Complex Problems

Complex variable solved problems Pavel Pyrih 11:03 May 29, 2012 ( public domain )

Contents 1 Residue theorem problems

2

2 Zero Sum theorem for residues problems 3 Power series problems

76 157

Acknowledgement. The following problems were solved using my own procedure in a program Maple V, release 5. All possible errors are my faults.

1

1

Residue theorem problems

We will solve several problems using the following theorem: Theorem. (Residue theorem) Suppose U is a simply connected open subset of the complex plane, and w1 , . . . ,wn are finitely many points of U and f is a function which is defined and holomorphic on U \ {w1 , . . . , wn }. If ϕ is a simply closed curve in U contaning the points wk in the interior, then I f (z) dz = 2πi ϕ

k X

res (f, wk ) .

k=1

The following rules can be used for residue counting: Theorem. (Rule 1) If f has a pole of order k at the point w then res (f, w) =


(k−1) 1 lim (z − w)k f (z) . (k − 1)! z→w

Theorem. (Rule 2) If f, g are holomorphic at the point w and f (w) 6= 0. If g(w) = 0, g 0 (w) 6= 0, then   f f (w) res ,w = 0 . g g (w)

Theorem. (Rule 3) If h is holomorphic at w and g has a simple pole at w, then res (gh, w) = h(w) res (g, w).

2

We will use special formulas for special types of problems: Theorem. ( TYPE I. Integral from a rational function in sin and cos.) If Q(a, b) is a rational function of two complex variables such that for real a, b, a2 + b2 = 1 is Q(a, b) finite, then the function   z + 1/z z − 1/z T (z) := Q , /(iz) 2 2i is rational, has no poles on the real line and Z

2π

Q(cos t, sin t) dt = 2πi · 0

X |a| 1 and Q has no real roots. P Then for the rational function f = Q holds Z

+∞

f (x) dx = 2πi −∞

X

res (f, wk ) ,

k

where all singularities of f with a positive imaginary part are considered in the above sum.

Example Z

+∞

−∞

1 dx 1 + x2

4

Theorem. ( TYPE III. Integral from a rational function multiplied by cos or sin ) If Q is a rational function such that has no pole at the real line and for z → ∞ is Q(z) = O(z −1 ). For b > 0 denote f (z) = Q(z) eibz . Then ! Z +∞ X Q(x) cos(bx) dx = Re 2πi · res (f, w) −∞

Z

w

!

+∞

Q(x) sin(bx) dx = Im

2πi ·

−∞

X

res (f, w)

w

where only w with a positive imaginary part are considered in the above sums.

Example Z

+∞

−∞

cos(x) dx 1 + x2

5

1.1

Problem.

Using the Residue theorem evaluate Z 2π

cos(x)2 dx 13 + 12 cos(x)

0

Hint. Type I Solution. We denote 1 1 1 2 I( z+ ) 2 2 z f(z) = − 1 (13 + 6 z + 6 ) z z We find singularities [{z = 0}, {z =

−3 −2 }, {z = }] 2 3

The singularity z=0 is in our region and we will add the following residue res(0, f(z)) =

13 I 144

The singularity −3 2 will be skipped because the singularity is not in our region. The singularity z=

−2 3 is in our region and we will add the following residue z=

res(

169 −2 , f(z)) = − I 3 720

Our sum is 2I π(

X

res(z, f(z))) =

13 π 45

The solution is Z 0

2π

13 cos(x)2 dx = π 13 + 12 cos(x) 45 6

We can try to solve it using real calculus and obtain the result Z 2π cos(x)2 13 dx = π 13 + 12 cos(x) 45 0 Info. not given Comment. no comment

7

1.2

Problem.

Using the Residue theorem evaluate Z 2π 0

cos(x)4 dx 1 + sin(x)2

Hint. Type I Solution. We denote 1 1 1 4 I( z+ ) 2 2 z f(z) = − 1 1 (1 − (z − )2 ) z 4 z We find singularities √ √ √ √ [{z = 0}, {z = 2+1}, {z = 1− 2}, {z = 2−1}, {z = −1− 2}, {z = ∞}, {z = −∞}] The singularity z=0 is in our region and we will add the following residue res(0, f(z)) =

5 I 2

The singularity z=

√

2+1

will be skipped because the singularity is not in our region. The singularity √ z =1− 2 is in our region and we will add the following residue √ √ −768 I 2 + 1088 I √ res(1 − 2, f(z)) = 768 − 544 2 The singularity √ z = 2−1 is in our region and we will add the following residue √ √ −768 I 2 + 1088 I √ res( 2 − 1, f(z)) = 768 − 544 2 The singularity √ z = −1 − 2 8

will be skipped because the singularity is not in our region. The singularity z=∞ will be skipped because only residues at finite singularities are counted. The singularity z = −∞ will be skipped because only residues at finite singularities are counted. Our sum is √ X −768 I 2 + 1088 I 5 √ ) 2I π( res(z, f(z))) = 2 I π ( I + 2 2 768 − 544 2 The solution is Z 2π 0

√ cos(x)4 5 −768 I 2 + 1088 I √ dx = 2 I π ( I + 2 ) 1 + sin(x)2 2 768 − 544 2

We can try to solve it using real calculus and obtain the result √ Z 2π cos(x)4 π (4 2 − 5) √ dx = √ 1 + sin(x)2 ( 2 + 1) ( 2 − 1) 0 Info. not given Comment. no comment

9

1.3

Problem.

Using the Residue theorem evaluate Z 2π 0

sin(x)2

5 − cos(x) 4

dx

Hint. Type I Solution. We denote 1 I (z − )2 1 z f(z) = 1 1 4 5 1 ( − z− )z 4 2 2 z We find singularities [{z = 0}, {z =

1 }, {z = 2}] 2

The singularity z=0 is in our region and we will add the following residue 5 res(0, f(z)) = − I 4 The singularity z=

1 2

is in our region and we will add the following residue 1 3 res( , f(z)) = I 2 4 The singularity z=2 will be skipped because the singularity is not in our region. Our sum is 2I π(

X

res(z, f(z))) = π

10

The solution is Z 0

2π

sin(x)2 5 − cos(x) 4

dx = π

We can try to solve it using real calculus and obtain the result Z 2π sin(x)2 dx = π 5 0 − cos(x) 4 Info. not given Comment. no comment

11

1.4

Problem.

Using the Residue theorem evaluate Z 2π 0

1 dx sin(x)2 + 4 cos(x)2

Hint. Type I Solution. We denote f(z) = −

I 1 1 2 1 1 1 2 (− (z − ) + 4 ( z + ) )z 4 z 2 2 z

We find singularities √ √ 1 √ 1 √ [{z = − I 3}, {z = I 3}, {z = I 3}, {z = −I 3}] 3 3 The singularity 1 √ z=− I 3 3 is in our region and we will add the following residue 1 √ 1 res(− I 3, f(z)) = − I 3 4 The singularity 1 √ I 3 3 is in our region and we will add the following residue z=

1 √ 1 res( I 3, f(z)) = − I 3 4 The singularity z=I

√

3

will be skipped because the singularity is not in our region. The singularity √ z = −I 3 will be skipped because the singularity is not in our region. Our sum is 2I π(

X

res(z, f(z))) = π

12

The solution is Z 0

2π

1 dx = π sin(x)2 + 4 cos(x)2

We can try to solve it using real calculus and obtain the result Z 2π 1 dx = π 2 sin(x) + 4 cos(x)2 0 Info. not given Comment. no comment

13

1.5

Problem.

Using the Residue theorem evaluate Z 2π 0

1 dx 13 + 12 sin(x)

Hint. Type I Solution. We denote f(z) = −

I 1 (13 − 6 I (z − )) z z

We find singularities 3 2 [{z = − I}, {z = − I}] 3 2 The singularity 2 z=− I 3 is in our region and we will add the following residue 2 1 res(− I, f(z)) = − I 3 5 The singularity 3 z=− I 2 will be skipped because the singularity is not in our region. Our sum is X 2 2I π( res(z, f(z))) = π 5 The solution is Z 0

2π

2 1 dx = π 13 + 12 sin(x) 5

We can try to solve it using real calculus and obtain the result Z 2π 1 2 dx = π 13 + 12 sin(x) 5 0 Info. not given 14

Comment. no comment

15

1.6

Problem.

Using the Residue theorem evaluate Z 2π

1 dx 2 + cos(x)

0

Hint. Type I Solution. We denote f(z) = −

I 1 1 1 )z (2 + z + 2 2 z

We find singularities [{z = −2 +

√

3}, {z = −2 −

√

3}]

The singularity z = −2 +

√

3

is in our region and we will add the following residue √ 1 √ res(−2 + 3, f(z)) = − I 3 3 The singularity √ z = −2 − 3 will be skipped because the singularity is not in our region. Our sum is X 2 √ 2I π( res(z, f(z))) = π 3 3 The solution is Z 0

2π

2 √ 1 dx = π 3 2 + cos(x) 3

We can try to solve it using real calculus and obtain the result Z 2π 1 2 √ dx = π 3 2 + cos(x) 3 0 Info. not given Comment. no comment 16

1.7

Problem.

Using the Residue theorem evaluate Z 2π 0

1 dx (2 + cos(x))2

Hint. Type I Solution. We denote f(z) = −

I 1 1 2 1 ) z (2 + z + 2 2 z

We find singularities [{z = −2 +

√

3}, {z = −2 −

√

3}]

The singularity z = −2 +

√

3

is in our region and we will add the following residue √ 1 1 2 1 √ √ res(−2 + 3, f(z)) = − I + (− I + I 3) 3 3 3 3 3 The singularity √ z = −2 − 3 will be skipped because the singularity is not in our region. Our sum is 2I π(

X

1 1 2 1 √ √ res(z, f(z))) = 2 I π (− I + (− I + I 3) 3) 3 3 3 3

The solution is Z 2π 0

1 1 2 1 √ √ 1 dx = 2 I π (− I + (− I + I 3) 3) (2 + cos(x))2 3 3 3 3

We can try to solve it using real calculus and obtain the result Z 2π 1 4 √ dx = π 3 2 (2 + cos(x)) 9 0 Info. not given Comment. no comment 17

1.8

Problem.

Using the Residue theorem evaluate Z 2π

1 dx 2 + sin(x)

0

Hint. Type I Solution. We denote f(z) = −

I 1 1 (2 − I (z − )) z 2 z

We find singularities [{z = −2 I + I

√

3}, {z = −2 I − I

√

3}]

The singularity z = −2 I + I

√

3

is in our region and we will add the following residue √ 1 √ res(−2 I + I 3, f(z)) = − I 3 3 The singularity √ z = −2 I − I 3 will be skipped because the singularity is not in our region. Our sum is X 2 √ 2I π( res(z, f(z))) = π 3 3 The solution is Z 0

2π

2 √ 1 dx = π 3 2 + sin(x) 3

We can try to solve it using real calculus and obtain the result Z 2π 1 2 √ dx = π 3 2 + sin(x) 3 0 Info. not given Comment. no comment 18

1.9

Problem.

Using the Residue theorem evaluate Z 2π 0

1 dx 5 + 4 cos(x)

Hint. Type I Solution. We denote f(z) = −

I 1 (5 + 2 z + 2 ) z z

We find singularities [{z = −2}, {z =

−1 }] 2

The singularity z = −2 will be skipped because the singularity is not in our region. The singularity −1 2 is in our region and we will add the following residue z=

res(

−1 1 , f(z)) = − I 2 3

Our sum is X 2 2I π( res(z, f(z))) = π 3 The solution is Z 0

2π

1 2 dx = π 5 + 4 cos(x) 3

We can try to solve it using real calculus and obtain the result Z 2π 1 2 dx = π 5 + 4 cos(x) 3 0 Info. not given Comment. no comment 19

1.10

Problem.

Using the Residue theorem evaluate Z 2π

1 dx 5 + 4 sin(x)

0

Hint. Type I Solution. We denote f(z) = −

I 1 (5 − 2 I (z − )) z z

We find singularities 1 [{z = −2 I}, {z = − I}] 2 The singularity z = −2 I will be skipped because the singularity is not in our region. The singularity 1 z=− I 2 is in our region and we will add the following residue 1 1 res(− I, f(z)) = − I 2 3 Our sum is X 2 2I π( res(z, f(z))) = π 3 The solution is Z 0

2π

1 2 dx = π 5 + 4 sin(x) 3

We can try to solve it using real calculus and obtain the result Z 2π 1 2 dx = π 5 + 4 sin(x) 3 0 Info. not given Comment. no comment 20

1.11

Problem.

Using the Residue theorem evaluate Z 2π 0

cos(x) dx 5 − cos(x) 4

Hint. Type I Solution. We denote 1 1 1 I( z+ ) 2 2 z f(z) = − 5 1 1 1 ( − z− )z 4 2 2 z We find singularities [{z = 0}, {z =

1 }, {z = 2}] 2

The singularity z=0 is in our region and we will add the following residue res(0, f(z)) = I The singularity 1 2 is in our region and we will add the following residue z=

5 1 res( , f(z)) = − I 2 3 The singularity z=2 will be skipped because the singularity is not in our region. Our sum is X 4 2I π( res(z, f(z))) = π 3 The solution is Z 0

2π

cos(x) 4 dx = π 5 3 − cos(x) 4 21

We can try to solve it using real calculus and obtain the result Z 2π cos(x) 4 dx = π 5 3 0 − cos(x) 4 Info. not given Comment. no comment

22

1.12

Problem.

Using the Residue theorem evaluate Z ∞

x2 + 1 dx 4 −∞ x + 1

Hint. Type II Solution. We denote f(z) =

z2 + 1 z4 + 1

We find singularities 1√ 1 √ 1√ 1 √ 1√ 1 √ 1√ 1 √ 2− I 2}, {z = 2+ I 2}, {z = − 2+ I 2}, {z = 2− I 2}] 2 2 2 2 2 2 2 2 The singularity

[{z = −

1√ 1 √ 2− I 2 2 2 will be skipped because the singularity is not in our region. The singularity z=−

1√ 1 √ 2+ I 2 2 2 is in our region and we will add the following residue z=

res(

1 √ 1+I 1√ √ √ 2 + I 2, f(z)) = 2 2 2I 2 − 2 2

The singularity 1 √ 1√ 2+ I 2 2 2 is in our region and we will add the following residue z=−

res(−

1√ 1 √ 1−I √ √ 2 + I 2, f(z)) = 2 2 2I 2 + 2 2

The singularity 1√ 1 √ 2− I 2 2 2 will be skipped because the singularity is not in our region. z=

Our sum is X 2I π( res(z, f(z))) = 2 I π (

2I

23

1+I 1−I √ √ + √ √ ) 2 − 2 2 2I 2 + 2 2

The solution is Z ∞

x2 + 1 1+I 1−I √ + √ √ ) dx = 2 I π ( √ 4 2I 2 − 2 2 2I 2 + 2 2 −∞ x + 1

We can try to solve it using real calculus and obtain the result Z ∞ 2 √ x +1 dx = 2 π 4 −∞ x + 1 Info. not given Comment. no comment

24

1.13

Problem.

Using the Residue theorem evaluate Z ∞

x2 − 1 dx 2 2 −∞ (x + 1)

Hint. Type II Solution. We denote f(z) =

z2 − 1 (z 2 + 1)2

We find singularities [{z = −I}, {z = I}] The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) = 0 Our sum is X 2I π( res(z, f(z))) = 0 The solution is ∞

x2 − 1 dx = 0 2 2 −∞ (x + 1)

Z

We can try to solve it using real calculus and obtain the result Z ∞ x2 − 1 dx = 0 2 2 −∞ (x + 1) Info. not given Comment. no comment 25

1.14

Problem.

Using the Residue theorem evaluate Z ∞ x2 − x + 2 dx 4 2 −∞ x + 10 x + 9 Hint. Type II Solution. We denote f(z) =

z2 − z + 2 z 4 + 10 z 2 + 9

We find singularities [{z = 3 I}, {z = −3 I}, {z = −I}, {z = I}] The singularity z = 3I is in our region and we will add the following residue res(3 I, f(z)) =

7 1 − I 16 48

The singularity z = −3 I will be skipped because the singularity is not in our region. The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) = −

1 1 − I 16 16

Our sum is 2I π(

X

res(z, f(z))) =

26

5 π 12

The solution is ∞

x2 − x + 2 5 dx = π 4 2 12 −∞ x + 10 x + 9

Z

We can try to solve it using real calculus and obtain the result Z ∞ x2 − x + 2 5 dx = π 4 2 12 −∞ x + 10 x + 9 Info. not given Comment. no comment

27

1.15

Problem.

Using the Residue theorem evaluate Z ∞ −∞ (x

2

1 dx + 1) (x2 + 4)2

Hint. Type II Solution. We denote f(z) =

1 (z 2 + 1) (z 2 + 4)2

We find singularities [{z = 2 I}, {z = −I}, {z = I}, {z = −2 I}] The singularity z = 2I is in our region and we will add the following residue res(2 I, f(z)) =

11 I 288

The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) = −

1 I 18

The singularity z = −2 I will be skipped because the singularity is not in our region. Our sum is 2I π(

X

res(z, f(z))) =

28

5 π 144

The solution is Z

∞

−∞

(x2

5 1 dx = π 2 2 + 1) (x + 4) 144

We can try to solve it using real calculus and obtain the result Z ∞ 1 5 dx = π 2 + 1) (x2 + 4)2 (x 144 −∞ Info. not given Comment. no comment

29

1.16

Problem.

Using the Residue theorem evaluate Z ∞ −∞ (x

2

1 dx + 1) (x2 + 9)

Hint. Type II Solution. We denote f(z) =

1 (z 2 + 1) (z 2 + 9)

We find singularities [{z = 3 I}, {z = −3 I}, {z = −I}, {z = I}] The singularity z = 3I is in our region and we will add the following residue res(3 I, f(z)) =

1 I 48

The singularity z = −3 I will be skipped because the singularity is not in our region. The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) = −

1 I 16

Our sum is 2I π(

X

res(z, f(z))) =

30

1 π 12

The solution is Z

∞

−∞

(x2

1 1 dx = π 2 + 1) (x + 9) 12

We can try to solve it using real calculus and obtain the result Z ∞ 1 1 dx = π 2 + 1) (x2 + 9) (x 12 −∞ Info. not given Comment. no comment

31

1.17

Problem.

Using the Residue theorem evaluate Z ∞

1 dx −∞ (x − I) (x − 2 I)

Hint. Type II Solution. We denote f(z) =

1 (z − I) (z − 2 I)

We find singularities [{z = 2 I}, {z = I}] The singularity z = 2I is in our region and we will add the following residue res(2 I, f(z)) = −I The singularity z=I is in our region and we will add the following residue res(I, f(z)) = I Our sum is X 2I π( res(z, f(z))) = 0 The solution is Z

∞

1 dx = 0 (x − I) (x − 2 I) −∞

We can try to solve it using real calculus and obtain the result Z ∞ 1 dx = 0 (x − I) (x − 2 I) −∞ Info. not given Comment. no comment 32

1.18

Problem.

Using the Residue theorem evaluate Z ∞ −∞ (x

1 dx − I) (x − 2 I) (x + 3 I)

Hint. Type II Solution. We denote 1 (z − I) (z − 2 I) (z + 3 I)

f(z) = We find singularities

[{z = 2 I}, {z = −3 I}, {z = I}] The singularity z = 2I is in our region and we will add the following residue res(2 I, f(z)) =

−1 5

The singularity z = −3 I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) =

1 4

Our sum is 2I π(

X

res(z, f(z))) =

1 Iπ 10

The solution is Z

∞

1 1 dx = Iπ (x − I) (x − 2 I) (x + 3 I) 10 −∞

33

We can try to solve it using real calculus and obtain the result Z ∞ 1 1 dx = Iπ 10 −∞ (x − I) (x − 2 I) (x + 3 I) Info. not given Comment. no comment

34

1.19

Problem.

Using the Residue theorem evaluate Z ∞

1 dx 2 −∞ 1 + x

Hint. Type II Solution. We denote f(z) =

1 z2 + 1

We find singularities [{z = −I}, {z = I}] The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue 1 res(I, f(z)) = − I 2 Our sum is 2I π(

X

res(z, f(z))) = π

The solution is Z

∞

1 dx = π 1 + x2 −∞

We can try to solve it using real calculus and obtain the result Z ∞ 1 dx = π 2 −∞ 1 + x Info. not given Comment. no comment 35

1.20

Problem.

Using the Residue theorem evaluate Z ∞ 3 −∞ x

1 dx +1

Hint. Type II Solution. We denote f(z) =

1 z3 + 1

We find singularities [{z = −1}, {z =

1 1 √ 1 1 √ − I 3}, {z = + I 3}] 2 2 2 2

The singularity z = −1 is on the real line and we will add one half of the following residue res(−1, f(z)) =

1 3

The singularity z=

1 1 √ − I 3 2 2

will be skipped because the singularity is not in our region. The singularity z=

1 1 √ + I 3 2 2

is in our region and we will add the following residue 1 1 √ 1 √ res( + I 3, f(z)) = 2 2 2 3I 3 − 3 Our sum is X 1 1 √ 2I π( res(z, f(z))) = 2 I π ( + 2 ) 6 3I 3 − 3 The solution is Z

∞

−∞

x3

1 1 1 √ dx = 2 I π ( + 2 ) +1 6 3I 3 − 3

36

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ 1 1 dx = dx 3 3 −∞ x + 1 −∞ x + 1 Info. not given Comment. no comment

37

1.21

Problem.

Using the Residue theorem evaluate Z ∞ −∞

x6

1 dx +1

Hint. Type II Solution. We denote f(z) =

1 z6 + 1

We find singularities 1 [{z = −I}, {z = I}, {z = 2 q √ 1 {z = − 2 − 2 I 3}] 2 The singularity

q q q √ √ √ 1 1 2 + 2 I 3}, {z = − 2 + 2 I 3}, {z = 2 − 2 I 3}, 2 2

z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue 1 res(I, f(z)) = − I 6 The singularity q √ 1 2 + 2I 3 2 is in our region and we will add the following residue q √ 1 1 16 √ res( 2 + 2 I 3, f(z)) = 2 3 (2 + 2 I 3)(5/2) The singularity q √ 1 z=− 2 + 2I 3 2 will be skipped because the singularity is not in our region. The singularity q √ 1 z= 2 − 2I 3 2 z=

38

will be skipped because the singularity is not in our region. The singularity q √ 1 z=− 2 − 2I 3 2 is in our region and we will add the following residue q √ 1 16 1 √ res(− 2 − 2 I 3, f(z)) = − 2 3 (2 − 2 I 3)(5/2) Our sum is 2I π(

X

1 16 16 1 1 √ √ res(z, f(z))) = 2 I π (− I+ − ) 6 3 (2 + 2 I 3)(5/2) 3 (2 − 2 I 3)(5/2)

The solution is Z ∞ 1 1 16 1 16 1 √ √ dx = 2 I π (− I + − ) 6+1 (5/2) x 6 3 3 (2 + 2 I 3) (2 − 2 I 3)(5/2) −∞ We can try to solve it using real calculus and obtain the result Z ∞ 1 2 dx = π 6+1 x 3 −∞ Info. not given Comment. no comment

39

1.22

Problem.

Using the Residue theorem evaluate Z ∞ −∞

(x2

x dx + 4 x + 13)2

Hint. Type II Solution. We denote f(z) =

(z 2

z + 4 z + 13)2

We find singularities [{z = −2 + 3 I}, {z = −2 − 3 I}] The singularity z = −2 + 3 I is in our region and we will add the following residue res(−2 + 3 I, f(z)) =

1 I 54

The singularity z = −2 − 3 I will be skipped because the singularity is not in our region. Our sum is X 1 2I π( res(z, f(z))) = − π 27 The solution is Z

∞

x 1 dx = − π 2 + 4 x + 13)2 (x 27 −∞

We can try to solve it using real calculus and obtain the result Z ∞ x 1 dx = − π 2 2 27 −∞ (x + 4 x + 13) Info. not given Comment. no comment 40

1.23

Problem.

Using the Residue theorem evaluate Z ∞ −∞

(x2

x2 dx + 1) (x2 + 9)

Hint. Type II Solution. We denote f(z) =

z2 (z 2 + 1) (z 2 + 9)

We find singularities [{z = 3 I}, {z = −3 I}, {z = −I}, {z = I}] The singularity z = 3I is in our region and we will add the following residue res(3 I, f(z)) = −

3 I 16

The singularity z = −3 I will be skipped because the singularity is not in our region. The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) =

1 I 16

Our sum is X 1 2I π( res(z, f(z))) = π 4

41

The solution is ∞

x2 1 dx = π 2 2 4 −∞ (x + 1) (x + 9)

Z

We can try to solve it using real calculus and obtain the result Z ∞ x2 1 dx = π 2 2 4 −∞ (x + 1) (x + 9) Info. not given Comment. no comment

42

1.24

Problem.

Using the Residue theorem evaluate Z ∞ −∞

(x2

x2 dx + 1)3

Hint. Type II Solution. We denote f(z) =

z2 (z 2 + 1)3

We find singularities [{z = −I}, {z = I}] The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) = −

1 I 16

Our sum is X 1 2I π( res(z, f(z))) = π 8 The solution is ∞

1 x2 dx = π 2 3 8 −∞ (x + 1)

Z

We can try to solve it using real calculus and obtain the result Z ∞ x2 1 dx = π 2 3 8 −∞ (x + 1) Info. not given Comment. no comment 43

1.25

Problem.

Using the Residue theorem evaluate Z ∞ −∞

(x2

x2 dx + 4)2

Hint. Type II Solution. We denote f(z) =

z2 (z 2 + 4)2

We find singularities [{z = 2 I}, {z = −2 I}] The singularity z = 2I is in our region and we will add the following residue 1 res(2 I, f(z)) = − I 8 The singularity z = −2 I will be skipped because the singularity is not in our region. Our sum is X 1 2I π( res(z, f(z))) = π 4 The solution is ∞

x2 1 dx = π 2 2 4 −∞ (x + 4)

Z

We can try to solve it using real calculus and obtain the result Z ∞ x2 1 dx = π 2 2 4 −∞ (x + 4) Info. not given Comment. no comment 44

1.26

Problem.

Using the Residue theorem evaluate Z ∞ 3 (x + 5 x) e(I x) dx 4 2 −∞ x + 10 x + 9 Hint. Type III Solution. We denote f(z) =

(z 3 + 5 z) e(I z) z 4 + 10 z 2 + 9

We find singularities [{z = 3 I}, {z = −3 I}, {z = −I}, {z = I}] The singularity z = 3I is in our region and we will add the following residue 1 (−3) e 4

res(3 I, f(z)) = The singularity z = −3 I

will be skipped because the singularity is not in our region. The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) =

1 (−1) e 4

Our sum is 2I π(

X

1 1 res(z, f(z))) = 2 I π ( e(−3) + e(−1) ) 4 4

45

The solution is Z

∞

(x3 + 5 x) e(I x) 1 1 dx = 2 I π ( e(−3) + e(−1) ) 4 + 10 x2 + 9 x 4 4 −∞

We can try to solve it using real calculus and obtain the result Z ∞ 3 1 (x + 5 x) e(I x) 1 1 1 dx = I π cosh(1)+ I π cosh(3)− I π sinh(3)− I π sinh(1) 4 + 10 x2 + 9 x 2 2 2 2 −∞ Info. not given Comment. no comment

46

1.27

Problem.

Using the Residue theorem evaluate Z ∞ (I x) 2 e (x − 1) dx 2 + 1) x (x −∞ Hint. Type III Solution. We denote f(z) =

e(I z) (z 2 − 1) z (z 2 + 1)

We find singularities [{z = 0}, {z = −I}, {z = I}] The singularity z=0 is on the real line and we will add one half of the following residue res(0, f(z)) = −1 The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) = e(−1) Our sum is 2I π(

X

1 res(z, f(z))) = 2 I π (− + e(−1) ) 2

The solution is ∞

e(I x) (x2 − 1) 1 dx = 2 I π (− + e(−1) ) 2 2 −∞ x (x + 1)

Z

We can try to solve it using real calculus and obtain the result Z ∞ (I x) 2 Z ∞ e (x − 1) (cos(x) + I sin(x)) (x2 − 1) dx = dx 2 x (x2 + 1) −∞ x (x + 1) −∞ 47

Info. not given Comment. no comment

48

1.28

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 2 −∞ (x + 4) (x − 1)

Hint. Type III Solution. We denote f(z) =

(z 2

e(I z) + 4) (z − 1)

We find singularities [{z = 2 I}, {z = −2 I}, {z = 1}] The singularity z = 2I is in our region and we will add the following residue res(2 I, f(z)) = (−

1 1 + I) e(−2) 10 20

The singularity z = −2 I will be skipped because the singularity is not in our region. The singularity z=1 is on the real line and we will add one half of the following residue res(1, f(z)) =

1 I e 5

Our sum is 2I π(

X

res(z, f(z))) = 2 I π ((−

The solution is Z ∞

1 1 1 I + I) e(−2) + e ) 10 20 10

e(I x) 1 1 1 I dx = 2 I π ((− + I) e(−2) + e ) 2 + 4) (x − 1) (x 10 20 10 −∞

49

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ e(I x) cos(x) + I sin(x) dx = dx 2 + 4) (x − 1) 2 (x −∞ −∞ (x + 4) (x − 1) Info. not given Comment. no comment

50

1.29

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 2 −∞ x (x + 1)

Hint. Type III Solution. We denote f(z) =

e(I z) z (z 2 + 1)

We find singularities [{z = 0}, {z = −I}, {z = I}] The singularity z=0 is on the real line and we will add one half of the following residue res(0, f(z)) = 1 The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue 1 res(I, f(z)) = − e(−1) 2 Our sum is 2I π(

X

1 1 res(z, f(z))) = 2 I π ( − e(−1) ) 2 2

The solution is ∞

e(I x) 1 1 dx = 2 I π ( − e(−1) ) 2 2 2 −∞ x (x + 1)

Z

51

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ e(I x) cos(x) + I sin(x) dx = dx 2 + 1) x (x x (x2 + 1) −∞ −∞ Info. not given Comment. no comment

52

1.30

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 2 2 −∞ x (x + 9)

Hint. Type III Solution. We denote f(z) =

e(I z) z (z 2 + 9)2

We find singularities [{z = 0}, {z = 3 I}, {z = −3 I}] The singularity z=0 is on the real line and we will add one half of the following residue res(0, f(z)) =

1 81

The singularity z = 3I is in our region and we will add the following residue res(3 I, f(z)) = −

5 (−3) e 324

The singularity z = −3 I will be skipped because the singularity is not in our region. Our sum is X 5 (−3) 1 − e ) 2I π( res(z, f(z))) = 2 I π ( 162 324 The solution is ∞

5 (−3) e(I x) 1 − e ) dx = 2 I π ( 2 2 162 324 −∞ x (x + 9)

Z

53

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ e(I x) cos(x) + I sin(x) dx = dx 2 + 9)2 x (x x (x2 + 9)2 −∞ −∞ Info. not given Comment. no comment

54

1.31

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 2 −∞ x (x − 2 x + 2)

Hint. Type III Solution. We denote f(z) =

z

(z 2

e(I z) − 2 z + 2)

We find singularities [{z = 0}, {z = 1 + I}, {z = 1 − I}] The singularity z=0 is on the real line and we will add one half of the following residue res(0, f(z)) =

1 2

The singularity z =1+I is in our region and we will add the following residue 1 1 res(1 + I, f(z)) = (− − I) eI e(−1) 4 4 The singularity z =1−I will be skipped because the singularity is not in our region. Our sum is 2I π(

X

1 1 1 res(z, f(z))) = 2 I π ( + (− − I) eI e(−1) ) 4 4 4

The solution is Z ∞

1 1 1 e(I x) dx = 2 I π ( + (− − I) eI e(−1) ) 2 4 4 4 −∞ x (x − 2 x + 2)

55

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ e(I x) cos(x) + I sin(x) dx = dx 2 − 2 x + 2) 2 x (x −∞ −∞ x (x − 2 x + 2) Info. not given Comment. no comment

56

1.32

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 2 −∞ x + 1

Hint. Type III Solution. We denote f(z) =

e(I z) z2 + 1

We find singularities [{z = −I}, {z = I}] The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue 1 res(I, f(z)) = − I e(−1) 2 Our sum is 2I π(

X

res(z, f(z))) = π e(−1)

The solution is ∞

e(I x) dx = π e(−1) 2 −∞ x + 1

Z

We can try to solve it using real calculus and obtain the result Z ∞ (I x) e dx = −π sinh(1) + π cosh(1) 2 −∞ x + 1 Info. not given Comment. no comment 57

1.33

Problem.

Using the Residue theorem evaluate Z ∞ −∞

x2

e(I x) dx + 4 x + 20

Hint. Type III Solution. We denote f(z) =

e(I z) z 2 + 4 z + 20

We find singularities [{z = −2 − 4 I}, {z = −2 + 4 I}] The singularity z = −2 − 4 I will be skipped because the singularity is not in our region. The singularity z = −2 + 4 I is in our region and we will add the following residue res(−2 + 4 I, f(z)) = −

1 I e(−4) 8 (eI )2

Our sum is 2I π(

X

res(z, f(z))) =

1 π e(−4) 4 (eI )2

The solution is Z

∞

−∞

x2

1 π e(−4) e(I x) dx = + 4 x + 20 4 (eI )2

We can try to solve it using real calculus and obtain the result Z ∞ e(I x) 1 1 dx = − I π sin(2 − 4 I) + π cos(2 − 4 I) 2 + 4 x + 20 x 4 4 −∞ Info. not given Comment. no comment 58

1.34

Problem.

Using the Residue theorem evaluate Z ∞ −∞

x2

e(I x) dx − 5x + 6

Hint. Type III Solution. We denote f(z) =

e(I z) z2 − 5 z + 6

We find singularities [{z = 2}, {z = 3}] The singularity z=2 is on the real line and we will add one half of the following residue res(2, f(z)) = −(eI )2 The singularity z=3 is on the real line and we will add one half of the following residue res(3, f(z)) = (eI )3 Our sum is X 1 1 2I π( res(z, f(z))) = 2 I π (− (eI )2 + (eI )3 ) 2 2 The solution is ∞

e(I x) 1 1 dx = 2 I π (− (eI )2 + (eI )3 ) 2 − 5x + 6 x 2 2 −∞

Z

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ cos(x) + I sin(x) e(I x) dx = dx 2 x2 − 5 x + 6 −∞ x − 5 x + 6 −∞ Info. not given Comment. no comment 59

1.35

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 3 −∞ x + 1

Hint. Type III Solution. We denote f(z) =

e(I z) z3 + 1

We find singularities [{z = −1}, {z =

1 1 √ 1 1 √ − I 3}, {z = + I 3}] 2 2 2 2

The singularity z = −1 is on the real line and we will add one half of the following residue res(−1, f(z)) =

1 1 3 eI

The singularity 1 1 √ − I 3 2 2 will be skipped because the singularity is not in our region. The singularity z=

1 1 √ + I 3 2 2 is in our region and we will add the following residue z=

1

1 1 √ (−1)(1/2 π ) p √ √ p √ res( + I 3, f(z)) = 2 2 2 3 I e( 3) 3 − 3 e( 3) Our sum is 1

X 2I π( res(z, f(z))) = 2 I π The solution is Z ∞ (I x) e dx = 2 I π 3+1 x −∞

1 1 (−1)(1/2 π ) p p √ + 2 √ √ 6 eI 3 I e( 3) 3 − 3 e( 3)

1

1 1 (−1)(1/2 π ) p √ √ p √ +2 I 6 e 3 I e( 3) 3 − 3 e( 3) 60

!

!

We can try to solve it using real calculus and obtain the result Z ∞ (I x) Z ∞ e cos(x) + I sin(x) dx = dx 3+1 x x3 + 1 −∞ −∞ Info. not given Comment. no comment

61

1.36

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 4 −∞ x − 1

Hint. Type III Solution. We denote f(z) =

e(I z) z4 − 1

We find singularities [{z = −1}, {z = −I}, {z = I}, {z = 1}] The singularity z = −1 is on the real line and we will add one half of the following residue res(−1, f(z)) = −

1 1 4 eI

The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) =

1 (−1) Ie 4

The singularity z=1 is on the real line and we will add one half of the following residue res(1, f(z)) =

1 I e 4

Our sum is X 1 1 1 1 2I π( res(z, f(z))) = 2 I π (− I + I e(−1) + eI ) 8 e 4 8

62

The solution is ∞

e(I x) 1 1 1 1 dx = 2 I π (− I + I e(−1) + eI ) 4−1 x 8 e 4 8 −∞

Z

We can try to solve it using real calculus and obtain the result Z ∞ (I x) Z ∞ cos(x) + I sin(x) e dx = dx 4−1 x x4 − 1 −∞ −∞ Info. not given Comment. no comment

63

1.37

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx −∞ x

Hint. Type III Solution. We denote f(z) =

e(I z) z

We find singularities [{z = 0}] The singularity z=0 is on the real line and we will add one half of the following residue res(0, f(z)) = 1 Our sum is 2I π(

X

res(z, f(z))) = I π

The solution is ∞

e(I x) dx = I π −∞ x

Z

We can try to solve it using real calculus and obtain the result Z ∞ (I x) Z ∞ e cos(x) + I sin(x) dx = dx x −∞ x −∞ Info. not given Comment. no comment

64

1.38

Problem.

Using the Residue theorem evaluate Z ∞

x e(I x) dx 4 −∞ 1 − x

Hint. Type III Solution. We denote f(z) =

z e(I z) 1 − z4

We find singularities [{z = −1}, {z = −I}, {z = I}, {z = 1}] The singularity z = −1 is on the real line and we will add one half of the following residue res(−1, f(z)) = −

1 1 4 eI

The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue res(I, f(z)) =

1 (−1) e 4

The singularity z=1 is on the real line and we will add one half of the following residue 1 res(1, f(z)) = − eI 4 Our sum is X 1 1 1 1 2I π( res(z, f(z))) = 2 I π (− I + e(−1) − eI ) 8 e 4 8

65

The solution is ∞

x e(I x) 1 1 1 1 dx = 2 I π (− I + e(−1) − eI ) 4 1 − x 8 e 4 8 −∞

Z

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ x (cos(x) + I sin(x)) x e(I x) dx = − dx 4 1 − x −1 + x4 −∞ −∞ Info. not given Comment. no comment

66

1.39

Problem.

Using the Residue theorem evaluate Z ∞

x e(I x) dx 2 −∞ x + 4 x + 20

Hint. Type III Solution. We denote f(z) =

z2

z e(I z) + 4 z + 20

We find singularities [{z = −2 − 4 I}, {z = −2 + 4 I}] The singularity z = −2 − 4 I will be skipped because the singularity is not in our region. The singularity z = −2 + 4 I is in our region and we will add the following residue res(−2 + 4 I, f(z)) = −

1 I (4 I e(−4) − 2 e(−4) ) 8 (eI )2

Our sum is 2I π(

X

res(z, f(z))) =

1 π (4 I e(−4) − 2 e(−4) ) 4 (eI )2

The solution is Z

∞

−∞

x2

x e(I x) 1 π (4 I e(−4) − 2 e(−4) ) dx = + 4 x + 20 4 (eI )2

We can try to solve it using real calculus and obtain the result Z ∞ x e(I x) dx = 2 −∞ x + 4 x + 20 1 1 π sin(2 − 4 I) + I π cos(2 − 4 I) + I π sin(2 − 4 I) − π cos(2 − 4 I) 2 2

67

Info. not given Comment. no comment

68

1.40

Problem.

Using the Residue theorem evaluate Z ∞

x e(I x) dx 2 −∞ x + 9

Hint. Type III Solution. We denote f(z) =

z e(I z) z2 + 9

We find singularities [{z = 3 I}, {z = −3 I}] The singularity z = 3I is in our region and we will add the following residue res(3 I, f(z)) =

1 (−3) e 2

The singularity z = −3 I will be skipped because the singularity is not in our region. Our sum is 2I π(

X

res(z, f(z))) = I π e(−3)

The solution is ∞

x e(I x) dx = I π e(−3) 2 −∞ x + 9

Z

We can try to solve it using real calculus and obtain the result Z ∞ x e(I x) dx = I π cosh(3) − I π sinh(3) 2 −∞ x + 9 Info. not given Comment. no comment 69

1.41

Problem.

Using the Residue theorem evaluate Z ∞ −∞

x2

x e(I x) dx − 2 x + 10

Hint. Type III Solution. We denote f(z) =

z e(I z) z 2 − 2 z + 10

We find singularities [{z = 1 − 3 I}, {z = 1 + 3 I}] The singularity z = 1 − 3I will be skipped because the singularity is not in our region. The singularity z = 1 + 3I is in our region and we will add the following residue 1 res(1 + 3 I, f(z)) = − I (3 I eI e(−3) + eI e(−3) ) 6 Our sum is 2I π( The solution is Z

X

res(z, f(z))) =

1 π (3 I eI e(−3) + eI e(−3) ) 3

∞

x e(I x) 1 dx = π (3 I eI e(−3) + eI e(−3) ) 2 3 −∞ x − 2 x + 10

We can try to solve it using real calculus and obtain the result Z ∞ x e(I x) dx = 2 −∞ x − 2 x + 10 1 1 −π sin(1 + 3 I) + I π cos(1 + 3 I) + I π sin(1 + 3 I) + π cos(1 + 3 I) 3 3 Info. not given 70

Comment. no comment

71

1.42

Problem.

Using the Residue theorem evaluate Z ∞ −∞

x2

x e(I x) dx − 5x + 6

Hint. Type III Solution. We denote f(z) =

z e(I z) z2 − 5 z + 6

We find singularities [{z = 2}, {z = 3}] The singularity z=2 is on the real line and we will add one half of the following residue res(2, f(z)) = −2 (eI )2 The singularity z=3 is on the real line and we will add one half of the following residue res(3, f(z)) = 3 (eI )3 Our sum is X 3 2I π( res(z, f(z))) = 2 I π (−(eI )2 + (eI )3 ) 2 The solution is ∞

x e(I x) 3 dx = 2 I π (−(eI )2 + (eI )3 ) 2 − 5x + 6 x 2 −∞

Z

We can try to solve it using real calculus and obtain the result Z ∞ Z ∞ x (cos(x) + I sin(x)) x e(I x) dx = dx 2 x2 − 5 x + 6 −∞ x − 5 x + 6 −∞ Info. not given Comment. no comment 72

1.43

Problem.

Using the Residue theorem evaluate Z ∞

e(I x) dx 2 −∞ x

Hint. Type III Solution. We denote f(z) =

e(I z) z2

We find singularities [{z = 0}] The singularity z=0 is on the real line and is not a simple pole, we cannot count the integral with the residue theorem ... Our sum is X 2I π( res(z, f(z))) = 2 I π ∞ The solution is ∞

e(I x) dx = 2 I π ∞ 2 −∞ x

Z

We can try to solve it using real calculus and obtain the result Z ∞ (I x) e dx = ∞ 2 −∞ x Info. not given Comment. no comment

73

1.44

Problem.

Using the Residue theorem evaluate Z ∞

x3 e(I x) dx 4 2 −∞ x + 5 x + 4

Hint. Type III Solution. We denote f(z) =

z4

z 3 e(I z) + 5 z2 + 4

We find singularities [{z = 2 I}, {z = −I}, {z = I}, {z = −2 I}] The singularity z = 2I is in our region and we will add the following residue res(2 I, f(z)) =

2 (−2) e 3

The singularity z = −I will be skipped because the singularity is not in our region. The singularity z=I is in our region and we will add the following residue 1 res(I, f(z)) = − e(−1) 6 The singularity z = −2 I will be skipped because the singularity is not in our region. Our sum is 2I π(

X

1 2 res(z, f(z))) = 2 I π ( e(−2) − e(−1) ) 3 6

74

The solution is Z

∞

−∞

x4

x3 e(I x) 2 1 dx = 2 I π ( e(−2) − e(−1) ) + 5 x2 + 4 3 6

We can try to solve it using real calculus and obtain the result Z ∞ 1 x3 e(I x) 4 4 1 dx = − I π cosh(1)+ I π cosh(2)− I π sinh(2)+ I π sinh(1) 4 2+4 x + 5 x 3 3 3 3 −∞ Info. not given Comment. no comment

75

2

Zero Sum theorem for residues problems

We recall the definition: Notation. For a function f holomorphic on some neighbourhood of infinity we define     1 −1 ·f ,0 . res (f, ∞) = res 2 z z We will solve several problems using the following theorem: Theorem. (Zero Sum theorem for residues) For a function f holomorphic in the extended complex plane C ∪ {∞} with at most finitely many exceptions the sum of residues is zero, i.e. X res (f, w) = 0 . w∈C∪{∞}

Notation.(Example f (z) = 1/z) Projecting on the Riemann sphere we observe the north pole on globe (i.e. ∞) with the residue -1 and at the south pole (the origin) with residue 1. Green is the zero level on the Riemann sphere while the blue gradually turning red is the imaginary part of log z demonstrating that Zero Sum theorem holds for 1/z.

76

2.1

Problem.

Check the Zero Sum theorem for the following function z2 + 1 z4 + 1 Hint. no hint Solution. We denote f(z) =

z2 + 1 z4 + 1

We find singularities [{z =

1√ 1 √ 1√ 1 √ 1√ 1 √ 1√ 1 √ 2+ I 2}, {z = − 2− I 2}, {z = 2− I 2}, {z = − 2+ I 2}] 2 2 2 2 2 2 2 2

The singularity z=

1 √ 1√ 2+ I 2 2 2

adds the following residue res(f(z),

1√ 1 √ 1+I √ √ 2 + I 2) = 2 2 2I 2 − 2 2

The singularity z=−

1 √ 1√ 2− I 2 2 2

adds the following residue res(f(z), −

1√ 1 √ −1 − I √ √ 2 − I 2) = 2 2 2I 2 − 2 2

The singularity z=

1√ 1 √ 2− I 2 2 2

adds the following residue res(f(z),

1√ 1 √ −1 + I √ √ 2 − I 2) = 2 2 2I 2 + 2 2

The singularity z=−

1√ 1 √ 2+ I 2 2 2

77

adds the following residue res(f(z), −

1√ 1 √ 1−I √ √ 2 + I 2) = 2 2 2I 2 + 2 2

At infinity we get the residue res(f(z), ∞) =

1+I −1 − I −1 + I 1−I √ √ + √ √ + √ √ + √ √ 2I 2 − 2 2 2I 2 − 2 2 2I 2 + 2 2 2I 2 + 2 2

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

78

2.2

Problem.

Check the Zero Sum theorem for the following function z2 − 1 (z 2 + 1)2 Hint. no hint Solution. We denote f(z) =

z2 − 1 (z 2 + 1)2

We find singularities [{z = −I}, {z = I}] The singularity z = −I adds the following residue res(f(z), −I) = 0 The singularity z=I adds the following residue res(f(z), I) = 0 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

79

2.3

Problem.

Check the Zero Sum theorem for the following function z2 − z + 2 + 10 z 2 + 9

z4 Hint.

no hint Solution. We denote f(z) =

z2 − z + 2 + 10 z 2 + 9

z4

We find singularities [{z = −3 I}, {z = −I}, {z = I}, {z = 3 I}] The singularity z = −3 I adds the following residue 1 7 + I 16 48

res(f(z), −3 I) = The singularity z = −I adds the following residue res(f(z), −I) = −

1 1 + I 16 16

The singularity z=I adds the following residue res(f(z), I) = −

1 1 − I 16 16

The singularity z = 3I adds the following residue res(f(z), 3 I) =

80

1 7 − I 16 48

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

81

2.4

Problem.

Check the Zero Sum theorem for the following function (z 2

1 + 1) (z 2 + 4)2

Hint. no hint Solution. We denote f(z) =

(z 2

1 + 1) (z 2 + 4)2

We find singularities [{z = −I}, {z = 2 I}, {z = −2 I}, {z = I}] The singularity z = −I adds the following residue 1 I 18

res(f(z), −I) = The singularity z = 2I adds the following residue res(f(z), 2 I) =

11 I 288

The singularity z = −2 I adds the following residue res(f(z), −2 I) = −

11 I 288

The singularity z=I adds the following residue res(f(z), I) = −

82

1 I 18

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

83

2.5

Problem.

Check the Zero Sum theorem for the following function (z 2

1 + 1) (z 2 + 9)

Hint. no hint Solution. We denote f(z) =

(z 2

1 + 1) (z 2 + 9)

We find singularities [{z = −3 I}, {z = −I}, {z = I}, {z = 3 I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) = −

1 I 48

The singularity z = −I adds the following residue res(f(z), −I) =

1 I 16

The singularity z=I adds the following residue res(f(z), I) = −

1 I 16

The singularity z = 3I adds the following residue res(f(z), 3 I) =

84

1 I 48

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

85

2.6

Problem.

Check the Zero Sum theorem for the following function 1 (z − I) (z − 2 I) Hint. no hint Solution. We denote f(z) =

1 (z − I) (z − 2 I)

We find singularities [{z = 2 I}, {z = I}] The singularity z = 2I adds the following residue res(f(z), 2 I) = −I The singularity z=I adds the following residue res(f(z), I) = I At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

86

2.7

Problem.

Check the Zero Sum theorem for the following function 1 (z − I) (z − 2 I) (z + 3 I) Hint. no hint Solution. We denote f(z) =

1 (z − I) (z − 2 I) (z + 3 I)

We find singularities [{z = −3 I}, {z = 2 I}, {z = I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) =

−1 20

The singularity z = 2I adds the following residue res(f(z), 2 I) =

−1 5

The singularity z=I adds the following residue res(f(z), I) =

1 4

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

87

Info. not given Comment. no comment

88

2.8

Problem.

Check the Zero Sum theorem for the following function z2

1 +1

Hint. no hint Solution. We denote f(z) =

1 z2 + 1

We find singularities [{z = −I}, {z = I}] The singularity z = −I adds the following residue res(f(z), −I) =

1 I 2

The singularity z=I adds the following residue 1 res(f(z), I) = − I 2 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

89

2.9

Problem.

Check the Zero Sum theorem for the following function z3

1 +1

Hint. no hint Solution. We denote f(z) =

1 z3 + 1

We find singularities [{z = −1}, {z =

1 1 √ 1 1 √ − I 3}, {z = + I 3}] 2 2 2 2

The singularity z = −1 adds the following residue res(f(z), −1) =

1 3

The singularity z=

1 1 √ − I 3 2 2

adds the following residue res(f(z),

1 1 1 √ √ − I 3) = −2 2 2 3I 3 + 3

The singularity z=

1 1 √ + I 3 2 2

adds the following residue res(f(z),

1 1 √ 1 √ + I 3) = 2 2 2 3I 3 − 3

At infinity we get the residue 1 1 1 √ √ res(f(z), ∞) = − + 2 −2 3 3I 3 + 3 3I 3 − 3

90

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

91

2.10

Problem.

Check the Zero Sum theorem for the following function 1 +1

z6 Hint.

no hint Solution. We denote f(z) =

1 z6 + 1

We find singularities q q q √ √ √ 1 1 1 [{z = −I}, {z = − 2 − 2 I 3}, {z = 2 − 2 I 3}, {z = − 2 + 2 I 3}, 2 2 2 q √ 1 {z = 2 + 2 I 3}, {z = I}] 2 The singularity z = −I adds the following residue 1 I 6

res(f(z), −I) = The singularity 1 z=− 2

q 2 − 2I

√

3

adds the following residue 1 res(f(z), − 2

q 2 − 2I

√

3) = −

16 1 √ 3 (2 − 2 I 3)(5/2)

The singularity 1 z= 2

q 2 − 2I

√

3

adds the following residue 1 res(f(z), 2

q 2 − 2I

√

3) =

92

1 16 √ 3 (2 − 2 I 3)(5/2)

The singularity z=−

1 2

q 2 + 2I

√

3

adds the following residue res(f(z), −

1 2

q 2 + 2I

√

3) = −

16 1 √ 3 (2 + 2 I 3)(5/2)

The singularity 1 z= 2

q 2 + 2I

√

3

adds the following residue 1 res(f(z), 2

q 2 + 2I

√

3) =

16 1 √ 3 (2 + 2 I 3)(5/2)

The singularity z=I adds the following residue 1 res(f(z), I) = − I 6 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

93

2.11

Problem.

Check the Zero Sum theorem for the following function z (z 2 + 4 z + 13)2 Hint. no hint Solution. We denote f(z) =

(z 2

z + 4 z + 13)2

We find singularities [{z = −2 − 3 I}, {z = −2 + 3 I}] The singularity z = −2 − 3 I adds the following residue res(f(z), −2 − 3 I) = −

1 I 54

The singularity z = −2 + 3 I adds the following residue res(f(z), −2 + 3 I) =

1 I 54

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

94

2.12

Problem.

Check the Zero Sum theorem for the following function (z 2

z2 + 1) (z 2 + 9)

Hint. no hint Solution. We denote f(z) =

z2 (z 2 + 1) (z 2 + 9)

We find singularities [{z = −3 I}, {z = −I}, {z = I}, {z = 3 I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) =

3 I 16

The singularity z = −I adds the following residue res(f(z), −I) = −

1 I 16

The singularity z=I adds the following residue res(f(z), I) =

1 I 16

The singularity z = 3I adds the following residue res(f(z), 3 I) = −

95

3 I 16

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

96

2.13

Problem.

Check the Zero Sum theorem for the following function (z 2

z2 + 1)3

Hint. no hint Solution. We denote f(z) =

z2 (z 2 + 1)3

We find singularities [{z = −I}, {z = I}] The singularity z = −I adds the following residue res(f(z), −I) =

1 I 16

The singularity z=I adds the following residue res(f(z), I) = −

1 I 16

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

97

2.14

Problem.

Check the Zero Sum theorem for the following function (z 2

z2 + 4)2

Hint. no hint Solution. We denote f(z) =

z2 (z 2 + 4)2

We find singularities [{z = 2 I}, {z = −2 I}] The singularity z = 2I adds the following residue 1 res(f(z), 2 I) = − I 8 The singularity z = −2 I adds the following residue res(f(z), −2 I) =

1 I 8

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

98

2.15

Problem.

Check the Zero Sum theorem for the following function (z 3 + 5 z) e(I z) z 4 + 10 z 2 + 9 Hint. no hint Solution. We denote f(z) =

(z 3 + 5 z) e(I z) z 4 + 10 z 2 + 9

We find singularities [{z = −3 I}, {z = −I}, {z = I}, {z = 3 I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) =

1 3 e 4

The singularity z = −I adds the following residue res(f(z), −I) =

1 e 4

The singularity z=I adds the following residue res(f(z), I) =

1 (−1) e 4

The singularity z = 3I adds the following residue res(f(z), 3 I) =

99

1 (−3) e 4

At infinity we get the residue 1 1 1 1 res(f(z), ∞) = − e3 − e − e(−1) − e(−3) 4 4 4 4 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

100

2.16

Problem.

Check the Zero Sum theorem for the following function e(I z) (z 2 − 1) z (z 2 + 1) Hint. no hint Solution. We denote f(z) =

e(I z) (z 2 − 1) z (z 2 + 1)

We find singularities [{z = 0}, {z = −I}, {z = I}] The singularity z=0 adds the following residue res(f(z), 0) = −1 The singularity z = −I adds the following residue res(f(z), −I) = e The singularity z=I adds the following residue res(f(z), I) = e(−1) At infinity we get the residue res(f(z), ∞) = 1 − e − e(−1) and finally we obtain the sum X

res(f(z), z) = 0

101

Info. not given Comment. no comment

102

2.17

Problem.

Check the Zero Sum theorem for the following function (z 2

e(I z) + 4) (z − 1)

Hint. no hint Solution. We denote f(z) =

(z 2

e(I z) + 4) (z − 1)

We find singularities [{z = 1}, {z = 2 I}, {z = −2 I}] The singularity z=1 adds the following residue res(f(z), 1) =

1 I e 5

The singularity z = 2I adds the following residue res(f(z), 2 I) = (−

1 1 + I) e(−2) 10 20

The singularity z = −2 I adds the following residue res(f(z), −2 I) = (−

1 1 − I) e2 10 20

At infinity we get the residue 1 1 1 1 1 I) e(−2) + ( + I) e2 res(f(z), ∞) = − eI + ( − 5 10 20 10 20 and finally we obtain the sum X res(f(z), z) = 0

103

Info. not given Comment. no comment

104

2.18

Problem.

Check the Zero Sum theorem for the following function e(I z) z (z 2 + 1) Hint. no hint Solution. We denote f(z) =

e(I z) z (z 2 + 1)

We find singularities [{z = 0}, {z = −I}, {z = I}] The singularity z=0 adds the following residue res(f(z), 0) = 1 The singularity z = −I adds the following residue 1 res(f(z), −I) = − e 2 The singularity z=I adds the following residue 1 res(f(z), I) = − e(−1) 2 At infinity we get the residue res(f(z), ∞) = −1 +

1 1 e + e(−1) 2 2

and finally we obtain the sum X

res(f(z), z) = 0

105

Info. not given Comment. no comment

106

2.19

Problem.

Check the Zero Sum theorem for the following function e(I z) z (z 2 + 9)2 Hint. no hint Solution. We denote f(z) =

e(I z) z (z 2 + 9)2

We find singularities [{z = −3 I}, {z = 0}, {z = 3 I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) =

1 3 e 324

The singularity z=0 adds the following residue res(f(z), 0) =

1 81

The singularity z = 3I adds the following residue res(f(z), 3 I) = −

5 (−3) e 324

At infinity we get the residue res(f(z), ∞) = −

1 5 (−3) 1 3 e − + e 324 81 324

and finally we obtain the sum X

res(f(z), z) = 0

107

Info. not given Comment. no comment

108

2.20

Problem.

Check the Zero Sum theorem for the following function z

(z 2

e(I z) − 2 z + 2)

Hint. no hint Solution. We denote f(z) =

z

(z 2

e(I z) − 2 z + 2)

We find singularities [{z = 0}, {z = 1 − I}, {z = 1 + I}] The singularity z=0 adds the following residue res(f(z), 0) =

1 2

The singularity z =1−I adds the following residue 1 1 res(f(z), 1 − I) = (− + I) eI e 4 4 The singularity z =1+I adds the following residue 1 1 res(f(z), 1 + I) = (− − I) eI e(−1) 4 4 At infinity we get the residue 1 1 1 1 1 res(f(z), ∞) = − + ( − I) eI e + ( + I) eI e(−1) 2 4 4 4 4 and finally we obtain the sum X res(f(z), z) = 0

109

Info. not given Comment. no comment

110

2.21

Problem.

Check the Zero Sum theorem for the following function e(I z) z2 + 1 Hint. no hint Solution. We denote f(z) =

e(I z) z2 + 1

We find singularities [{z = −I}, {z = I}] The singularity z = −I adds the following residue res(f(z), −I) =

1 Ie 2

The singularity z=I adds the following residue 1 res(f(z), I) = − I e(−1) 2 At infinity we get the residue 1 1 res(f(z), ∞) = − I e + I e(−1) 2 2 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment 111

2.22

Problem.

Check the Zero Sum theorem for the following function z2

e(I z) + 4 z + 20

Hint. no hint Solution. We denote f(z) =

z2

e(I z) + 4 z + 20

We find singularities [{z = −2 − 4 I}, {z = −2 + 4 I}] The singularity z = −2 − 4 I adds the following residue res(f(z), −2 − 4 I) =

1 I e4 8 (eI )2

The singularity z = −2 + 4 I adds the following residue res(f(z), −2 + 4 I) = −

1 I e(−4) 8 (eI )2

At infinity we get the residue res(f(z), ∞) = −

1 I e4 1 I e(−4) + 8 (eI )2 8 (eI )2

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment 112

2.23

Problem.

Check the Zero Sum theorem for the following function e(I z) z2 − 5 z + 6 Hint. no hint Solution. We denote f(z) =

e(I z) z2 − 5 z + 6

We find singularities [{z = 2}, {z = 3}] The singularity z=2 adds the following residue res(f(z), 2) = −(eI )2 The singularity z=3 adds the following residue res(f(z), 3) = (eI )3 At infinity we get the residue res(f(z), ∞) = (eI )2 − (eI )3 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

113

2.24

Problem.

Check the Zero Sum theorem for the following function e(I z) z3 + 1 Hint. no hint Solution. We denote f(z) =

e(I z) z3 + 1

We find singularities [{z = −1}, {z =

1 1 √ 1 1 √ − I 3}, {z = + I 3}] 2 2 2 2

The singularity z = −1 adds the following residue res(f(z), −1) =

1 1 3 eI

The singularity z=

1 1 √ − I 3 2 2

adds the following residue p √ 1 1 1 √ (−1)(1/2 π ) e( 3) √ res(f(z), − I 3) = −2 2 2 3I 3 + 3 The singularity z=

1 1 √ + I 3 2 2

adds the following residue 1

1 1 √ (−1)(1/2 π ) p √ √ p √ res(f(z), + I 3) = 2 2 2 3 I e( 3) 3 − 3 e( 3) At infinity we get the residue p √ 1 1 1 1 (−1)(1/2 π ) e( 3) (−1)(1/2 π ) √ p √ √ p √ res(f(z), ∞) = − I + 2 −2 3 e 3I 3 + 3 3 I e( 3) 3 − 3 e( 3) 114

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

115

2.25

Problem.

Check the Zero Sum theorem for the following function e(I z) z4 − 1 Hint. no hint Solution. We denote f(z) =

e(I z) z4 − 1

We find singularities [{z = 1}, {z = −1}, {z = −I}, {z = I}] The singularity z=1 adds the following residue res(f(z), 1) =

1 I e 4

The singularity z = −1 adds the following residue res(f(z), −1) = −

1 1 4 eI

The singularity z = −I adds the following residue 1 res(f(z), −I) = − I e 4 The singularity z=I adds the following residue res(f(z), I) =

116

1 (−1) Ie 4

At infinity we get the residue 1 1 1 1 1 + I e − I e(−1) res(f(z), ∞) = − eI + 4 4 eI 4 4 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

117

2.26

Problem.

Check the Zero Sum theorem for the following function e(I z) z Hint. no hint Solution. We denote f(z) =

e(I z) z

We find singularities [{z = 0}] The singularity z=0 adds the following residue res(f(z), 0) = 1 At infinity we get the residue res(f(z), ∞) = −1 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

118

2.27

Problem.

Check the Zero Sum theorem for the following function z e(I z) 1 − z4 Hint. no hint Solution. We denote f(z) =

z e(I z) 1 − z4

We find singularities [{z = 1}, {z = −1}, {z = −I}, {z = I}] The singularity z=1 adds the following residue 1 res(f(z), 1) = − eI 4 The singularity z = −1 adds the following residue res(f(z), −1) = −

1 1 4 eI

The singularity z = −I adds the following residue res(f(z), −I) =

1 e 4

The singularity z=I adds the following residue res(f(z), I) =

119

1 (−1) e 4

At infinity we get the residue 1 1 I 1 1 1 − e − e(−1) e + 4 4 eI 4 4

res(f(z), ∞) = and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

120

2.28

Problem.

Check the Zero Sum theorem for the following function z2

z e(I z) + 4 z + 20

Hint. no hint Solution. We denote f(z) =

z2

z e(I z) + 4 z + 20

We find singularities [{z = −2 − 4 I}, {z = −2 + 4 I}] The singularity z = −2 − 4 I adds the following residue res(f(z), −2 − 4 I) = −

1 I (4 I e4 + 2 e4 ) 8 (eI )2

The singularity z = −2 + 4 I adds the following residue res(f(z), −2 + 4 I) = −

1 I (4 I e(−4) − 2 e(−4) ) 8 (eI )2

At infinity we get the residue res(f(z), ∞) =

1 I (4 I e4 + 2 e4 ) 1 I (4 I e(−4) − 2 e(−4) ) + 8 (eI )2 8 (eI )2

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment 121

2.29

Problem.

Check the Zero Sum theorem for the following function z e(I z) z2 + 9 Hint. no hint Solution. We denote f(z) =

z e(I z) z2 + 9

We find singularities [{z = −3 I}, {z = 3 I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) =

1 3 e 2

The singularity z = 3I adds the following residue res(f(z), 3 I) =

1 (−3) e 2

At infinity we get the residue 1 1 res(f(z), ∞) = − e3 − e(−3) 2 2 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment 122

2.30

Problem.

Check the Zero Sum theorem for the following function z e(I z) z 2 − 2 z + 10 Hint. no hint Solution. We denote f(z) =

z e(I z) z 2 − 2 z + 10

We find singularities [{z = 1 − 3 I}, {z = 1 + 3 I}] The singularity z = 1 − 3I adds the following residue 1 res(f(z), 1 − 3 I) = − I (3 I eI e3 − eI e3 ) 6 The singularity z = 1 + 3I adds the following residue 1 res(f(z), 1 + 3 I) = − I (3 I eI e(−3) + eI e(−3) ) 6 At infinity we get the residue res(f(z), ∞) =

1 1 I (3 I eI e3 − eI e3 ) + I (3 I eI e(−3) + eI e(−3) ) 6 6

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment 123

2.31

Problem.

Check the Zero Sum theorem for the following function z e(I z) z2 − 5 z + 6 Hint. no hint Solution. We denote f(z) =

z e(I z) z2 − 5 z + 6

We find singularities [{z = 2}, {z = 3}] The singularity z=2 adds the following residue res(f(z), 2) = −2 (eI )2 The singularity z=3 adds the following residue res(f(z), 3) = 3 (eI )3 At infinity we get the residue res(f(z), ∞) = 2 (eI )2 − 3 (eI )3 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

124

2.32

Problem.

Check the Zero Sum theorem for the following function e(I z) z2 Hint. no hint Solution. We denote f(z) =

e(I z) z2

We find singularities [{z = 0}] The singularity z=0 adds the following residue res(f(z), 0) = I At infinity we get the residue res(f(z), ∞) = −I and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

125

2.33

Problem.

Check the Zero Sum theorem for the following function z 3 e(I z) z4 + 5 z2 + 4 Hint. no hint Solution. We denote f(z) =

z 3 e(I z) z4 + 5 z2 + 4

We find singularities [{z = −I}, {z = 2 I}, {z = −2 I}, {z = I}] The singularity z = −I adds the following residue 1 res(f(z), −I) = − e 6 The singularity z = 2I adds the following residue res(f(z), 2 I) =

2 (−2) e 3

The singularity z = −2 I adds the following residue res(f(z), −2 I) =

2 2 e 3

The singularity z=I adds the following residue 1 res(f(z), I) = − e(−1) 6 126

At infinity we get the residue res(f(z), ∞) =

1 2 2 1 e − e(−2) − e2 + e(−1) 6 3 3 6

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

127

2.34

Problem.

Check the Zero Sum theorem for the following function z2 + 1 ez Hint. no hint Solution. We denote f(z) =

z2 + 1 ez

We find singularities [{z = −∞}] At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

128

2.35

Problem.

Check the Zero Sum theorem for the following function z2 + z − 1 z 2 (z − 1) Hint. no hint Solution. We denote f(z) =

z2 + z − 1 z 2 (z − 1)

We find singularities [{z = 1}, {z = 0}] The singularity z=1 adds the following residue res(f(z), 1) = 1 The singularity z=0 adds the following residue res(f(z), 0) = 0 At infinity we get the residue res(f(z), ∞) = −1 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

129

2.36

Problem.

Check the Zero Sum theorem for the following function z2 − 2 z + 5 (z − 2) (z 2 + 1) Hint. no hint Solution. We denote f(z) =

z2 − 2 z + 5 (z − 2) (z 2 + 1)

We find singularities [{z = 2}, {z = −I}, {z = I}] The singularity z=2 adds the following residue res(f(z), 2) = 1 The singularity z = −I adds the following residue res(f(z), −I) = −I The singularity z=I adds the following residue res(f(z), I) = I At infinity we get the residue res(f(z), ∞) = −1 and finally we obtain the sum X

res(f(z), z) = 0

130

Info. not given Comment. no comment

131

2.37

Problem.

Check the Zero Sum theorem for the following function 1 (z + 1) (z − 1) Hint. no hint Solution. We denote f(z) =

1 (z + 1) (z − 1)

We find singularities [{z = 1}, {z = −1}] The singularity z=1 adds the following residue res(f(z), 1) =

1 2

The singularity z = −1 adds the following residue res(f(z), −1) =

−1 2

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

132

2.38

Problem.

Check the Zero Sum theorem for the following function 1 z (1 − z 2 ) Hint. no hint Solution. We denote f(z) =

1 z (1 − z 2 )

We find singularities [{z = 1}, {z = −1}, {z = 0}] The singularity z=1 adds the following residue res(f(z), 1) =

−1 2

The singularity z = −1 adds the following residue res(f(z), −1) =

−1 2

The singularity z=0 adds the following residue res(f(z), 0) = 1 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

133

Info. not given Comment. no comment

134

2.39

Problem.

Check the Zero Sum theorem for the following function z

(z 2

1 + 4)2

Hint. no hint Solution. We denote f(z) =

z

(z 2

1 + 4)2

We find singularities [{z = 0}, {z = 2 I}, {z = −2 I}] The singularity z=0 adds the following residue res(f(z), 0) =

1 16

The singularity z = 2I adds the following residue res(f(z), 2 I) =

−1 32

The singularity z = −2 I adds the following residue res(f(z), −2 I) =

−1 32

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

135

Info. not given Comment. no comment

136

2.40

Problem.

Check the Zero Sum theorem for the following function 1 z (z − 1) Hint. no hint Solution. We denote f(z) =

1 z (z − 1)

We find singularities [{z = 1}, {z = 0}] The singularity z=1 adds the following residue res(f(z), 1) = 1 The singularity z=0 adds the following residue res(f(z), 0) = −1 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

137

2.41

Problem.

Check the Zero Sum theorem for the following function 1 z (z − 1) Hint. no hint Solution. We denote f(z) =

1 z (z − 1)

We find singularities [{z = 1}, {z = 0}] The singularity z=1 adds the following residue res(f(z), 1) = 1 The singularity z=0 adds the following residue res(f(z), 0) = −1 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

138

2.42

Problem.

Check the Zero Sum theorem for the following function (z 2

1 + 1)2

Hint. no hint Solution. We denote f(z) =

(z 2

1 + 1)2

We find singularities [{z = −I}, {z = I}] The singularity z = −I adds the following residue res(f(z), −I) =

1 I 4

The singularity z=I adds the following residue 1 res(f(z), I) = − I 4 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

139

2.43

Problem.

Check the Zero Sum theorem for the following function z3

1 − z5

Hint. no hint Solution. We denote f(z) =

1 z3 − z5

We find singularities [{z = 1}, {z = −1}, {z = 0}] The singularity z=1 adds the following residue res(f(z), 1) =

−1 2

The singularity z = −1 adds the following residue res(f(z), −1) =

−1 2

The singularity z=0 adds the following residue res(f(z), 0) = 1 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

140

Info. not given Comment. no comment

141

2.44

Problem.

Check the Zero Sum theorem for the following function z4

1 +1

Hint. no hint Solution. We denote f(z) =

1 z4 + 1

We find singularities [{z =

1√ 1 √ 1√ 1 √ 1√ 1 √ 1√ 1 √ 2+ I 2}, {z = − 2− I 2}, {z = 2− I 2}, {z = − 2+ I 2}] 2 2 2 2 2 2 2 2

The singularity z=

1√ 1 √ 2+ I 2 2 2

adds the following residue 1√ 1 √ 1 √ √ 2 + I 2) = 2 2 2I 2 − 2 2

res(f(z), The singularity

z=−

1√ 1 √ 2− I 2 2 2

adds the following residue res(f(z), −

1 √ 1 1√ √ 2 − I 2) = − √ 2 2 2I 2 − 2 2

The singularity z=

1√ 1 √ 2− I 2 2 2

adds the following residue res(f(z),

1√ 1 √ 1 √ 2 − I 2) = − √ 2 2 2I 2 + 2 2

The singularity z=−

1√ 1 √ 2+ I 2 2 2

142

adds the following residue res(f(z), −

1√ 1 √ 1 √ √ 2 + I 2) = 2 2 2I 2 + 2 2

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

143

2.45

Problem.

Check the Zero Sum theorem for the following function 1 z − z3 Hint. no hint Solution. We denote f(z) =

1 z − z3

We find singularities [{z = 1}, {z = −1}, {z = 0}] The singularity z=1 adds the following residue res(f(z), 1) =

−1 2

The singularity z = −1 adds the following residue res(f(z), −1) =

−1 2

The singularity z=0 adds the following residue res(f(z), 0) = 1 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

144

Info. not given Comment. no comment

145

2.46

Problem.

Check the Zero Sum theorem for the following function 1 z Hint. no hint Solution. We denote f(z) =

1 z

We find singularities [{z = 0}] The singularity z=0 adds the following residue res(f(z), 0) = 1 At infinity we get the residue res(f(z), ∞) = −1 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

146

2.47

Problem.

Check the Zero Sum theorem for the following function ez (z 2 + 9) z 2 Hint. no hint Solution. We denote f(z) =

(z 2

ez + 9) z 2

We find singularities [{z = ∞}, {z = −3 I}, {z = 0}, {z = 3 I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) = −

1 I 54 (eI )3

The singularity z=0 adds the following residue res(f(z), 0) =

1 9

The singularity z = 3I adds the following residue res(f(z), 3 I) =

1 I (eI )3 54

At infinity we get the residue res(f(z), ∞) =

1 I 1 1 − − I (eI )3 I 3 54 (e ) 9 54

and finally we obtain the sum X

res(f(z), z) = 0

147

Info. not given Comment. no comment

148

2.48

Problem.

Check the Zero Sum theorem for the following function ez z2 + 1 Hint. no hint Solution. We denote f(z) =

ez z2 + 1

We find singularities [{z = ∞}, {z = −I}, {z = I}] The singularity z = −I adds the following residue res(f(z), −I) =

1 I 2 eI

The singularity z=I adds the following residue 1 res(f(z), I) = − I eI 2 At infinity we get the residue res(f(z), ∞) = −

1 I 1 + I eI 2 eI 2

and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

149

2.49

Problem.

Check the Zero Sum theorem for the following function ez (z 2 + 9) z 2 Hint. no hint Solution. We denote f(z) =

(z 2

ez + 9) z 2

We find singularities [{z = ∞}, {z = −3 I}, {z = 0}, {z = 3 I}] The singularity z = −3 I adds the following residue res(f(z), −3 I) = −

1 I 54 (eI )3

The singularity z=0 adds the following residue res(f(z), 0) =

1 9

The singularity z = 3I adds the following residue res(f(z), 3 I) =

1 I (eI )3 54

At infinity we get the residue res(f(z), ∞) =

1 I 1 1 − − I (eI )3 I 3 54 (e ) 9 54

and finally we obtain the sum X

res(f(z), z) = 0

150

Info. not given Comment. no comment

151

2.50

Problem.

Check the Zero Sum theorem for the following function z (z − 1) (z − 2)2 Hint. no hint Solution. We denote f(z) =

z (z − 1) (z − 2)2

We find singularities [{z = 1}, {z = 2}] The singularity z=1 adds the following residue res(f(z), 1) = 1 The singularity z=2 adds the following residue res(f(z), 2) = −1 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

152

2.51

Problem.

Check the Zero Sum theorem for the following function (z 2

z2 + 1)2

Hint. no hint Solution. We denote f(z) =

z2 (z 2 + 1)2

We find singularities [{z = −I}, {z = I}] The singularity z = −I adds the following residue res(f(z), −I) =

1 I 4

The singularity z=I adds the following residue 1 res(f(z), I) = − I 4 At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

153

2.52

Problem.

Check the Zero Sum theorem for the following function z4 +1

z4 Hint.

no hint Solution. We denote f(z) =

z4 +1

z4

We find singularities [{z =

1√ 1 √ 1√ 1 √ 1√ 1 √ 1√ 1 √ 2+ I 2}, {z = − 2− I 2}, {z = 2− I 2}, {z = − 2+ I 2}] 2 2 2 2 2 2 2 2

The singularity z=

1 √ 1√ 2+ I 2 2 2

adds the following residue res(f(z),

1√ 1 √ 1 √ 2 + I 2) = − √ 2 2 2I 2 − 2 2

The singularity z=−

1√ 1 √ 2− I 2 2 2

adds the following residue res(f(z), −

1√ 1 √ 1 √ √ 2 − I 2) = 2 2 2I 2 − 2 2

The singularity z=

1√ 1 √ 2− I 2 2 2

adds the following residue res(f(z),

1√ 1 √ 1 √ √ 2 − I 2) = 2 2 2I 2 + 2 2

The singularity z=−

1√ 1 √ 2+ I 2 2 2

154

adds the following residue res(f(z), −

1√ 1 √ 1 √ 2 + I 2) = − √ 2 2 2I 2 + 2 2

At infinity we get the residue res(f(z), ∞) = 0 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

155

2.53

Problem.

Check the Zero Sum theorem for the following function z5 (1 − z)2 Hint. no hint Solution. We denote f(z) =

z5 (1 − z)2

We find singularities [{z = ∞}, {z = 1}, {z = −∞}] The singularity z=1 adds the following residue res(f(z), 1) = 5 At infinity we get the residue res(f(z), ∞) = −5 and finally we obtain the sum X

res(f(z), z) = 0

Info. not given Comment. no comment

156

3

Power series problems

Example ∞ X zn n2 n=1

157

3.1

Problem.

Sum the following power series ∞ X (−1)(n+1) z n n n=1

Hint. derive once and sum Solution. We denote the n-th term in the series by a(n) =

(−1)(n+1) z n n

For the ratio test we need the term a(n + 1) a(n + 1) =

(−1)(n+2) z (n+1) n+1

Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n+1) z n lim = |z| n→∞ (n + 1) z n Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case lim [

n→∞

(−1)(n+1) z n ( 1 ) ] n = |z| n

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X (−1)(n+1) z n = ln(1 + z) n n=1

Info. not given Comment. no comment 158

3.2

Problem.

Sum the following power series ∞ X

(−1)n n2 z n

n=1

Hint. divide by z and integrate Solution. We denote the n-th term in the series by a(n) = (−1)n n2 z n For the ratio test we need the term a(n + 1) a(n + 1) = (−1)(n+1) (n + 1)2 z (n+1) Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n + 1)2 z (n+1) = |z| lim n→∞ n2 z n Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case 1

lim [(−1)n n2 z n ]( n ) = |z|

n→∞

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X

(−1)n n2 z n = −

n=1

z (−z + 1) (1 + z)3

Info. not given Comment. no comment

159

3.3

Problem.

Sum the following power series ∞ X (−1)n n3 z n (n + 1)! n=0

Hint. manipulate the numerator to cancel Solution. We denote the n-th term in the series by a(n) =

(−1)n n3 z n (n + 1)!

For the ratio test we need the term a(n + 1) a(n + 1) =

(−1)(n+1) (n + 1)3 z (n+1) (n + 2)!

Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n + 1)3 z (n+1) (n + 1)! =0 lim n→∞ (n + 2)! n3 z n Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case lim [

n→∞

(−1)n n3 z n ( 1 ) (−1)n n3 z n ( 1 ) ] n = lim ( ) n n→∞ (n + 1)! (n + 1)!

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X (−1)n n3 z n = (n + 1)! n=0  (−z)

1  1−e (1 + z) − z 2 − 12 2 2 z

1 − e(−z) (1 + z + z2

160

 1 1 2 1 z ) 1 − e(−z) (1 + z + z 2 + z 3 )  2 2 6 + 12  2 z

Info. not given Comment. no comment

161

3.4

Problem.

Sum the following power series ∞ X (−1)n z n n (2 n − 1) n=1

Hint. make the power 2n and derive twice Solution. We denote the n-th term in the series by (−1)n z n n (2 n − 1)

a(n) =

For the ratio test we need the term a(n + 1) a(n + 1) =

(−1)(n+1) z (n+1) (n + 1) (2 n + 1)

Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n+1) z n (2 n − 1) lim = |z| n→∞ (n + 1) (2 n + 1) z n Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case lim [

n→∞

(−1)n z n ( 1 ) ] n = |z| n (2 n − 1)

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X (−1)n z n 1 3 = −z hypergeom([1, 1, ], [ , 2], −z) n (2 n − 1) 2 2 n=1

Info. not given Comment. no comment 162

3.5

Problem.

Sum the following power series ∞ X (2 n + 1) z n n! n=0

Hint. prepare a combination of exponentials Solution. We denote the n-th term in the series by a(n) =

(2 n + 1) z n n!

For the ratio test we need the term a(n + 1) (2 n + 3) z (n+1) (n + 1)!

a(n + 1) = Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (2 n + 3) z (n+1) n! =0 lim n→∞ (n + 1)! (2 n + 1) z n Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case (2 n + 1) z n ( 1 ) (2 n + 1) z n ( 1 ) n n lim [ ] = lim ( ) n→∞ n→∞ n! n! From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X (2 n + 1) z n 1 = 2 ez ( + z) n! 2 n=0

Info. not given Comment. no comment 163

3.6

Problem.

Sum the following power series ∞ X (n2 − 2) z n 2n n! n=0

Hint. prepare a combination of exponentials Solution. We denote the n-th term in the series by (n2 − 2) z n 2n n! For the ratio test we need the term a(n + 1) a(n) =

a(n + 1) =

((n + 1)2 − 2) z (n+1) 2(n+1) (n + 1)!

Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case ((n + 1)2 − 2) z (n+1) n! 2 (n + 1)! (n2 − 2) z n lim =0 n→∞ 2(n+1) Moreover we can check the root test computing p lim n |a(n)| n

n→∞

and obtain in our case lim [

n→∞

(n2 − 2) z n ( 1 ) (n2 − 2) z n ( 1 ) n n ] = lim ( ) n→∞ 2n n! 2n n!

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X (n2 − 2) z n 1 √ (1/2 z) 1 √ 1 = −2 e(1/2 z) + 2e z+ (− 2+1) e(1/2 z) z+ e(1/2 z) z 2 n 2 n! 2 2 4 n=0

Info. not given Comment. no comment 164

3.7

Problem.

Sum the following power series ∞ X

n (n + 1) z n

n=0

Hint. divide by z and integrate twice Solution. We denote the n-th term in the series by a(n) = n (n + 1) z n For the ratio test we need the term a(n + 1) a(n + 1) = (n + 1) (n + 2) z (n+1) Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n + 2) z (n+1) = |z| lim n→∞ n zn Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case 1

lim [n (n + 1) z n ]( n ) = |z|

n→∞

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X

n (n + 1) z n = −2

n=0

Info. not given Comment. no comment

165

z (z − 1)3

3.8

Problem.

Sum the following power series ∞ X n zn 5n n=0

Hint. divide by z and integrate Solution. We denote the n-th term in the series by a(n) =

n zn 5n

For the ratio test we need the term a(n + 1) a(n + 1) =

(n + 1) z (n+1) 5(n+1)

Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n + 1) z (n+1) 5 1 n zn lim = |z| (n+1) n→∞ 5 5 Moreover we can check the root test computing p lim n |a(n)| n

n→∞

and obtain in our case lim [

n→∞

n z n ( 1 ) n z n ( 1 ) n n ] lim ( ) = n→∞ 5n 5n

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X n zn z =5 n 5 (z − 5)2 n=0

Info. not given Comment. no comment 166

3.9

Problem.

Sum the following power series ∞ X

n2 z (n−1)

n=1

Hint. integrate divide by z and integrate Solution. We denote the n-th term in the series by a(n) = n2 z (n−1) For the ratio test we need the term a(n + 1) a(n + 1) = (n + 1)2 z n Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n + 1)2 z n = |z| lim n→∞ n2 z (n−1) Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case 1

lim [n2 z (n−1) ]( n ) = |z|

n→∞

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X

n2 z (n−1) =

n=1

1+z (−z + 1)3

Info. not given Comment. no comment

167

3.10

Problem.

Sum the following power series ∞ X z (2 n) (2 n)! n=0

Hint. manipulate to exponentials Solution. We denote the n-th term in the series by a(n) =

z (2 n) (2 n)!

For the ratio test we need the term a(n + 1) a(n + 1) =

z (2 n+2) (2 n + 2)!

Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (2 n+2) z (2 n)! =0 lim n→∞ (2 n + 2)! z (2 n) Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case lim [

n→∞

z (2 n) ( 1 ) ] n =0 (2 n)!

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X z (2 n) = cosh(z) (2 n)! n=0

Info. not given Comment. no comment 168

3.11

Problem.

Sum the following power series ∞ X z (2 n−1) (2 n − 1)! n=0

Hint. manipulate to exponentials Solution. We denote the n-th term in the series by a(n) =

z (2 n−1) (2 n − 1)!

For the ratio test we need the term a(n + 1) a(n + 1) =

z (2 n+1) (2 n + 1)!

Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (2 n+1) z (2 n − 1)! lim =0 n→∞ (2 n + 1)! z (2 n−1) Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case z (2 n−1) ( 1 ) z (2 n−1) ( 1 ) lim [ ] n = lim ( ) n n→∞ (2 n − 1)! n→∞ (2 n − 1)! From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X z (2 n−1) = sinh(z) (2 n − 1)! n=0

Info. not given Comment. no comment 169

3.12

Problem.

Sum the following power series ∞ X

zn n (n + 1) n=1 Hint. multiple by z and derive twice Solution. We denote the n-th term in the series by a(n) =

zn n (n + 1)

For the ratio test we need the term a(n + 1) z (n+1) (n + 1) (n + 2)

a(n + 1) = Ratio test computes lim

n→∞

|a(n + 1)| |a(n)|

and obtains in our case (n+1) z n = |z| lim n→∞ (n + 2) z n Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case lim [

n→∞

1 zn ]( n ) = |z| n (n + 1)

From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series   ln(−z + 1) ∞ (− − 1) (z − 1) n X 1  −2 z + 2 z z = z 2 −  n (n + 1) 2 z z−1 n=1 Info. not given Comment. no comment 170

3.13

Problem.

Sum the following power series ∞ X zn n! n=0

Hint. manipulate to exponentials Solution. We denote the n-th term in the series by a(n) =

zn n!

For the ratio test we need the term a(n + 1) a(n + 1) =

z (n+1) (n + 1)!

Ratio test computes |a(n + 1)| n→∞ |a(n)| lim

and obtains in our case (n+1) z n! =0 lim n→∞ (n + 1)! z n Moreover we can check the root test computing p lim n |a(n)| n→∞

and obtain in our case z n ( 1 ) z n ( 1 ) n n = lim ( ) lim [ ] n→∞ n! n→∞ n! From this we conclude the radius of convergence R. For |z| < R we sum the series using common tricks for power series ∞ X zn = ez n! n=0

Info. not given Comment. no comment 171