Complex Problems

7/21/2019 Complex Problems

1/171

Complex variable

solvedproblemsPavel Pyrih

11:03 May 29, 2012

( public domain )

Contents

1 Residue theorem problems 2

2 Zero Sum theorem for residues problems 76

3 Power series problems 157

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

1


2/171

1 Residue theorem problems

We will solve several problems using the following theorem:

Theorem. (Residue theorem) Suppose U is a simply connected open subsetof the complex plane, and w1, . . . ,wn are finitely many points ofUand f is afunction which is defined and holomorphic on U\{w1, . . . , wn}. If is a simplyclosed curve in U contaning the points wk in the interior, then

f(z) dz = 2ik

k=1

res(f, wk) .

The following rules can be used for residue counting:

Theorem. (Rule 1) Iffhas a pole of order k at the point w then

res(f, w) = 1

(k 1)! limzw

(z w)kf(z)(k1) .

Theorem. (Rule 2) Iff, g are holomorphic at the point w and f(w)= 0. Ifg(w) = 0, g(w)= 0, then

res f

g

, w = f(w)

g(w)

.

Theorem. (Rule 3) Ifh is holomorphic atw and g has a simple pole atw, thenres(gh,w) = h(w)res(g, w).

2


3/171

We will use special formulas for special types of problems:

Theorem. ( TYPE I. Integral from a rational function in sin and cos.) IfQ(a, b) is a rational function of two complex variables such that for real a, b,a2 + b2 = 1 isQ(a, b) finite, then the function

T(z) := Q

z+ 1/z

2 ,

z 1/z2i

/(iz)

is rational, has no poles on the real line and 20

Q(cos t, sin t) dt= 2i

|a|


4/171

Theorem. ( TYPE II. Integral from a rational function. ) SupposeP, Q arepolynomial of order m, n respectively, and n

m >1 and Q has no real roots.

Then for the rational functionf= PQ holds +

f(x) dx= 2i

k

res(f, wk) ,

where all singularities offwith a positive imaginary part are considered in theabove sum.

Example

+

1

1 + x2dx

4


5/171

Theorem. ( TYPE III. Integral from a rational function multiplied by cos orsin ) IfQ is a rational function such that has no pole at the real line and for

z is Q(z) = O(z1). Forb >0 denote f(z) = Q(z) eibz. Then +

Q(x)cos(bx) dx= Re

2i

w

res(f, w)

+

Q(x)sin(bx) dx= Im

2i

w

res(f, w)

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

Example

+

cos(x)1 + x2

dx

5


6/171

1.1 Problem.

Using the Residue theorem evaluate 20

cos(x)2

13 + 12 cos(x)dx

Hint.

Type I

Solution.We denote

f(z) =I(

1

2z +

1

2

1

z)2

(13 + 6 z+ 61

z

) z

We find singularities

[{z= 0},{z =32},{z =2

3}]

The singularity

z= 0

is in our region and we will add the following residue

res(0, f(z)) = 13

144I

The singularity

z =3

2

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

z =2

3


res(23

, f(z)) =169720

I

Our sum is

2 I (

res(z, f(z))) =

13

45

The solution is 20

cos(x)2

13 + 12 cos(x)dx =

13

45

6


7/171

We can try to solve it using real calculus and obtain the result

20

cos(x)2

13 + 12 cos(x)dx = 13

45

Info.

not given

Comment.

no comment

7


8/171

1.2 Problem.


cos(x)4

1 + sin(x)2dx

Hint.

Type I

Solution.We denote

f(z) =I(

1

2z +

1

2

1

z)4

(1

1

4(z

1

z)2) z


[{z = 0},{z=

2+1},{z= 1

2},{z =

21},{z =1

2},{z=},{z=}]The singularity

z= 0


res(0, f(z)) =5

2I

The singularity

z =

2 + 1


z = 1

2


res(1

2, f(z)) =768 I2 + 1088 I

768 544 2The singularity

z =

2 1is in our region and we will add the following residue

res(

2 1, f(z)) =768 I

2 + 1088 I


z =1

2

8


9/171


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

2 I (

res(z, f(z))) = 2 I (5

2I+ 2

768 I2 + 1088 I768 544 2 )

The solution is 2 0

cos(x)4

1 + sin(x)2dx = 2 I (

5

2I+ 2

768 I2 + 1088 I768 544 2 )

We can try to solve it using real calculus and obtain the result 2 0

cos(x)4

1 + sin(x)2dx =

(4

2 5)(

2 + 1) (

2 1)Info.

not given

Comment.

no comment

9


10/171

1.3 Problem.


sin(x)2

5

4 cos(x)

dx

Hint.

Type I

Solution.We denote

f(z) =1

4

I(z1z

)2

( 54 1

2z 1

21z

) z


[{z= 0},{z = 12},{z= 2}]

The singularity

z= 0


res(0, f(z)) =54

I

The singularity

z =1

2


res(1

2, f(z)) =

3

4I

The singularity

z= 2

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

Our sum is

2 I (

res(z, f(z))) =

10


11/171

The solution is

20

sin(x)25

4 cos(x)

dx=

We can try to solve it using real calculus and obtain the result 20

sin(x)2

5

4 cos(x)

dx=

Info.

not given

Comment.

no comment

11


12/171

1.4 Problem.

Using the Residue theorem evaluate 2 0

1

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

Hint.

Type I

Solution.We denote

f(z) = I(1

4(z1

z)2 + 4 (

1

2z +

1

2

1

z)2) z


[{z=13

I

3},{z = 13

I

3},{z = I

3},{z =I

3}]The singularity

z =13

I

3


res(13

I

3, f(z)) =14

I

The singularity

z= 13

I3


res(1

3I

3, f(z)) =1

4I

The singularity

z = I

3


z =I

3


Our sum is

2 I (

res(z, f(z))) =

12


13/171

The solution is

20

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

dx =


1

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

Info.

not given

Comment.

no comment

13


14/171

1.5 Problem.


1

13 + 12 sin(x)dx

Hint.

Type I

Solution.We denote

f(z) = I(13 6 I(z1

z)) z


[{z=23

I},{z=32

I}]The singularity

z =23

I


res(23

I , f(z)) =15

I

The singularity

z =3

2I


Our sum is

2 I (

res(z, f(z))) =2

5

The solution is 2 0

1

13 + 12 sin(x)dx =

2

5


0

1

13 + 12 sin(x)dx =

2

5

Info.

not given

14


15/171

Comment.

no comment

15


16/171

1.6 Problem.


1

2 + cos(x)dx

Hint.

Type I

Solution.We denote

f(z) = I(2 +

1

2z +

1

2

1

z) z


[{z=2 +

3},{z=2

3}]The singularity

z =2 +

3


res(2 +

3, f(z)) =13

I

3

The singularity

z =2

3


Our sum is

2 I (

res(z, f(z))) =2

3

3

The solution is 2 0

1

2 + cos(x)dx =

2

3

3


1

2 + cos(x)dx =

2

3

3

Info.

not given

Comment.

no comment

16


17/171

1.7 Problem.


1

(2 + cos(x))2dx

Hint.

Type I

Solution.We denote

f(z) = I(2 +

1

2z +

1

2

1

z)2 z


[{z=2 +

3},{z=2

3}]The singularity

z =2 +

3


res(2 +

3, f(z)) =13

I+1

3(2

3I+

1

3I

3)

3

The singularity

z =2

3


Our sum is

2 I (

res(z, f(z))) = 2 I (13

I+1

3(2

3I+

1

3I

3)

3)

The solution is 20

1

(2 + cos(x))2dx = 2 I (1

3I+

1

3(2

3I+

1

3I

3)

3)


1

(2 + cos(x))2dx =

4

9

3

Info.

not given

Comment.

no comment

17


18/171

1.8 Problem.


1

2 + sin(x)dx

Hint.

Type I

Solution.We denote

f(z) = I(2 1

2I(z 1

z)) z


[{z =2 I+ I

3},{z=2 I I

3}]The singularity

z =2 I+ I

3


res(2 I+ I

3, f(z)) =13

I

3

The singularity

z =2 I I

3


Our sum is

2 I (

res(z, f(z))) =2

3

3

The solution is 20

1

2 + sin(x)dx =

2

3

3


1

2 + sin(x)dx =

2

3

3

Info.

not given

Comment.

no comment

18


19/171

1.9 Problem.

Using the Residue theorem evaluate 2 0

1

5 + 4 cos(x)dx

Hint.

Type I

Solution.We denote

f(z) = I(5 + 2 z+ 2

1

z) z


[{z=2},{z =12}]

The singularity

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

z =1

2


res(12

, f(z)) =

1

3I

Our sum is

2 I (

res(z, f(z))) =2

3

The solution is 20

1

5 + 4 cos(x)dx =

2

3


2

0

1

5 + 4 cos(x)dx =

2

3

Info.

not given

Comment.

no comment

19


20/171

1.10 Problem.


1

5 + 4 sin(x)dx

Hint.

Type I

Solution.We denote

f(z) = I(5 2 I(z 1

z)) z


[{z =2 I},{z =12

I}]The singularity

z =2 Iwill be skipped because the singularity is not in our region.The singularity

z =12

I


res(

1

2I , f(z)) =

1

3I

Our sum is

2 I (

res(z, f(z))) =2

3

The solution is 20

1

5 + 4 sin(x)dx =

2

3


2

0

1

5 + 4 sin(x)dx =

2

3

Info.

not given

Comment.

no comment

20


21/171

1.11 Problem.


cos(x)5

4 cos(x)

dx

Hint.

Type I

Solution.We denote

f(z) =I(

1

2z +

1

2

1

z)

(5

41

2z1

2

1

z ) z


[{z= 0},{z = 12},{z= 2}]

The singularity

z= 0


res(0, f(z)) = I

The singularity

z =1

2


res(1

2, f(z)) =5

3I

The singularity

z= 2


Our sum is

2 I ( res(z, f(z))) =4

3

The solution is 20

cos(x)5

4 cos(x)

dx=4

3

21


22/171


20

cos(x)5

4 cos(x)

dx= 43

Info.

not given

Comment.

no comment

22


23/171

1.12 Problem.

Using the Residue theorem evaluate

x2 + 1

x4 + 1dx

Hint.

Type II

Solution.We denote

f(z) =z2 + 1

z4 + 1


[{z =12

21

2I

2},{z = 1

2

2+

1

2I

2},{z =1

2

2+

1

2I

2},{z = 1

2

21

2I

2}]

The singularity

z =12

2 1

2I

2


z=1

2

2 +

1

2I

2


res( 12

2 +12

I2, f(z)) = 1 + I2 I

2 2 2

The singularity

z =12

2 +

1

2I

2


res(12

2 +

1

2I

2, f(z)) =

1 I2 I

2 + 2

2

The singularity

z=1

2

2 1

2I

2


Our sum is

2 I (

res(z, f(z))) = 2 I ( 1 + I

2 I

2 2 2 + 1 I

2 I

2 + 2

2)

23


24/171

The solution is

x2

+ 1x4 + 1dx = 2 I ( 1 + I2 I

2 2 2 + 1 I2 I2 + 2 2 )


x2 + 1

x4 + 1dx =

2

Info.

not given

Comment.

no comment

24


25/171

1.13 Problem.


x2 1(x2 + 1)2

dx

Hint.

Type II

Solution.We denote

f(z) = z2 1(z2 + 1)2

We find singularities[{z=I},{z = I}]

The singularity

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

z = I


res(I, f(z)) = 0

Our sum is

2 I (

res(z, f(z))) = 0

The solution is

x2 1(x2 + 1)2

dx = 0


x2 1(x2 + 1)2

dx = 0

Info.

not given

Comment.

no comment

25


26/171

1.14 Problem.


x2 x + 2x4 + 10 x2 + 9

dx

Hint.

Type II

Solution.We denote

f(z) = z2 z+ 2z4 + 10 z2 + 9

We find singularities[{z= 3 I},{z =3 I},{z=I},{z = I}]

The singularity

z= 3 I


res(3 I, f(z)) = 1

16 7

48I

The singularity

z =3 I



z = I


res(I, f(z)) =116

116

I

Our sum is

2 I (

res(z, f(z))) = 5

12

26


27/171

The solution is

x2

x + 2x4 + 10 x2 + 9dx = 512


x2 x + 2x4 + 10 x2 + 9

dx = 5

12

Info.

not given

Comment.

no comment

27


28/171

1.15 Problem.


1

(x2 + 1) (x2 + 4)2dx

Hint.

Type II

Solution.We denote

f(z) = 1

(z2 + 1) (z2 + 4)2

We find singularities[{z= 2 I},{z =I},{z= I},{z =2 I}]

The singularity

z= 2 I


res(2 I, f(z)) = 11

288I

The singularity

z=I


z = I


res(I, f(z)) =118

I

The singularity

z =2 Iwill be skipped because the singularity is not in our region.

Our sum is

2 I (

res(z, f(z))) = 5

144

28


29/171

The solution is

1(x2 + 1) (x2 + 4)2 dx = 5144


1

(x2 + 1) (x2 + 4)2dx =

5

144

Info.

not given

Comment.

no comment

29


30/171

1.16 Problem.


1

(x2 + 1) (x2 + 9)dx

Hint.

Type II

Solution.We denote

f(z) = 1

(z2 + 1) (z2 + 9)

We find singularities[{z= 3 I},{z =3 I},{z=I},{z = I}]

The singularity

z= 3 I


res(3 I, f(z)) = 1

48I

The singularity

z =3 I



z = I


res(I, f(z)) =116

I

Our sum is

2 I (

res(z, f(z))) = 1

12

30


31/171

The solution is

1(x2 + 1) (x2 + 9)dx = 112


1

(x2 + 1) (x2 + 9)dx =

1

12

Info.

not given

Comment.

no comment

31


32/171

1.17 Problem.


1

(x I) (x 2 I)dx

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= 2 I


res(2 I, f(z)) =IThe singularity

z = I


res(I, f(z)) = I

Our sum is

2 I (

res(z, f(z))) = 0

The solution is

1

(x I) (x 2 I)dx = 0


1

(x I) (x 2 I)dx = 0

Info.

not given

Comment.

no comment

32


33/171

1.18 Problem.


1

(x I) (x 2 I) (x + 3 I)dx

Hint.

Type II

Solution.We denote

f(z) = 1

(z I) (z 2 I) (z+ 3 I)

We find singularities[{z = 2 I},{z=3 I},{z = I}]

The singularity

z= 2 I


res(2 I, f(z)) =1

5The singularity

z =3 I


z = I


res(I, f(z)) =1

4

Our sum is

2 I (

res(z, f(z))) = 1

10I

The solution is

1

(x I) (x 2 I) (x + 3 I)dx = 1

10I

33


34/171


1(x I) (x 2 I) (x + 3 I)dx = 110I

Info.

not given

Comment.

no comment

34


35/171

1.19 Problem.


1

1 + x2dx

Hint.

Type II

Solution.We denote

f(z) = 1

z2 + 1


[{z=I},{z = I}]The singularity


z = I


res(I, f(z)) =12

I

Our sum is

2 I (

res(z, f(z))) =

The solution is

1

1 + x2dx =


1

1 + x2dx =

Info.

not given

Comment.

no comment

35


36/171

1.20 Problem.


1

x3 + 1dx

Hint.

Type II

Solution.We denote

f(z) = 1

z3 + 1


[{z=1},{z = 12 1

2I

3},{z= 1

2+

1

2I

3}]

The singularity

z =1is on the real line and we will add one half of the following residue

res(1, f(z)) = 13

The singularity

z=1

2 1

2I

3


z=1

2+

1

2I

3


res(1

2+

1

2I

3, f(z)) = 2

1

3 I

3 3Our sum is

2 I ( res(z, f(z))) = 2 I (1

6+ 2

1

3 I

3 3

)

The solution is

1

x3 + 1dx = 2 I (

1

6+ 2

1

3 I

3 3 )

36


37/171


1x3 + 1 dx =

1x3 + 1dx

Info.

not given

Comment.

no comment

37


38/171

1.21 Problem.


1

x6 + 1dx

Hint.

Type II

Solution.We denote

f(z) = 1

z6 + 1


[{z =I},{z = I},{z= 12

2 + 2 I

3},{z=1

2

2 + 2 I

3},{z= 1

2

2 2 I

3},

{z=12

2 2 I

3}]

The singularity


z = I


res(I, f(z)) =16

I

The singularity

z =1

2

2 + 2 I

3


res(1

2

2 + 2 I

3, f(z)) =

16

3

1

(2 + 2 I

3)(5/2)

The singularity

z=

1

22 + 2 I3


z =1

2

2 2 I

3

38


39/171


z=12

2 2 I

3


res(12

2 2 I

3, f(z)) =16

3

1

(2 2 I3)(5/2)Our sum is

2 I (

res(z, f(z))) = 2 I (16

I+16

3

1

(2 + 2 I

3)(5/2)16

3

1

(2 2 I3)(5/2) )

The solution is

1

x6 + 1dx = 2 I (1

6I+

16

3

1

(2 + 2 I

3)(5/2) 16

3

1

(2 2 I3)(5/2) )


1

x6 + 1dx =

2

3

Info.

not given

Comment.

no comment

39


40/171

1.22 Problem.


x

(x2 + 4 x + 13)2dx

Hint.

Type II

Solution.We denote

f(z) = z

(z2 + 4 z+ 13)2


[{z=2 + 3 I},{z=2 3 I}]The singularity

z=2 + 3 Iis in our region and we will add the following residue

res(2 + 3 I, f(z)) = 154

I

The singularity

z=2 3 Iwill be skipped because the singularity is not in our region.

Our sum is

2 I (

res(z, f(z))) =127

The solution is

x

(x2 + 4 x + 13)2dx =1

27


x

(x2 + 4 x + 13)2dx =1

27

Info.

not given

Comment.

no comment

40


41/171

1.23 Problem.


x2

(x2 + 1) (x2 + 9)dx

Hint.

Type II

Solution.We denote

f(z) = z2

(z2 + 1) (z2 + 9)


[{z= 3 I},{z =3 I},{z=I},{z = I}]The singularity

z= 3 I


res(3 I, f(z)) =316

I

The singularity

z =3 I



z = I


res(I, f(z)) = 1

16I

Our sum is

2 I (

res(z, f(z))) =1

4

41


42/171

The solution is

x2

(x2 + 1) (x2 + 9)dx = 14


x2

(x2 + 1) (x2 + 9)dx =

1

4

Info.

not given

Comment.

no comment

42


43/171

1.24 Problem.


x2

(x2 + 1)3dx

Hint.

Type II

Solution.We denote

f(z) = z2

(z2 + 1)3




z = I


res(I, f(z)) =116

I

Our sum is

2 I (

res(z, f(z))) =1

8

The solution is

x2

(x2 + 1)3dx =

1

8


x2

(x2 + 1)3dx =

1

8

Info.

not given

Comment.

no comment

43


44/171

1.25 Problem.


x2

(x2 + 4)2dx

Hint.

Type II

Solution.We denote

f(z) = z2

(z2 + 4)2


[{z= 2 I},{z =2 I}]The singularity

z= 2 I


res(2 I, f(z)) =18

I

The singularity


Our sum is

2 I (

res(z, f(z))) =1

4

The solution is

x2

(x2 + 4)2dx =

1

4


x2

(x2 + 4)2dx =

1

4

Info.

not given

Comment.

no comment

44


45/171

1.26 Problem.


(x3 + 5 x) e(I x)

x4 + 10 x2 + 9 dx

Hint.

Type III

Solution.We denote

f(z) =(z3 + 5 z) e(I z)

z4 + 10 z2 + 9


[{z= 3 I},{z =3 I},{z=I},{z = I}]The singularity

z= 3 I


res(3 I, f(z)) =1

4e(3)

The singularity

z =3 I



z = I


res(I, f(z)) =1

4e(1)

Our sum is

2 I (

res(z, f(z))) = 2 I (1

4e(3) +

1

4e(1))

45


46/171

The solution is

(x3

+ 5 x) e(I x)

x4 + 10 x2 + 9 dx= 2 I ( 1

4e(3) +1

4e(1))


(x3 + 5 x) e(I x)

x4 + 10 x2 + 9 dx=

1

2I cosh(1)+

1

2I cosh(3)1

2I sinh(3)1

2I sinh(1)

Info.

not given

Comment.

no comment

46


47/171

1.27 Problem.


e(I x) (x2 1)x (x2 + 1)

dx

Hint.

Type III

Solution.We denote

f(z) =e(I z) (z2 1)

z (z2 + 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)) =1The singularity

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

The singularityz = I


res(I, f(z)) = e(1)

Our sum is

2 I (

res(z, f(z))) = 2 I (12

+ e(1))

The solution is

e(I x) (x2

1)

x (x2 + 1) dx= 2 I (1

2+ e(

1)

)


e(I x) (x2 1)x (x2 + 1)

dx=

(cos(x) + Isin(x)) (x2 1)x (x2 + 1)

dx

47


48/171

Info.

not givenComment.

no comment

48


49/171

1.28 Problem.


e(I x)

(x2 + 4) (x 1)dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

(z2 + 4) (z 1)


[{z= 2 I},{z =2 I},{z= 1}]The singularity

z= 2 I


res(2 I, f(z)) = ( 110

+ 1

20I) e(2)

The singularity

z =2 I


z= 1


res(1, f(z)) =1

5eI

Our sum is

2 I (

res(z, f(z))) = 2 I (( 110

+ 1

20I) e(2) +

1

10eI)

The solution is

e(I x)

(x2 + 4) (x 1)dx = 2 I ((1

10+

1

20I) e(2) +

1

10eI)

49


50/171


e(I x)

(x2 + 4) (x 1)dx =

cos(x) + Isin(x)(x2 + 4) (x 1) dx

Info.

not given

Comment.

no comment

50


51/171

1.29 Problem.


e(I x)

x (x2 + 1)dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z (z2 + 1)


[{z= 0},{z=I},{z= I}]The singularity

z= 0


res(0, f(z)) = 1

The singularity

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

The singularity

z = I


res(I, f(z)) =12

e(1)

Our sum is

2 I (

res(z, f(z))) = 2 I (1

2 1

2e(1))

The solution is

e(I x)

x (x2 + 1)dx = 2 I (

1

2 1

2e(1))

51


52/171


e(I x)

x (x2 + 1)dx =

cos(x) + Isin(x)x (x2 + 1)

dx

Info.

not given

Comment.

no comment

52


53/171

1.30 Problem.


e(I x)

x (x2 + 9)2dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z (z2 + 9)2


[{z= 0},{z = 3 I},{z =3 I}]The singularity

z= 0


res(0, f(z)) = 1

81The singularity

z= 3 I


res(3 I, f(z)) = 5324

e(3)

The singularity


Our sum is

2 I (

res(z, f(z))) = 2 I ( 1

162 5

324e(3))

The solution is

e(I x)

x (x2 + 9)2dx = 2 I (

1

162 5

324e(3))

53


54/171


e(I x)

x (x2 + 9)2dx =

cos(x) + Isin(x)x (x2 + 9)2

dx

Info.

not given

Comment.

no comment

54


55/171

1.31 Problem.


e(I x)

x (x2 2 x + 2)dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z (z2 2 z+ 2)


[{z= 0},{z = 1 + I},{z = 1 I}]The singularity

z= 0


res(0, f(z)) =1

2The singularity

z= 1 + I


res(1 + I, f(z)) = (14 1

4I) eI e(1)

The singularity

z= 1 Iwill be skipped because the singularity is not in our region.

Our sum is

2 I (

res(z, f(z))) = 2 I (1

4+ (1

4 1

4I) eI e(1))

The solution is

e(I x)

x (x2 2 x + 2)dx = 2 I (1

4+ (1

4 1

4I) eI e(1))

55


56/171


e(I x)

x (x2 2 x + 2)dx =

cos(x) + Isin(x)x (x2 2 x + 2) dx

Info.

not given

Comment.

no comment

56


57/171

1.32 Problem.


e(I x)

x2 + 1dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z2 + 1




z = I


res(I, f(z)) =12

I e(1)

Our sum is

2 I (

res(z, f(z))) = e(1)

The solution is

e(I x)

x2 + 1dx = e(1)


e(I x)

x2 + 1dx = sinh(1) + cosh(1)

Info.

not given

Comment.

no comment

57


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1.33 Problem.


e(I x)

x2 + 4 x + 20dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z2 + 4 z+ 20


[{z=2 4 I},{z=2 + 4 I}]The singularity

z=2 4 Iwill be skipped because the singularity is not in our region.The singularity


res(2 + 4 I, f(z)) =18

I e(4)

(eI)2

Our sum is

2 I (

res(z, f(z))) =1

4

e(4)

(eI)2

The solution is

e(I x)

x2 + 4 x + 20dx =

1

4

e(4)

(eI)2


e(I x)

x2 + 4 x + 20dx =1

4I sin(2 4 I) +1

4 cos(2 4 I)

Info.

not given

Comment.

no comment

58


59/171

1.34 Problem.


e(I x)

x2 5 x + 6dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z2 5 z+ 6We find singularities

[{z= 2},{z= 3}]The singularity

z= 2


res(2, f(z)) =(eI)2The singularity

z= 3


res(3, f(z)) = (eI)3

Our sum is

2 I (

res(z, f(z))) = 2 I (12

(eI)2 +1

2(eI)3)

The solution is

e(I x)

x2 5 x + 6 dx = 2 I (1

2(eI)2 +

1

2(eI)3)


e(I x)

x2 5 x + 6dx =

cos(x) + Isin(x)

x2 5 x + 6 dx

Info.

not given

Comment.

no comment

59


60/171

1.35 Problem.


e(I x)

x3 + 1dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z3 + 1


[{z=1},{z = 12 1

2I

3},{z= 1

2+

1

2I

3}]

The singularity


res(1, f(z)) = 13

1

eIThe singularity

z=1

2 1

2I

3


z=1

2+

1

2I

3


res(1

2+

1

2I

3, f(z)) = 2

(1)(1/2 1 )3 I

e(3)

3 3

e(3)

Our sum is

2 I ( res(z, f(z))) = 2 I 16

1

eI

+ 2 (1)(1/2 1 )

3 I

e(3) 3 3e(3)The solution is

e(I x)

x3 + 1dx = 2 I

1

6

1

eI + 2

(1)(1/2 1 )3 I

e(3)

3 3

e(3)

60


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e(I x)

x3 + 1dx =

cos(x) + Isin(x)x3 + 1

dx

Info.

not given

Comment.

no comment

61


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1.36 Problem.


e(I x)

x4 1dx

Hint.

Type III

Solution.We denote

f(z) = e(I z)

z4 1


[{z =1},{z =I},{z = I},{z= 1}]The singularity


res(1, f(z)) =14

1

eIThe singularity

z=I


z = I


res(I, f(z)) =1

4I e(1)

The singularity

z= 1


res(1, f(z)) =

1

4eI

Our sum is

2 I (

res(z, f(z))) = 2 I (18

1

eI +

1

4I e(1) +

1

8eI)

62


63/171

The solution is

e(I x)

x4 1dx = 2 I (18 1eI +14I e(1) +18eI)


e(I x)

x4 1dx =

cos(x) + Isin(x)

x4 1 dx

Info.

not given

Comment.

no comment

63


64/171

1.37 Problem.


e(I x)

x dx

Hint.

Type III

Solution.We denote

f(z) =e(I z)

z

We find singularities[{z= 0}]

The singularity

z= 0


res(0, f(z)) = 1

Our sum is

2 I (

res(z, f(z))) =I

The solution is

e(I x)

x dx= I


e(I x)

x dx=

cos(x) + Isin(x)

x dx

Info.

not given

Comment.no comment

64


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1.38 Problem.


x e(I x)

1 x4 dx

Hint.

Type III

Solution.We denote

f(z) =z e(I z)

1 z4


[{z =1},{z =I},{z = I},{z= 1}]The singularity


res(1, f(z)) =14

1

eIThe singularity

z=I


z = I


res(I, f(z)) =1

4e(1)

The singularity

z= 1


res(1, f(z)) =1

4eI

Our sum is

2 I (

res(z, f(z))) = 2 I (18

1

eI +

1

4e(1) 1

8eI)

65


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The solution is

x e(I x)

1 x4 dx= 2 I (18 1eI +14e(1) 18eI)


x e(I x)

1 x4 dx=

x (cos(x) + Isin(x))1 + x4 dx

Info.

not given

Comment.

no comment

66


67/171

1.39 Problem.


x e(I x)

x2 + 4 x + 20dx

Hint.

Type III

Solution.We denote

f(z) = z e(I z)

z2 + 4 z+ 20



z=2 4 Iwill be skipped because the singularity is not in our region.The singularity


res(2 + 4 I, f(z)) =1

8

I(4 I e(4)

2 e(4))

(eI)2

Our sum is

2 I (

res(z, f(z))) =1

4

(4 I e(4) 2 e(4))(eI)2

The solution is

x e(I x)

x2 + 4 x + 20dx =

1

4

(4 I e(4) 2 e(4))(eI)2


x e(I x)

x2 + 4 x + 20dx =

sin(2 4 I) + I cos(2 4 I) +12

I sin(2 4 I) 12

cos(2 4 I)

67


68/171

Info.

not givenComment.

no comment

68


69/171

1.40 Problem.


x e(I x)

x2 + 9dx

Hint.

Type III

Solution.We denote

f(z) =z e(I z)

z2 + 9



z= 3 I


res(3 I, f(z)) =1

2e(3)

The singularity


Our sum is

2 I (

res(z, f(z))) =I e(3)

The solution is

x e(I x)

x2 + 9dx= I e(3)


x e(I x)

x2 + 9dx = I cosh(3) I sinh(3)

Info.

not given

Comment.

no comment

69


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1.41 Problem.


x e(I x)

x2 2 x + 10dx

Hint.

Type III

Solution.We denote

f(z) = z e(I z)

z2 2 z+ 10

We find singularities[{z = 1 3 I},{z= 1 + 3 I}]

The singularity

z = 1 3 Iwill be skipped because the singularity is not in our region.The singularity

z = 1 + 3 I


res(1 + 3 I, f(z)) =

1

6

I(3 I eI e(3) + eI e(3))

Our sum is

2 I (

res(z, f(z))) =1

3 (3 I eI e(3) + eI e(3))

The solution is

x e(I x)

x2 2 x + 10dx =1

3 (3 I eI e(3) + eI e(3))


x e(I x)

x2

2 x + 10

dx =

sin(1 + 3 I) + I cos(1 + 3 I) +13

I sin(1 + 3 I) +1

3 cos(1 + 3 I)

Info.

not given

70


71/171

Comment.

no comment

71


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1.42 Problem.


x e(I x)

x2 5 x + 6dx

Hint.

Type III

Solution.We denote

f(z) = z e(I z)

z2 5 z+ 6We find singularities


z= 2


res(2, f(z)) =2 (eI)2The singularity

z= 3


res(3, f(z)) = 3 (eI)3

Our sum is

2 I (

res(z, f(z))) = 2 I ((eI)2 +32

(eI)3)

The solution is

x e(I x)

x2 5 x + 6dx = 2 I ((eI)2 +

3

2(eI)3)


x e(I x)

x2 5 x + 6 dx =

x (cos(x) + Isin(x))

x2 5 x + 6 dx

Info.

not given

Comment.

no comment

72


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1.43 Problem.


e(I x)

x2 dx

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 withthe residue theorem ...

Our sum is

2 I (

res(z, f(z))) = 2 I

The solution is

e(I x)

x2 dx= 2 I


e(I x)

x2 dx=

Info.

not given

Comment.

no comment

73


74/171

1.44 Problem.


x3 e(I x)

x4 + 5 x2 + 4dx

Hint.

Type III

Solution.We denote

f(z) = z3 e(I z)

z4 + 5 z2 + 4


[{z= 2 I},{z =I},{z= I},{z =2 I}]The singularity

z= 2 I


res(2 I, f(z)) =2

3e(2)

The singularity

z=I


z = I


res(I, f(z)) =16

e(1)

The singularity


Our sum is

2 I (

res(z, f(z))) = 2 I (2

3e(2) 1

6e(1))

74


75/171

The solution is

x3

e(I x)

x4 + 5 x2 + 4dx = 2 I ( 2

3e(2) 1

6e(1))


x3 e(I x)

x4 + 5 x2 + 4dx =1

3I cosh(1)+

4

3I cosh(2)4

3I sinh(2)+

1

3I sinh(1)

Info.

not given

Comment.

no comment

75


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2 Zero Sum theorem for residues problems

We recall the definition:

Notation. For a functionfholomorphic on some neighbourhood of infinity wedefine

res(f, ) = res1

z2 f

1

z

, 0

.

We will solve several problems using the following theorem:

Theorem. (Zero Sum theorem for residues) For a function f holomorphic inthe extended complex plane C {}with at most finitely many exceptions thesum of residues is zero, i.e.

wC{}

res(f, w) = 0 .

Notation.(Examplef(z) = 1/z) Projecting on the Riemann sphere we observethe north pole on globe (i.e.) with the residue -1 and at the south pole (theorigin) with residue 1. Green is the zero level on the Riemann sphere while theblue gradually turning red is the imaginary part of log z demonstrating thatZero Sum theorem holds for 1/z.

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2.1 Problem.

Check the Zero Sum theorem for the following functionz2 + 1

z4 + 1

Hint.

no hint

Solution.We denote

f(z) =z2 + 1

z4 + 1


[{z = 12

2+

1

2I

2},{z =1

2

21

2I

2},{z = 1

2

21

2I

2},{z =1

2

2+

1

2I

2}]

The singularity

z=1

2

2 +

1

2I

2

adds the following residue

res(f(z), 1

2

2 +

1

2I

2) =

1 + I

2 I


z =12

2 12

I2


res(f(z),12

2 1

2I

2) =

1 I2 I

2 2 2

The singularity

z=1

2

2 1

2I

2


res(f(z), 1

2 2 1

2I2) = 1 + I

2 I2 + 2 2The singularity

z =12

2 +

1

2I

2

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res(f(z),1

2 2 +1

2I2) = 1

I

2 I2 + 2 2At infinity we get the residue

res(f(z),) = 1 + I2 I

2 2 2 +

1 I2 I

2 2 2 +

1 + I2 I

2 + 2

2

+ 1 I

2 I

2 + 2

2

and finally we obtain the sumres(f(z), z) = 0

Info.

not given

Comment.

no comment

78


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2.2 Problem.

Check the Zero Sum theorem for the following functionz2 1

(z2 + 1)2

Hint.

no hint

Solution.We denote

f(z) = z2 1(z2 + 1)2



z=Iadds the following residue

res(f(z),I) = 0The singularity

z = I


res(f(z), I) = 0

At infinity we get the residue

res(f(z),) = 0and finally we obtain the sum

res(f(z), z) = 0

Info.

not given

Comment.

no comment

79


80/171

2.3 Problem.

Check the Zero Sum theorem for the following functionz2 z+ 2

z4 + 10 z2 + 9

Hint.

no hint

Solution.We denote

f(z) = z2 z+ 2z4 + 10 z2 + 9


[{z=3 I},{z =I},{z = I},{z= 3 I}]The singularity

z =3 Iadds the following residue

res(f(z),3 I) = 116

+ 7

48I

The singularity


res(f(z),I) =116

+ 1

16I

The singularity

z = I


res(f(z), I) =116

116

I

The singularity

z= 3 I


res(f(z), 3 I) = 1

16 7

48I

80


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res(f(z), z) = 0

Info.

not given

Comment.

no comment

81


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2.4 Problem.

Check the Zero Sum theorem for the following function1

(z2 + 1) (z2 + 4)2

Hint.

no hint

Solution.We denote

f(z) = 1

(z2 + 1) (z2 + 4)2


[{z=I},{z = 2 I},{z=2 I},{z = I}]The singularity


res(f(z),I) = 118

I

The singularity

z= 2 I


res(f(z), 2 I) = 11

288I

The singularity


res(f(z),2 I) =11288

I

The singularity

z = I


res(f(z), I) =118

I

82


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res(f(z), z) = 0

Info.

not given

Comment.

no comment

83


84/171

2.5 Problem.


(z2 + 1) (z2 + 9)

Hint.

no hint

Solution.We denote

f(z) = 1

(z2 + 1) (z2 + 9)


[{z=3 I},{z =I},{z = I},{z= 3 I}]The singularity


res(f(z),3 I) =148

I

The singularity


res(f(z),I) = 116

I

The singularity

z = I


res(f(z), I) =116

I

The singularity

z= 3 I


res(f(z), 3 I) = 1

48I

84


85/171



res(f(z), z) = 0

Info.

not given

Comment.

no comment

85


86/171

2.6 Problem.


(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= 2 I


res(f(z), 2 I) =IThe singularity

z = I


res(f(z), I) = I



res(f(z), z) = 0

Info.

not given

Comment.

no comment

86


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2.7 Problem.


(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


res(f(z),3 I) =120

The singularity

z= 2 I


res(f(z), 2 I) =1

5

The singularity

z = I


res(f(z), I) =1

4



res(f(z), z) = 0

87


88/171

Info.

not given

Comment.

no comment

88


89/171

2.8 Problem.


z2 + 1

Hint.

no hint

Solution.We denote

f(z) = 1

z2 + 1




res(f(z),I) = 12

I

The singularity

z = I


res(f(z), I) =12

I



res(f(z), z) = 0

Info.

not given

Comment.

no comment

89


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2.9 Problem.


z3 + 1

Hint.

no hint

Solution.We denote

f(z) = 1

z3 + 1


[{z=1},{z = 12 1

2I

3},{z= 1

2+

1

2I

3}]

The singularity

z =1adds the following residue

res(f(z),1) = 13

The singularity

z=

1

21

2I

3


res(f(z), 1

2 1

2I

3) =2 1

3 I

3 + 3

The singularity

z=1

2+

1

2I

3


res(f(z), 1

2+

1

2I

3) = 2

1

3 I

3

3


res(f(z),) =13

+ 2 1

3 I

3 + 3 2 1

3 I

3 3

90


91/171

and finally we obtain the sum

res(f(z), z) = 0

Info.

not given

Comment.

no comment

91


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2.10 Problem.


z6 + 1

Hint.

no hint

Solution.We denote

f(z) = 1

z6 + 1


[{z =I},{z=12

2 2 I

3},{z = 1

2

2 2 I

3},{z =1

2

2 + 2 I

3},

{z= 12

2 + 2 I

3},{z= I}]

The singularity


res(f(z),I) = 16

I

The singularity

z=12

2 2 I

3


res(f(z),12

2 2 I

3) =16

3

1

(2 2 I3)(5/2)The singularity

z =1

2

2 2 I

3


res(f(z), 1

2

2 2 I

3) =

16

3

1

(2 2 I3)(5/2)

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The singularity

z=12

2 + 2 I3adds the following residue

res(f(z),12

2 + 2 I

3) =16

3

1

(2 + 2 I

3)(5/2)

The singularity

z =1

2

2 + 2 I

3


res(f(z),

1

2

2 + 2 I3) =16

3

1

(2 + 2 I3)(5/2)The singularity

z = I


res(f(z), I) =16

I


res(f(z),) = 0


Info.

not given

Comment.

no comment

93


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2.11 Problem.

Check the Zero Sum theorem for the following functionz

(z2 + 4 z+ 13)2

Hint.

no hint

Solution.We denote

f(z) = z

(z2 + 4 z+ 13)2



z=2 3 Iadds the following residue

res(f(z),2 3 I) =154

I

The singularity

z=2 + 3 I


res(f(z),2 + 3 I) = 154

I



res(f(z), z) = 0

Info.

not given

Comment.

no comment

94


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2.12 Problem.

Check the Zero Sum theorem for the following functionz2

(z2 + 1) (z2 + 9)

Hint.

no hint

Solution.We denote

f(z) = z2

(z2 + 1) (z2 + 9)

We find singularities[{z=3 I},{z =I},{z = I},{z= 3 I}]

The singularity


res(f(z),3 I) = 316

I

The singularity

z=

I


res(f(z),I) =116

I

The singularity

z = I


res(f(z), I) = 1

16I

The singularity

z= 3 I


res(f(z), 3 I) =316

I

95


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res(f(z), z) = 0

Info.

not given

Comment.

no comment

96


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2.13 Problem.


(z2 + 1)3

Hint.

no hint

Solution.We denote

f(z) = z2

(z2 + 1)3




res(f(z),I) = 116

I

The singularity

z = I


res(f(z), I) =116

I



res(f(z), z) = 0

Info.

not given

Comment.

no comment

97


98/171

2.14 Problem.


(z2 + 4)2

Hint.

no hint

Solution.We denote

f(z) = z2

(z2 + 4)2



z= 2 I


res(f(z), 2 I) =18

I

The singularity

z =

2 I


res(f(z),2 I) = 18

I



res(f(z), z) = 0

Info.

not given

Comment.

no comment

98


99/171

2.15 Problem.

Check the Zero Sum theorem for the following function

(z3 + 5 z) e(I z)

z4 + 10 z2 + 9

Hint.

no hint

Solution.We denote

f(z) =(z3 + 5 z) e(I z)

z4 + 10 z2 + 9

We find singularities[{z=3 I},{z =I},{z = I},{z= 3 I}]

The singularity


res(f(z),3 I) = 14

e3

The singularity

z=

I


res(f(z),I) = 14

e

The singularity

z = I


res(f(z), I) =1

4e(1)

The singularity

z= 3 I


res(f(z), 3 I) =1

4e(3)

99


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res(f(z),) =14

e3 14

e 14

e(1) 14

e(3)


Info.

not given

Comment.

no comment

100


101/171

2.16 Problem.


e(I z) (z2 1)z (z2 + 1)

Hint.

no hint

Solution.We denote

f(z) =e(I z) (z2 1)

z (z2 + 1)



z= 0


res(f(z), 0) =1The singularity


res(f(z),I) = eThe singularity

z = I


res(f(z), I) = e(1)


res(f(z),

) = 1

e

e(1)


101


102/171

Info.

not givenComment.

no comment

102


103/171

2.17 Problem.


e(I z)

(z2 + 4) (z 1)Hint.

no hint

Solution.We denote

f(z) = e(I z)

(z2 + 4) (z 1)

We find singularities[{z= 1},{z = 2 I},{z =2 I}]

The singularity

z= 1


res(f(z), 1) =1

5eI

The singularity

z= 2 I


res(f(z), 2 I) = ( 110

+ 1

20I) e(2)

The singularity


res(f(z),2 I) = ( 110

120

I) e2


res(f(z),) =15eI + ( 110 120I) e(2) + ( 110+ 120I) e2and finally we obtain the sum

res(f(z), z) = 0

103


104/171

Info.

not givenComment.

no comment

104


105/171

2.18 Problem.


e(I z)

z (z2 + 1)

Hint.

no hint

Solution.We denote

f(z) = e(I z)

z (z2 + 1)

We find singularities[{z= 0},{z=I},{z= I}]

The singularity

z= 0


res(f(z), 0) = 1

The singularity

z=I


res(f(z),I) =12

e

The singularity

z = I


res(f(z), I) =12

e(1)


res(f(z),) =1 +12e +12e(1)and finally we obtain the sum

res(f(z), z) = 0

105


106/171

Info.

not given

Comment.

no comment

106


107/171

2.19 Problem.


e(I z)

z (z2 + 9)2

Hint.

no hint

Solution.We denote

f(z) = e(I z)

z (z2 + 9)2

We find singularities[{z=3 I},{z = 0},{z= 3 I}]

The singularity


res(f(z),3 I) = 1324

e3

The singularity

z= 0


res(f(z), 0) = 1

81

The singularity

z= 3 I


res(f(z), 3 I) = 5324

e(3)


res(f(z),) = 1324e3 181+ 5324e(3)and finally we obtain the sum

res(f(z), z) = 0

107


108/171

Info.

not givenComment.

no comment

108


109/171

2.20 Problem.


e(I z)

z (z2 2 z+ 2)Hint.

no hint

Solution.We denote

f(z) = e(I z)

z (z2 2 z+ 2)

We find singularities[{z= 0},{z = 1 I},{z = 1 + I}]

The singularity

z= 0


res(f(z), 0) =1

2

The singularity

z= 1 Iadds the following residue

res(f(z), 1 I) = (14

+1

4I) eI e

The singularity

z= 1 + I


res(f(z), 1 + I) = (14 1

4I) eI e(1)


res(f(z),) =12+ ( 14 14I) eI e + ( 14+ 14I) eI e(1)and finally we obtain the sum

res(f(z), z) = 0

109


110/171

Info.

not givenComment.

no comment

110


111/171

2.21 Problem.


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


res(f(z),I) = 12

I e

The singularity

z = I


res(f(z), I) =12

I e(1)


res(f(z),) =12

I e +1

2I e(1)


Info.not given

Comment.

no comment

111


112/171

2.22 Problem.


e(I z)

z2 + 4 z+ 20

Hint.

no hint

Solution.We denote

f(z) = e(I z)

z2 + 4 z+ 20




res(f(z),2 4 I) = 18

I e4

(eI)2

The singularity

z=2 + 4 Iadds the following residue

res(f(z),2 + 4 I) =18

I e(4)

(eI)2


res(f(z),) =18

I e4

(eI)2+

1

8

I e(4)

(eI)2


Info.

not given

Comment.

no comment

112


113/171

2.23 Problem.


e(I z)

z2 5 z+ 6Hint.

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


res(f(z), 2) =(eI)2

The singularity

z= 3


res(f(z), 3) = (eI)3


res(f(z),) = (eI)2 (eI)3


Info.

not given

Comment.

no comment

113


114/171

2.24 Problem.


e(I z)

z3 + 1

Hint.

no hint

Solution.We denote

f(z) = e(I z)

z3 + 1


[{z=1},{z = 12 1

2I

3},{z= 1

2+

1

2I

3}]

The singularity


res(f(z),1) = 13

1

eI

The singularity

z=

1

21

2I

3


res(f(z), 1

2 1

2I

3) =2(1)

(1/2 1)

e(3)

3 I

3 + 3

The singularity

z=1

2+

1

2I

3


res(f(z), 1

2

+1

2

I

3) = 2 (1)(1/2 1 )

3 I

e(3) 3 3e(3)At infinity we get the residue

res(f(z),) =13

1

eI + 2

(1)(1/2 1 )

e(3)

3 I

3 + 3 2 (1)

(1/2 1)

3 I

e(3)

3 3

e(3)

114


115/171

and finally we obtain the sum

res(f(z), z) = 0

Info.

not given

Comment.

no comment

115


116/171

2.25 Problem.


e(I z)

z4 1Hint.

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


res(f(z), 1) =1

4eI

The singularity

z =

1


res(f(z),1) =14

1

eI

The singularity


res(f(z),I) =14

I e

The singularity

z = I


res(f(z), I) =1

4I e(1)

116


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res(f(z),) =14

eI +14

1eI

+14

I e 14

I e(1)


Info.

not given

Comment.

no comment

117


118/171

2.26 Problem.


e(I z)

z

Hint.

no hint

Solution.We denote

f(z) =e(I z)

z


[{z= 0}]The singularity

z= 0


res(f(z), 0) = 1


res(f(z),) =1


Info.

not given

Comment.

no comment

118


119/171

2.27 Problem.


z e(I z)

1 z4Hint.

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


res(f(z), 1) =14

eI

The singularity

z =

1


res(f(z),1) =14

1

eI

The singularity


res(f(z),I) = 14

e

The singularity

z = I


res(f(z), I) =1

4e(1)

119


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res(f(z),) = 14

eI +14

1eI

14

e 14

e(1)


Info.

not given

Comment.

no comment

120


121/171

2.28 Problem.


z e(I z)

z2 + 4 z+ 20

Hint.

no hint

Solution.We denote

f(z) = z e(I z)

z2 + 4 z+ 20




res(f(z),2 4 I) =18

I(4 I e4 + 2 e4)

(eI)2

The singularity

z=2 + 4 Iadds the following residue

res(f(z),2 + 4 I) =18

I(4 I e(4) 2 e(4))(eI)2


res(f(z),) = 18

I(4 I e4 + 2 e4)

(eI)2 +

1

8

I(4 I e(4) 2 e(4))(eI)2


Info.

not given

Comment.

no comment

121


122/171

2.29 Problem.


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


res(f(z),3 I) = 12

e3

The singularity

z= 3 I


res(f(z), 3 I) =1

2e(3)


res(f(z),) =12

e3 12

e(3)


Info.not given

Comment.

no comment

122


123/171

2.30 Problem.


z e(I z)

z2 2 z+ 10Hint.

no hint

Solution.We denote

f(z) = z e(I z)

z2 2 z+ 10

We find singularities[{z = 1 3 I},{z= 1 + 3 I}]

The singularity

z = 1 3 Iadds the following residue

res(f(z), 1 3 I) =16

I(3 I eI e3 eI e3)

The singularity

z = 1 + 3 I


res(f(z), 1 + 3 I) =16

I(3 I eI e(3) + eI e(3))


res(f(z),) = 16

I(3 I eI e3 eI e3) +16

I(3 I eI e(3) + eI e(3))


Info.not given

Comment.

no comment

123


124/171

2.31 Problem.


z e(I z)

z2 5 z+ 6Hint.

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


res(f(z), 2) =2 (eI)2

The singularity

z= 3


res(f(z), 3) = 3 (eI)3


res(f(z),) = 2 (eI)2 3 (eI)3


Info.

not given

Comment.

no comment

124


125/171

2.32 Problem.


e(I z)

z2

Hint.

no hint

Solution.We denote

f(z) =e(I z)

z2



z= 0


res(f(z), 0) =I


res(f(z),) =I


Info.

not given

Comment.

no comment

125


126/171

2.33 Problem.


z3 e(I z)

z4 + 5 z2 + 4

Hint.

no hint

Solution.We denote

f(z) = z3 e(I z)

z4 + 5 z2 + 4

We find singularities[{z=I},{z = 2 I},{z=2 I},{z = I}]

The singularity


res(f(z),I) =16

e

The singularity

z= 2 I


res(f(z), 2 I) =2

3e(2)

The singularity


res(f(z),2 I) = 23

e2

The singularity

z = I


res(f(z), I) =16

e(1)

126


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res(f(z),) = 16

e 23

e(2) 23

e2 +16

e(1)


Info.

not given

Comment.

no comment

127


128/171

2.34 Problem.

Check the Zero Sum theorem for the following functionz2 + 1

ez

Hint.

no hint

Solution.We denote

f(z) =z2 + 1

ez


[{z=}]At infinity we get the residue


res(f(z), z) = 0

Info.

not given

Comment.

no comment

128


129/171

2.35 Problem.

Check the Zero Sum theorem for the following functionz2 + z 1z2 (z 1)

Hint.

no hint

Solution.We denote

f(z) =z2 + z 1

z2 (z 1)



z= 1


res(f(z), 1) = 1

The singularity

z= 0


res(f(z), 0) = 0


res(f(z),) =1and finally we obtain the sum

res(f(z), z) = 0

Info.

not given

Comment.

no comment

129


130/171

2.36 Problem.

Check the Zero Sum theorem for the following functionz2 2 z+ 5

(z 2) (z2 + 1)Hint.

no hint

Solution.We denote

f(z) = z2 2 z+ 5(z 2) (z2 + 1)



z= 2


res(f(z), 2) = 1

The singularity

z=I


res(f(z),I) =IThe singularity

z = I


res(f(z), I) = I



res(f(z), z) = 0

130


131/171

Info.

not givenComment.

no comment

131


132/171

2.37 Problem.


(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


res(f(z), 1) =1

2

The singularity


res(f(z),1) =12



res(f(z), z) = 0

Info.

not given

Comment.

no comment

132


133/171

2.38 Problem.


z (1 z2)Hint.

no hint

Solution.We denote

f(z) = 1

z (1 z2)We find singularities

[{z = 1},{z =1},{z= 0}]The singularity

z= 1


res(f(z), 1) =1

2

The singularity


res(f(z),1) =12

The singularity

z= 0


res(f(z), 0) = 1



res(f(z), z) = 0

133


134/171

Info.

not givenComment.

no comment

134


135/171

2.39 Problem.


z (z2 + 4)2

Hint.

no hint

Solution.We denote

f(z) = 1

z (z2 + 4)2


[{z= 0},{z = 2 I},{z =2 I}]The singularity

z= 0


res(f(z), 0) = 1

16

The singularity

z= 2 I


res(f(z), 2 I) =1

32

The singularity


res(f(z),2 I) =132



res(f(z), z) = 0

135


136/171

Info.

not given

Comment.

no comment

136


137/171

2.40 Problem.


z (z 1)Hint.

no hint

Solution.We denote

f(z) = 1

z (z 1)We find singularities


z= 1


res(f(z), 1) = 1

The singularity

z= 0


res(f(z), 0) =1At infinity we get the residue


res(f(z), z) = 0

Info.

not given

Comment.

no comment

137


138/171

2.41 Problem.


z (z 1)Hint.

no hint

Solution.We denote

f(z) = 1

z (z 1)We find singularities


z= 1


res(f(z), 1) = 1

The singularity

z= 0


res(f(z), 0) =1At infinity we get the residue


res(f(z), z) = 0

Info.

not given

Comment.

no comment

138


139/171

2.42 Problem.


(z2 + 1)2

Hint.

no hint

Solution.We denote

f(z) = 1

(z2 + 1)2




res(f(z),I) = 14

I

The singularity

z = I


res(f(z), I) =14

I



res(f(z), z) = 0

Info.

not given

Comment.

no comment

139


140/171

2.43 Problem.


z3 z5Hint.

no hint

Solution.We denote

f(z) = 1

z3 z5We find singularities


z= 1


res(f(z), 1) =1

2

The singularity

z =1


res(f(z),1) =12

The singularity

z= 0


res(f(z), 0) = 1


res(f(z),

) = 0


140


141/171

Info.

not givenComment.

no comment

141


142/171

2.44 Problem.


z4 + 1

Hint.

no hint

Solution.We denote

f(z) = 1

z4 + 1


[{z = 12

2+

1

2I

2},{z =1

2

21

2I

2},{z = 1

2

21

2I

2},{z =1

2

2+

1

2I

2}]

The singularity

z=1

2

2 +

1

2I

2


res(f(z), 1

2

2 +

1

2I

2) =

1

2 I


z =12

2 12

I2


res(f(z),12

2 1

2I

2) = 1

2 I


z=1

2

2 1

2I

2


res(f(z), 1

2

2

1

2I

2) =

1

2 I2 + 2 2The singularity

z =12

2 +

1

2I

2

142


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res(f(z),12

2 +12

I2) = 12 I

2 + 2

2



res(f(z), z) = 0

Info.

not given

Comment.

no comment

143


144/171

2.45 Problem.


z z3Hint.

no hint

Solution.We denote

f(z) = 1

z z3We find singularities


z= 1


res(f(z), 1) =1

2

The singularity

z =1


res(f(z),1) =12

The singularity

z= 0


res(f(z), 0) = 1


res(f(z),

) = 0


144


145/171

Info.

not givenComment.

no comment

145


146/171

2.46 Problem.


z

Hint.

no hint

Solution.We denote

f(z) =1

z



z= 0


res(f(z), 0) = 1



res(f(z), z) = 0

Info.

not given

Comment.

no comment

146


147/171

2.47 Problem.

Check the Zero Sum theorem for the following functionez

(z2 + 9) z2

Hint.

no hint

Solution.We denote

f(z) = ez

(z2 + 9) z2


[{z =},{z =3 I},{z= 0},{z = 3 I}]The singularity


res(f(z),3 I) =154

I

(eI)3

The singularity

z= 0


res(f(z), 0) =1

9

The singularity

z= 3 I


res(f(z

Complex Problems

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Transcript of Complex Problems