p(2iy0 ) is a real number, p'(2iy0 ) is a purely imaginary number). (From (•) it follows that p' (2m0 )d3:0 = ip' (2iy0 ) dy0 • Butfor the analytic function p(z) the derivative p'(z) =
apa
a:
=.;. i
apa
y
and it follows from the preceding that
dp(2111o) = dp(2iy0 ), consequently, p(2111o)-p(2iyo) =
a.
(••)
Such are the trajectories of the vortex within the rectangle. In order to investigate them, let us note, that when a:0 increases from 0 to a, the value of p(2m0 ) decreases from +oo to 8 1 = p(2a), then increases to +oo, and when y0 increases from 0 to b, the value of p(2iy0 ) increases from - oo to 8 1 = p(2bi), then decreases to -oo. We conclude from this that for C > 8 1 - 88 the trajectories (••) represent closed curves symmetrical about the point a+bi; for 0 = 8 1 -e1 the curve shrinks to a point - the stationary case. In particular, if 1110 = a+,, Yo = b+17 and E, 17 are small, then p(2a + 2E) AS
e8 +
;!
(2E}1 p''(2a), as p'(2a) = O; p(2bi+2Ei) ""'e8 +
;!
(217i)1 p"(2bi),
as
p'(2bi) = 0 and the equation (.. ) assumes the form p"(2a),1 +p"(2bi)17'
""' !
[0-(e1 -e8 )] that is the trajectories of the vortex are close to ellipses.
1451. u = ru:-{Jy, "= {Ja:+rx,y, E = -iii; a dipole (oo; -iC). 1452. u = 2q, "= 2q log.!.; E =
r
(co; -2q).
~ el; r
point charges (a; 2q) and
lz -a1
z
-b 1458. u=2qargz-a• t1=2qlog z-b; E=
2q(b-ii) _ _ _ (z-'ii)(z-b)
point charges (b; 2q) and (a; -2q). 1454. u = -2q arg (zl-a8), " = 2q log lz1 -a1 1; E = -4qi/(r-a8); point charges (a;-2q), (-a; -2q) and (oo; 4q) (see Fig. 122). 1455, u =
J!1 sin (-cx)," = E1 cos (-«); ~
a dipole (O; p) (Fig. 144). 1456. u = ( r dipoles (O; =F
± ~1 )
iB8)
cos .,,, " = ( r
± ~a )
sin 4>; E = - i ( 1 =i=
~8
e•l4>) ;
and (-o; -i) (see Fig. 119, 120).
1457, u= -py+2q4>, t1=pa:+2qlog_!._; E =
r
-p-f-~e14'; r
point charges (O; 2q) and (oo;-2q); a dipole (oo; p) (see Fig. 127). 1458. u = -py+
1:" 2q1;4>1;, k=l
"=pa:+
" l 2 2q1; log-; k=l r1;
E = -p
+ 2"
2q1; el4'1; '
k-1 r1;
where z-a1; = r1; e1"'"; point sources (ak; 2q1;); a dipole (oo; p) (Fig. 145).
420
PROBLEMS ON COMPLEX ANALYSIS
i
i
/
FIG. 144
Fm. 145
421
ANSWERS AND SOLUTIONS
1469. (1) The value of the point charge is preserved; the law of variation of the moment of the dipole is the same as in problem 1893; (2) The sign of the charge is reversed; the law of variation of the moment of the dipole is the same as in problem 1896 on continuation across the equipotential line. 1460. " ""' 2q g(z,a). 1461. w
= 2qi log z-Ro +c. z-zo
. RI-Roz 1462. (1) and (2) w = 2qi log R(z-Zg) 1
1463. w
= 2qi log /(z)
1464. w
= 2qi log f;z) +c,
1466. w
= 2qi log/(~) , 1
.. M
2d
--
B
f
(.!..2 • .!.) t 4
+c • z-y~-~
a-b
+ const, where /(z) =
,
where f(z) = [z- y(z1 -R1 )]/R.
where t
= /(z)
is determined from the equation
y(l-1') ... .:I•+ _d (see problem 1838 for ei 2
n
=
4 and
1374 for k .... l/y2).
1-cn(!z,k) 1 1466. w = 2qi log /(z) +c, where /(z) = and k is desn (
! z, 1c)
termined from the equation K' /K = b/a (see problem 1871).
1467. w
=
1 2qi log/-() +c, where f(z) =
z
6 ( Z-Zo) 1
4a
(J ( Z-Z9 ) 1
4a
61 ( z;az1) 61 ( z;aza)
and w = 2a, w' = 2ib, z1 = (4a-a:0 )+iy0 , z1 = a:0 +i(4b-y0 ) (see problem 1460).
z1 = (4a-a:0 )+i(4b-Yo),
.RI ~) , 1468. w = -pi - + - -~ . + c ( a+ 0, a*= -,p* =-::a z-a
w =
~
z-a
q
a
_ii.z!. z+c (a =0), c is a real number. Compare with problem 1868, (2). RB)
pi p*i ( R• 1469. w = - - + - - . + c a+ oo, a*=-, p* =-=a p ,
z-a
z-a
q
pi pi w = z--Ri"z+c (a= oo). See problem 1400, (2). 1470. w
= e(z cos a.+i sin a. y(z8-R1))+const.
q
422
PROBLEMS ON COMPLEX ANALYSIS
1471. w =
where o1
~b [(ae-by'(z1 -o1)) cos ix-i(bz-ay'(z1 -o1)) sin ix]+ const a-
== a 1 -b1•
2Ke K K' b 1472. w = - - (cos ix+isin tXCnu), where u = - z and-= - (see asnu a K a problem 1466). 1473. (1) If pi= eela., then /(z) = et'(a>{[,;z)+t(z)]oosix+i[ ,;z) -t{z)]sinix}+o;
R•pi
(2)/(z)=pil(z)- t(z) +o=e
{[t{z)+t(z) RI] oosix+i [ t(z)-t(z) R•] sinix}+o,
where pi= 0 and o is a real number. 1474. w
I-
, if
Imz~ 0,
- log-1- 1- 1 , if z-a
Imz~O;
log
1476. v(z, a) =
lz~'iil
{J nI (a:-ix)•+(JI
e(a:, a) = -
(a= ix+i{J).
1477. (1) Inside the cirole
v(z, a)
RI-_ azl R
= log I
=
1
I
.Rl +log l;I] if a¢ O,
,-[log/
z- i.i logR,
ifa = 0,
Outside the circle v(z, a) = - log-1- 1- 1 • The density is z-a I
e(Rel", a)=
-"'iiR
R 1 -lal 1 .Rl-2Rlal cos (8-ix)+lal1
In particular for a = 0 it has the constant value -
(a= falel«).
2~
and is identical
t See the appendix by M. Schiffer to the book: R. Cc•URANT, Diriohlet'a Principle, ConJormal Mappings and Minimal Surfaoea, § 1, sec. 2, especially pages 242-3, Interscience, New York, 1950.
423
.ANSWEBS AND SOLUTION'S
with the potential of a layer having the constant value log R inside the circle and the value log lzl outside the circle.
I
log
(2) v(z, a)=
IR1 -iiil
R
, a of:. oo, lzl ;;;;;. R,
- log-1- 1- 1 , a of:. oo, lzl z-a
v(z, oo) = {
e(Re19, a)= -
~ R;
log lzl, if lzl ;;;;;. R, .
log R, if lzl ~ R; l !all-RI 2nR Rl-2Rlal cos (8-0t)+lal1 (a"" lalef« and a of:. oo).
U a = oo the same potential is induced all in the preceding case for a = O. lz+y(z1-R1 )1 1 1478. v(z, oo) =log , e(a:, oo) = - 23tf(R•-a;I) (la:I< R). 2 lz+y(z1-ol)l , outside the ellipse, "(z, oo) 1479. v(z, oo) =log 2 .... - log 2(0t-/1) inside the ellipse. The density is e(z, oo) = -
CC
2ny(l~-c•I)
is on the ellipse, e' = 0t•-f18).
1480. e(C) = - 12n 1 1481. e(C) = 2d
og(C, oo)
em=
1485, l/2R.
•
1 1482. e(a:) = 2ny(R8-a:I) (la:I
1483.
1484. R.
on
<
R).
1 2ny(IC8-0•1> (el= a 1 -b1 ).
1486. l/2(a-b). 1487. a.
1490. If c.o(z; LI) is the harmonic measure of the interval LI of the reel axis at the point z with respect to the upper bslf·plane: c.o(z; LI) = -1 - 1 dt (see problems 1091 and 1489), then (omitting 0n. log1t-z n
°
.!.. J.d
a real additive constant): (1) w = ±..1og (z-a)," =
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