Chapter 6
Mathematics
CHAPTER 6 Complex Variables Complex Variable , where x & y are real numbers and they are called real and imaginary part of Z. | |
√
( )
. /
Function of a Complex Variable I
wh ch ‘ ’
c
p
c
p
‘w’
D,
then , w = f (z) Where, z = x + iy, w = f (z) = u(x, y) + i v(x, y) Limits and Continuity Let w = f(z) be any function of z defined in a bounded or closed domain D, then, ( ) =
, if for every real
we can find real
Such that | ( ) | |< for | Basically it means: Single value for all values of z in the neighbourhood of z = possible exception of z = itself
Y
with the
v
f(z) )
z
z - Plane
pPlanePl ane PVT.LTD. THE GATE ACADEMY
X
w - Plane
u
H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 173
Chapter 6
Mathematics
Note: In real calculus x approaches only along a line, whereas in complex calculus, z approaches from any direction in the z- plane [i.e. limit is independent of the manner in which ] Continuity of f (z)
( ) A function = f (z) is said to be continuous at if ( ) Further f (z) is said to be continuous in any region R of the z-plane, if it is continuous at every point of that region. Also is w = f (z) = u(x, y) + i v(x, y) is continuous at , then u(x, y) and v(x, y) are also continuous at x = xo & y = yo.
Q (x +
,y+
)
( )
( )
P(x, y)
Derivative of f(z) =
(
( ) (
, ’( )
2 =
+i
=
) )
( ) ( )
, 3
–i
Provided the limit exists and has the same value for all the different ways in which approaches zero. Differential rule are same as real calculus . / =
( )
Theorem The necessary and sufficient conditions for the derivative of the function f( ) to exist for all values of in a region R. i) ii)
,
,
, ,
, are continuous functions of x and y in R. ,
Cauchy-Riemann equations (CR Equations)
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 174
Chapter 6
Mathematics
Example ̅
Evaluate
, if it exists.
Solution If limit exists, it is independent of the manner in which Now let
along x-axis, then y = 0
z = x + iy =x ̅ = x – iy = x ̅
= =1
Now let
___________ (i)
along y-axis, then x = 0
z = iy ̅ = – iy ̅
=
= -1
___________ (ii)
Hence limit do not exists.
Example f(z) = ̅
Is it differentiable?
Path 2
y
z+
z Solution
Path 1
z = x + iy
x
=
( )=
. =
(
)
( )
̅̅̅̅̅̅̅
/=.
0
(
̅)
/ )
1
= If
( ) = 1 (path 1)
If
( ) = 1 (path 2) Not differentiable
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 175
Chapter 6
Mathematics
Analytic Functions (or Regular Functions) or Holomorphic Functions A single valued function which is defined and differentiable at each point of a domain D is said to be analytic in that domain. A point at which an analytic function causes to posses a derivative is called Singular point. Thus if u and v are real Single – valued functions of x and y such that
,
,
,
are
continuous throughout a region R , then C–R equations = h“
c
= c
”c
h
c
( )
c
R
Real and imaginary part i.e. u, v of the function is called conjugate function. Analytic function posses derivatives of all order and these are themselves analytic.
Harmonic functions If f(z) = u + iv be an analytic function in some region of the z – plane then the C –R equations are satisfied. =
,
Differentiating with respect to x and y respectively, =
, =0
(Laplace Equation)
Methods of constructing Analytic functions 1.
If real part of a function is given then, f ’( )
-i
Integrate with points at (z, 0) f (z) = ∫ . /
( , )
dz
i ∫. /
dz + c
( , )
Similarly in case v(x, y) is known, then ’ (z) =
+i
f (z) = ∫ . /
( , )
dz + i ∫ . /
( , )
dz + c
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 176
Chapter 6
2.
Mathematics
If u (x, y) is known, then to find v(x, y) we have dv =
dx +
dv =
dy
dx +
dy
Integrate this equation to find v. 3.
f (z) = u(x, y) + i v(x, y) If a real part of the analytic function f(z) is a given harmonic function u (x, y), then f(z) = 2u . , / u(0, 0)
Example Find analytic function of u = Solution Approach 1 = = - 6xy ’( )
.
/ ( , )
(
=
)
= f(z) = =
z+ci i
Approach 2: f(z) = u + iv where, u = u . , / = . / – 3 . / . / + 3. / + 1 =
+
+
=
+
+1
+1
u (0, 0) = 1 f(z) = 2u . , /
u (0, 0)
f(z) =
+ci
Approach 3: dv = =
dx +
dy
( 6xy) dx + (
= (6xy) dx + (
) dy ) dy
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 177
Chapter 6
V=∫
+ ∫(
V = 6y. =3
+ (-3
y
)
[
(
)
]
+ 3y) + c
+ 3y + c f(z) = u + iv =
– 3x
=(
) + 3x + 1 + 3yi + c i
= (
) + 3(x + iy) +1 + c i
=
Mathematics
+ 3x + 1 + i (3
y-
+3y) + c i
+ 3z + 1 +c i
Complex integration Line integral = ∫ ( )
, C need not be closed path Here, f(z) = integrand curve C = path of integration
Contour integral = ∮ ( )
, if C is closed path
If f(z) = u(x, y) + i v(x, y) and dz = dx + i dy ∫ ( )
=∫(
)
∫(
)
Theorem f(z) is analytic in a simple connected domain then ∫
( )
= f( )
( )
Integration is independent of the path Dependence on path I “C p ” p h p (However analytic function in simple connected domain is independent of path.)
C
ch ’
h p h
theorem
If f(z) is analytic in a simple connected domain D, then for every simple closed path C in D,
∮
( )
C
=0 -------(A)
D
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 178
D
Chapter 6
Mathematics
Note I h w , C ch ’ h (z) is analytic on a simple closed path C and everywhere inside C (with no exception, not even a single point) then ∫ ( ) C
ch ’ I
If f(z) is analytic within and on a closed curve and if a is any point within C, then f(a) =
∫
( )
( )
f ’( )
∫
f ”( )
∫
. f n(a) =
∫
(
) ( )
(
) ( )
(
)
Note complex analytic function
has derivative of all order.
in real calculus if a real function is differentiated once, nothing follows, about the existence of second or higher derivative Example Find the complex integral of ∮
(C : circle of radius 3 )
Solution 2π
∮
, because
is analytic over entire region
Example = ? ( C : circle of radius 3 )
∮
Solution =0
∮
Example If g(z) = ∮ ( )
=?
where, C is circle shown as,
0
1
2
Here C : |z – 1| = 1 THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 179
Chapter 6
Mathematics
Solution g(z) =
=
(
)(
)
g(z) is not analytic at z = +1 & z = 1 (Hence watch these points ) Hence circle C, |z – 1| = 1 encloses
= 1,
Where g(z) is not analytic
Hence we write g(z) = (∮ ( )
= ( )
) =∮
=
( )
2π ( )
2π
Example In the above problem, if the circle shown is 1
=?
∮ Solution
2π
∮
Example ∮
(
)(
)
=? If circle C is as shown as
Solution ∮
(
= ∮
)(
)
( )
=∮ 2π ,
-
2π ,
] = - 2π
Example ∮
(
)(
)
, if the circle is as shown
Solution ∮
-1 (
)(
1
)
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 180
Chapter 6
Mathematics
Example ∮
(
)(
=?
)
where C is circle |z| = 3
Solution c
0
(
)(
Hence ∮
(
2π ,
)(
π(2)
2π (
1 , [ note: 3>2, 3>1] inside circle
)
∮
)
c
π(2) ]
(
∮
)
2π ,
π( )
(
c
)
π( ) ]
) - 2π ( – 1) = 4π
Example ∮
(
)
where C is circle |z| = 3 Solution f(z) = ∮
(
at a = 1,
( ) )
(a) =
∮
(
( ) )
Comparing n = 3 ( )
2π =
(
= )
(-1) 1
3
= π {Since f(z) = f ’( )
2
f ”( ) f ”’( ) f ”’(-1) =
}
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 181
Chapter 6
Mathematics
Example
i =?
∮
around a rectangle with vertices 2
1
-2
, 2
Solution
2
-i =
∮
∮(
)c
π
=
∮
=
,2π c (π )-
∮
= [ 2π -
,2π c (π (
)-
[ 2π -
’ h If f(z) is continuous in a region and ∫ ( ) analytic in that region. T
= 0 around every simple closed C then f(z) is
’ S
If f(z) is analytic inside a circle C with centre at a then for z inside C f(z) = f(a) + f ’( ) ( f(z) = ∑ Where
( =
a) +
( )
(z-a) + - - - - - - -
) ∫
() )
(
Other form, put z = a + h f(a+h) = f(a) + h f ’( )
L
f ”( )
-------
’ S
If f(z) in analytic in the ring shaped region R bounded by two concentric circles c and c of radii and ( ) and with centre at a then for all z in R (
f(z) = where,
=
) ∫
(
(
)
(
)
(
)
() )
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 182
Chapter 6
If f(z) is analytic inside the curve then
Mathematics
L
c
T
’
’ of Analytic Function The value of z for which f(z) = 0 I ( )
c (
f(z) = =∑
(
Where, if
=
h )
h
h
(
p
h
)
(
T
’ h
)
) =
=
( )
=------
= 0, then f(z) is said to have a zero of order n at z =a.
Singularities Of An Analytic Function “
p
”
c
h p
wh ch h
c
c
c
1.
Isolated Singularity If z =a is a singularity of f(z) such that f(z) is analytic at each point in its neighborhood (i.e. there exists a circle with centre a which has no other singularity 1, then z =a is called an isolated singularity).
2.
Removable Singularity If all the negative powers of (z-a) in Laurent series are zero then f(z) = ∑ ( ) Singularity can be removed by defining f(z) at z = a is such a way that it becomes analytic at z =a ( ) exists finitely, then z = a is a removable singularity. Example: f(z) = , then z = 0 is a removable singularity.
3.
Essential singularity If the numbers of negative power of ( an essential singularity. ( ) does not exist in this case
4.
)
L
’
Poles If all the negative power of (z ) L ’ singularity at z = a is called a pole of order n. p c “ p p ”
, h
c
are missing then, the
Example F
T
’
p
( )
(
)
about the point z = i
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 183
Chapter 6
Mathematics
Solution To expand about z = i, i.e. power of z +i, put t =z + i, ( )
=(
)
.
=(
)
.
∑
= .
(
)
(
) 0
1
/ (
(
)
(
( (
)
) )
)
(
/
)
/
Residue Theorem If f(z) is analytic in and on a closed curve C except at a finite number of singular point within C then ∫ ( )
2π (
h
h
p
w h
C)
Calculation of Residues 1. If f(x) has a simple pole at z=a , then Res f(a) =
,( ( ) ( )
2. If ( )
) ( )-
where ( )
(
) ( ), ( )
( ) ( )
Res ( )
C
3. If ( ) has a pole of order n at z=a , then ( )
(
)
2
,(
( )-3
)
Here n =order of singularity Example Solve using Residue Theorem ∮
(
)(
w h
C+
, if the circle is as shown
)
Solution ∮
(
)(
)
2π *R 2π R 2π
p
-1
(-1) ,(
) ( )-
2π
,(
)
(
)(
)
- = 2π
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 184
Chapter 6
Mathematics
Note C
p
h
R
h
C
ch ’
Basic concepts of complex number if z = x + iy r =√
z r
y x
In trigonometric from C
c
S c (c
I
)
Modulus of complex number |z| = r .√ Argument of complex Number = tan
1
=. /
In exponential form ,
c
+
=1 [ |
] |
√c
=1 |z| = r =√
If the any pole is outside the closed contour
|z| = a
|z| = a Its residue at this pole is always zero. THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 185
Chapter 6
Mathematics
We always find the residue at the poles, where poles are inside the closed contour, and for any outside point residue is zero. Cube root of unity ω
√
(
)
(1, 0)
√
ω2 = (
)
Point to remember 1. ω 2. 3.
ω
ω2 = 0 , ω, ω2
THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,
[email protected] © Copyright reserved. Web: www.thegateacademy.com Page 186
