complex problems
TRANSCRIPT
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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.
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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).
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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|
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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
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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
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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
is in our region and we will add the following residue
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
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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
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1.2 Problem.
Using the Residue theorem evaluate 20
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
We find singularities
[{z = 0},{z=
2+1},{z= 1
2},{z =
21},{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
2I
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
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
768 544 2The singularity
z =1
2
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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
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
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1.3 Problem.
Using the Residue theorem evaluate 20
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
We find singularities
[{z= 0},{z = 12},{z= 2}]
The singularity
z= 0
is in our region and we will add the following residue
res(0, f(z)) =54
I
The singularity
z =1
2
is in our region and we will add the following residue
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))) =
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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
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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
We find singularities
[{z=13
I
3},{z = 13
I
3},{z = I
3},{z =I
3}]The singularity
z =13
I
3
is in our region and we will add the following residue
res(13
I
3, f(z)) =14
I
The singularity
z= 13
I3
is in our region and we will add the following residue
res(1
3I
3, f(z)) =1
4I
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
2 I (
res(z, f(z))) =
12
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The solution is
20
1sin(x)2 + 4 cos(x)2
dx =
We can try to solve it using real calculus and obtain the result 20
1
sin(x)2 + 4 cos(x)2dx =
Info.
not given
Comment.
no comment
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1.5 Problem.
Using the Residue theorem evaluate 20
1
13 + 12 sin(x)dx
Hint.
Type I
Solution.We denote
f(z) = I(13 6 I(z1
z)) z
We find singularities
[{z=23
I},{z=32
I}]The singularity
z =23
I
is in our region and we will add the following residue
res(23
I , f(z)) =15
I
The singularity
z =3
2I
will be skipped because the singularity is not in our region.
Our sum is
2 I (
res(z, f(z))) =2
5
The solution is 2 0
1
13 + 12 sin(x)dx =
2
5
We can try to solve it using real calculus and obtain the result 2
0
1
13 + 12 sin(x)dx =
2
5
Info.
not given
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Comment.
no comment
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1.6 Problem.
Using the Residue theorem evaluate 20
1
2 + cos(x)dx
Hint.
Type I
Solution.We denote
f(z) = I(2 +
1
2z +
1
2
1
z) 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
res(2 +
3, f(z)) =13
I
3
The singularity
z =2
3
will be skipped because the singularity is not in our region.
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
We can try to solve it using real calculus and obtain the result 2 0
1
2 + cos(x)dx =
2
3
3
Info.
not given
Comment.
no comment
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1.7 Problem.
Using the Residue theorem evaluate 20
1
(2 + cos(x))2dx
Hint.
Type I
Solution.We denote
f(z) = I(2 +
1
2z +
1
2
1
z)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
res(2 +
3, f(z)) =13
I+1
3(2
3I+
1
3I
3)
3
The singularity
z =2
3
will be skipped because the singularity is not in our region.
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)
We can try to solve it using real calculus and obtain the result 2 0
1
(2 + cos(x))2dx =
4
9
3
Info.
not given
Comment.
no comment
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1.8 Problem.
Using the Residue theorem evaluate 20
1
2 + sin(x)dx
Hint.
Type I
Solution.We denote
f(z) = I(2 1
2I(z 1
z)) 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
res(2 I+ I
3, f(z)) =13
I
3
The singularity
z =2 I I
3
will be skipped because the singularity is not in our region.
Our sum is
2 I (
res(z, f(z))) =2
3
3
The solution is 20
1
2 + sin(x)dx =
2
3
3
We can try to solve it using real calculus and obtain the result 20
1
2 + sin(x)dx =
2
3
3
Info.
not given
Comment.
no comment
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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
We find singularities
[{z=2},{z =12}]
The singularity
z =2will 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
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
We can try to solve it using real calculus and obtain the result
2
0
1
5 + 4 cos(x)dx =
2
3
Info.
not given
Comment.
no comment
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1.10 Problem.
Using the Residue theorem evaluate 20
1
5 + 4 sin(x)dx
Hint.
Type I
Solution.We denote
f(z) = I(5 2 I(z 1
z)) z
We find singularities
[{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
is in our region and we will add the following residue
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
We can try to solve it using real calculus and obtain the result
2
0
1
5 + 4 sin(x)dx =
2
3
Info.
not given
Comment.
no comment
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1.11 Problem.
Using the Residue theorem evaluate 20
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
We find singularities
[{z= 0},{z = 12},{z= 2}]
The singularity
z= 0
is in our region and we will add the following residue
res(0, f(z)) = I
The singularity
z =1
2
is in our region and we will add the following residue
res(1
2, f(z)) =5
3I
The singularity
z= 2
will be skipped because the singularity is not in our region.
Our sum is
2 I ( res(z, f(z))) =4
3
The solution is 20
cos(x)5
4 cos(x)
dx=4
3
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We can try to solve it using real calculus and obtain the result
20
cos(x)5
4 cos(x)
dx= 43
Info.
not given
Comment.
no comment
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1.12 Problem.
Using the Residue theorem evaluate
x2 + 1
x4 + 1dx
Hint.
Type II
Solution.We denote
f(z) =z2 + 1
z4 + 1
We find singularities
[{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
will be skipped because the singularity is not in our region.The singularity
z=1
2
2 +
1
2I
2
is in our region and we will add the following residue
res( 12
2 +12
I2, f(z)) = 1 + I2 I
2 2 2
The singularity
z =12
2 +
1
2I
2
is in our region and we will add the following residue
res(12
2 +
1
2I
2, f(z)) =
1 I2 I
2 + 2
2
The singularity
z=1
2
2 1
2I
2
will be skipped because the singularity is not in our region.
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)
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The solution is
x2
+ 1x4 + 1dx = 2 I ( 1 + I2 I
2 2 2 + 1 I2 I2 + 2 2 )
We can try to solve it using real calculus and obtain the result
x2 + 1
x4 + 1dx =
2
Info.
not given
Comment.
no comment
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1.13 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
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
We can try to solve it using real calculus and obtain the result
x2 1(x2 + 1)2
dx = 0
Info.
not given
Comment.
no comment
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1.14 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
res(3 I, f(z)) = 1
16 7
48I
The singularity
z =3 I
will be skipped because the singularity is not in our region.The singularity
z=Iwill 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)) =116
116
I
Our sum is
2 I (
res(z, f(z))) = 5
12
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The solution is
x2
x + 2x4 + 10 x2 + 9dx = 512
We can try to solve it using real calculus and obtain the result
x2 x + 2x4 + 10 x2 + 9
dx = 5
12
Info.
not given
Comment.
no comment
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1.15 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
res(2 I, f(z)) = 11
288I
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)) =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
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The solution is
1(x2 + 1) (x2 + 4)2 dx = 5144
We can try to solve it using real calculus and obtain the result
1
(x2 + 1) (x2 + 4)2dx =
5
144
Info.
not given
Comment.
no comment
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1.16 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
res(3 I, f(z)) = 1
48I
The singularity
z =3 I
will be skipped because the singularity is not in our region.The singularity
z=Iwill 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)) =116
I
Our sum is
2 I (
res(z, f(z))) = 1
12
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The solution is
1(x2 + 1) (x2 + 9)dx = 112
We can try to solve it using real calculus and obtain the result
1
(x2 + 1) (x2 + 9)dx =
1
12
Info.
not given
Comment.
no comment
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1.17 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
res(2 I, f(z)) =IThe singularity
z = I
is in our region and we will add the following residue
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
We can try to solve it using real calculus and obtain the result
1
(x I) (x 2 I)dx = 0
Info.
not given
Comment.
no comment
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1.18 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
res(2 I, f(z)) =1
5The 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
2 I (
res(z, f(z))) = 1
10I
The solution is
1
(x I) (x 2 I) (x + 3 I)dx = 1
10I
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We can try to solve it using real calculus and obtain the result
1(x I) (x 2 I) (x + 3 I)dx = 110I
Info.
not given
Comment.
no comment
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1.19 Problem.
Using the Residue theorem evaluate
1
1 + x2dx
Hint.
Type II
Solution.We denote
f(z) = 1
z2 + 1
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
is in our region and we will add the following residue
res(I, f(z)) =12
I
Our sum is
2 I (
res(z, f(z))) =
The solution is
1
1 + x2dx =
We can try to solve it using real calculus and obtain the result
1
1 + x2dx =
Info.
not given
Comment.
no comment
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1.20 Problem.
Using the Residue theorem evaluate
1
x3 + 1dx
Hint.
Type II
Solution.We denote
f(z) = 1
z3 + 1
We find singularities
[{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
will be skipped because the singularity is not in our region.The singularity
z=1
2+
1
2I
3
is in our region and we will add the following residue
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 )
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We can try to solve it using real calculus and obtain the result
1x3 + 1 dx =
1x3 + 1dx
Info.
not given
Comment.
no comment
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1.21 Problem.
Using the Residue theorem evaluate
1
x6 + 1dx
Hint.
Type II
Solution.We denote
f(z) = 1
z6 + 1
We find singularities
[{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=Iwill 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)) =16
I
The singularity
z =1
2
2 + 2 I
3
is in our region and we will add the following residue
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
will be skipped because the singularity is not in our region.The singularity
z =1
2
2 2 I
3
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will be skipped because the singularity is not in our region.The singularity
z=12
2 2 I
3
is in our region and we will add the following residue
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) )
We can try to solve it using real calculus and obtain the result
1
x6 + 1dx =
2
3
Info.
not given
Comment.
no comment
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1.22 Problem.
Using the Residue theorem evaluate
x
(x2 + 4 x + 13)2dx
Hint.
Type II
Solution.We denote
f(z) = z
(z2 + 4 z+ 13)2
We find singularities
[{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
We can try to solve it using real calculus and obtain the result
x
(x2 + 4 x + 13)2dx =1
27
Info.
not given
Comment.
no comment
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1.23 Problem.
Using the Residue theorem evaluate
x2
(x2 + 1) (x2 + 9)dx
Hint.
Type II
Solution.We denote
f(z) = z2
(z2 + 1) (z2 + 9)
We find singularities
[{z= 3 I},{z =3 I},{z=I},{z = I}]The singularity
z= 3 I
is in our region and we will add the following residue
res(3 I, f(z)) =316
I
The singularity
z =3 I
will be skipped because the singularity is not in our region.The singularity
z=Iwill 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
16I
Our sum is
2 I (
res(z, f(z))) =1
4
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The solution is
x2
(x2 + 1) (x2 + 9)dx = 14
We can try to solve it using real calculus and obtain the result
x2
(x2 + 1) (x2 + 9)dx =
1
4
Info.
not given
Comment.
no comment
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1.24 Problem.
Using the Residue theorem evaluate
x2
(x2 + 1)3dx
Hint.
Type II
Solution.We denote
f(z) = z2
(z2 + 1)3
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
is in our region and we will add the following residue
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
We can try to solve it using real calculus and obtain the result
x2
(x2 + 1)3dx =
1
8
Info.
not given
Comment.
no comment
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1.25 Problem.
Using the Residue theorem evaluate
x2
(x2 + 4)2dx
Hint.
Type II
Solution.We denote
f(z) = z2
(z2 + 4)2
We find singularities
[{z= 2 I},{z =2 I}]The singularity
z= 2 I
is in our region and we will add the following residue
res(2 I, f(z)) =18
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))) =1
4
The solution is
x2
(x2 + 4)2dx =
1
4
We can try to solve it using real calculus and obtain the result
x2
(x2 + 4)2dx =
1
4
Info.
not given
Comment.
no comment
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1.26 Problem.
Using the Residue theorem evaluate
(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
We find singularities
[{z= 3 I},{z =3 I},{z=I},{z = I}]The singularity
z= 3 I
is in our region and we will add the following residue
res(3 I, f(z)) =1
4e(3)
The singularity
z =3 I
will be skipped because the singularity is not in our region.The singularity
z=Iwill 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
4e(1)
Our sum is
2 I (
res(z, f(z))) = 2 I (1
4e(3) +
1
4e(1))
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The solution is
(x3
+ 5 x) e(I x)
x4 + 10 x2 + 9 dx= 2 I ( 1
4e(3) +1
4e(1))
We can try to solve it using real calculus and obtain the result
(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
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1.27 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
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)
)
We can try to solve it using real calculus and obtain the result
e(I x) (x2 1)x (x2 + 1)
dx=
(cos(x) + Isin(x)) (x2 1)x (x2 + 1)
dx
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Info.
not givenComment.
no comment
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1.28 Problem.
Using the Residue theorem evaluate
e(I x)
(x2 + 4) (x 1)dx
Hint.
Type III
Solution.We denote
f(z) = e(I z)
(z2 + 4) (z 1)
We find singularities
[{z= 2 I},{z =2 I},{z= 1}]The singularity
z= 2 I
is in our region and we will add the following residue
res(2 I, f(z)) = ( 110
+ 1
20I) e(2)
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
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)
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We can try to solve it using real calculus and obtain the result
e(I x)
(x2 + 4) (x 1)dx =
cos(x) + Isin(x)(x2 + 4) (x 1) dx
Info.
not given
Comment.
no comment
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1.29 Problem.
Using the Residue theorem evaluate
e(I x)
x (x2 + 1)dx
Hint.
Type III
Solution.We denote
f(z) = e(I z)
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)) = 1
The singularity
z=Iwill 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)) =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))
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We can try to solve it using real calculus and obtain the result
e(I x)
x (x2 + 1)dx =
cos(x) + Isin(x)x (x2 + 1)
dx
Info.
not given
Comment.
no comment
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1.30 Problem.
Using the Residue theorem evaluate
e(I x)
x (x2 + 9)2dx
Hint.
Type III
Solution.We denote
f(z) = e(I z)
z (z2 + 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
81The singularity
z= 3 I
is in our region and we will add the following residue
res(3 I, f(z)) = 5324
e(3)
The singularity
z =3 Iwill be skipped because the singularity is not in our region.
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))
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We can try to solve it using real calculus and obtain the result
e(I x)
x (x2 + 9)2dx =
cos(x) + Isin(x)x (x2 + 9)2
dx
Info.
not given
Comment.
no comment
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1.31 Problem.
Using the Residue theorem evaluate
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)
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
2The singularity
z= 1 + I
is in our region and we will add the following residue
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))
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We can try to solve it using real calculus and obtain the result
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
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1.32 Problem.
Using the Residue theorem evaluate
e(I x)
x2 + 1dx
Hint.
Type III
Solution.We denote
f(z) = e(I z)
z2 + 1
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
is in our region and we will add the following residue
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)
We can try to solve it using real calculus and obtain the result
e(I x)
x2 + 1dx = sinh(1) + cosh(1)
Info.
not given
Comment.
no comment
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1.33 Problem.
Using the Residue theorem evaluate
e(I x)
x2 + 4 x + 20dx
Hint.
Type III
Solution.We denote
f(z) = e(I z)
z2 + 4 z+ 20
We find singularities
[{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
z=2 + 4 Iis in our region and we will add the following residue
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
We can try to solve it using real calculus and obtain the result
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
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1.34 Problem.
Using the Residue theorem evaluate
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
is on the real line and we will add one half of the following residue
res(2, f(z)) =(eI)2The 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
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)
We can try to solve it using real calculus and obtain the result
e(I x)
x2 5 x + 6dx =
cos(x) + Isin(x)
x2 5 x + 6 dx
Info.
not given
Comment.
no comment
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1.35 Problem.
Using the Residue theorem evaluate
e(I x)
x3 + 1dx
Hint.
Type III
Solution.We denote
f(z) = e(I z)
z3 + 1
We find singularities
[{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
1
eIThe singularity
z=1
2 1
2I
3
will be skipped because the singularity is not in our region.The singularity
z=1
2+
1
2I
3
is in our region and we will add the following residue
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)
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We can try to solve it using real calculus and obtain the result
e(I x)
x3 + 1dx =
cos(x) + Isin(x)x3 + 1
dx
Info.
not given
Comment.
no comment
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1.36 Problem.
Using the Residue theorem evaluate
e(I x)
x4 1dx
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 =1is on the real line and we will add one half of the following residue
res(1, f(z)) =14
1
eIThe 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
4I e(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
4eI
Our sum is
2 I (
res(z, f(z))) = 2 I (18
1
eI +
1
4I e(1) +
1
8eI)
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The solution is
e(I x)
x4 1dx = 2 I (18 1eI +14I e(1) +18eI)
We can try to solve it using real calculus and obtain the result
e(I x)
x4 1dx =
cos(x) + Isin(x)
x4 1 dx
Info.
not given
Comment.
no comment
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1.37 Problem.
Using the Residue theorem evaluate
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
is on the real line and we will add one half of the following residue
res(0, f(z)) = 1
Our sum is
2 I (
res(z, f(z))) =I
The solution is
e(I x)
x dx= I
We can try to solve it using real calculus and obtain the result
e(I x)
x dx=
cos(x) + Isin(x)
x dx
Info.
not given
Comment.no comment
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1.38 Problem.
Using the Residue theorem evaluate
x e(I x)
1 x4 dx
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 =1is on the real line and we will add one half of the following residue
res(1, f(z)) =14
1
eIThe 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
4e(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
4eI
Our sum is
2 I (
res(z, f(z))) = 2 I (18
1
eI +
1
4e(1) 1
8eI)
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The solution is
x e(I x)
1 x4 dx= 2 I (18 1eI +14e(1) 18eI)
We can try to solve it using real calculus and obtain the result
x e(I x)
1 x4 dx=
x (cos(x) + Isin(x))1 + x4 dx
Info.
not given
Comment.
no comment
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1.39 Problem.
Using the Residue theorem evaluate
x e(I x)
x2 + 4 x + 20dx
Hint.
Type III
Solution.We denote
f(z) = z e(I z)
z2 + 4 z+ 20
We find singularities
[{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
z=2 + 4 Iis in our region and we will add the following residue
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
We can try to solve it using real calculus and obtain the result
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)
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Info.
not givenComment.
no comment
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1.40 Problem.
Using the Residue theorem evaluate
x e(I x)
x2 + 9dx
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= 3 I
is in our region and we will add the following residue
res(3 I, f(z)) =1
2e(3)
The singularity
z =3 Iwill be skipped because the singularity is not in our region.
Our sum is
2 I (
res(z, f(z))) =I e(3)
The solution is
x e(I x)
x2 + 9dx= I e(3)
We can try to solve it using real calculus and obtain the result
x e(I x)
x2 + 9dx = I cosh(3) I sinh(3)
Info.
not given
Comment.
no comment
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1.41 Problem.
Using the Residue theorem evaluate
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
is in our region and we will add the following residue
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))
We can try to solve it using real calculus and obtain the result
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
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Comment.
no comment
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1.42 Problem.
Using the Residue theorem evaluate
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},{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)2The 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
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)
We can try to solve it using real calculus and obtain the result
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
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1.43 Problem.
Using the Residue theorem evaluate
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
We can try to solve it using real calculus and obtain the result
e(I x)
x2 dx=
Info.
not given
Comment.
no comment
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1.44 Problem.
Using the Residue theorem evaluate
x3 e(I x)
x4 + 5 x2 + 4dx
Hint.
Type III
Solution.We denote
f(z) = z3 e(I z)
z4 + 5 z2 + 4
We find singularities
[{z= 2 I},{z =I},{z= I},{z =2 I}]The singularity
z= 2 I
is in our region and we will add the following residue
res(2 I, f(z)) =2
3e(2)
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)) =16
e(1)
The singularity
z =2 Iwill be skipped because the singularity is not in our region.
Our sum is
2 I (
res(z, f(z))) = 2 I (2
3e(2) 1
6e(1))
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The solution is
x3
e(I x)
x4 + 5 x2 + 4dx = 2 I ( 2
3e(2) 1
6e(1))
We can try to solve it using real calculus and obtain the result
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
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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
We find singularities
[{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
2 2 2The singularity
z =12
2 12
I2
adds the following residue
res(f(z),12
2 1
2I
2) =
1 I2 I
2 2 2
The singularity
z=1
2
2 1
2I
2
adds the following residue
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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adds the following residue
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
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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
We find singularities
[{z=I},{z = I}]The singularity
z=Iadds the following residue
res(f(z),I) = 0The singularity
z = I
adds the following residue
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
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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
We find singularities
[{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
z=Iadds the following residue
res(f(z),I) =116
+ 1
16I
The singularity
z = I
adds the following residue
res(f(z), I) =116
116
I
The singularity
z= 3 I
adds the following residue
res(f(z), 3 I) = 1
16 7
48I
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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
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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
We find singularities
[{z=I},{z = 2 I},{z=2 I},{z = I}]The singularity
z=Iadds the following residue
res(f(z),I) = 118
I
The singularity
z= 2 I
adds the following residue
res(f(z), 2 I) = 11
288I
The singularity
z =2 Iadds the following residue
res(f(z),2 I) =11288
I
The singularity
z = I
adds the following residue
res(f(z), I) =118
I
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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
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2.5 Problem.
Check the Zero Sum theorem for the following function1
(z2 + 1) (z2 + 9)
Hint.
no hint
Solution.We denote
f(z) = 1
(z2 + 1) (z2 + 9)
We find singularities
[{z=3 I},{z =I},{z = I},{z= 3 I}]The singularity
z =3 Iadds the following residue
res(f(z),3 I) =148
I
The singularity
z=Iadds the following residue
res(f(z),I) = 116
I
The singularity
z = I
adds the following residue
res(f(z), I) =116
I
The singularity
z= 3 I
adds the following residue
res(f(z), 3 I) = 1
48I
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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
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2.6 Problem.
Check the Zero Sum theorem for the following function1
(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
adds the following residue
res(f(z), 2 I) =IThe singularity
z = I
adds the following residue
res(f(z), I) = I
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
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2.7 Problem.
Check the Zero Sum theorem for the following function1
(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 Iadds the following residue
res(f(z),3 I) =120
The singularity
z= 2 I
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),) = 0and finally we obtain the sum
res(f(z), z) = 0
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Info.
not given
Comment.
no comment
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2.8 Problem.
Check the Zero Sum theorem for the following function1
z2 + 1
Hint.
no hint
Solution.We denote
f(z) = 1
z2 + 1
We find singularities
[{z=I},{z = I}]The singularity
z=Iadds the following residue
res(f(z),I) = 12
I
The singularity
z = I
adds the following residue
res(f(z), I) =12
I
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
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2.9 Problem.
Check the Zero Sum theorem for the following function1
z3 + 1
Hint.
no hint
Solution.We denote
f(z) = 1
z3 + 1
We find singularities
[{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
adds the following residue
res(f(z), 1
2 1
2I
3) =2 1
3 I
3 + 3
The singularity
z=1
2+
1
2I
3
adds the following residue
res(f(z), 1
2+
1
2I
3) = 2
1
3 I
3
3
At infinity we get the residue
res(f(z),) =13
+ 2 1
3 I
3 + 3 2 1
3 I
3 3
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and finally we obtain the sum
res(f(z), z) = 0
Info.
not given
Comment.
no comment
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2.10 Problem.
Check the Zero Sum theorem for the following function1
z6 + 1
Hint.
no hint
Solution.We denote
f(z) = 1
z6 + 1
We find singularities
[{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
z=Iadds the following residue
res(f(z),I) = 16
I
The singularity
z=12
2 2 I
3
adds the following residue
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
adds the following residue
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
adds the following residue
res(f(z),
1
2
2 + 2 I3) =16
3
1
(2 + 2 I3)(5/2)The singularity
z = I
adds the following residue
res(f(z), I) =16
I
At infinity we get the residue
res(f(z),) = 0
and finally we obtain the sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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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
We find singularities
[{z=2 3 I},{z=2 + 3 I}]The singularity
z=2 3 Iadds the following residue
res(f(z),2 3 I) =154
I
The singularity
z=2 + 3 I
adds the following residue
res(f(z),2 + 3 I) = 154
I
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
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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
z =3 Iadds the following residue
res(f(z),3 I) = 316
I
The singularity
z=
I
adds the following residue
res(f(z),I) =116
I
The singularity
z = I
adds the following residue
res(f(z), I) = 1
16I
The singularity
z= 3 I
adds the following residue
res(f(z), 3 I) =316
I
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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
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2.13 Problem.
Check the Zero Sum theorem for the following functionz2
(z2 + 1)3
Hint.
no hint
Solution.We denote
f(z) = z2
(z2 + 1)3
We find singularities
[{z=I},{z = I}]The singularity
z=Iadds the following residue
res(f(z),I) = 116
I
The singularity
z = I
adds the following residue
res(f(z), I) =116
I
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
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2.14 Problem.
Check the Zero Sum theorem for the following functionz2
(z2 + 4)2
Hint.
no hint
Solution.We denote
f(z) = z2
(z2 + 4)2
We find singularities
[{z= 2 I},{z =2 I}]The singularity
z= 2 I
adds the following residue
res(f(z), 2 I) =18
I
The singularity
z =
2 I
adds the following residue
res(f(z),2 I) = 18
I
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
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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
z =3 Iadds the following residue
res(f(z),3 I) = 14
e3
The singularity
z=
I
adds the following residue
res(f(z),I) = 14
e
The singularity
z = I
adds the following residue
res(f(z), I) =1
4e(1)
The singularity
z= 3 I
adds the following residue
res(f(z), 3 I) =1
4e(3)
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At infinity we get the residue
res(f(z),) =14
e3 14
e 14
e(1) 14
e(3)
and finally we obtain the sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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2.16 Problem.
Check the Zero Sum theorem for the following function
e(I z) (z2 1)z (z2 + 1)
Hint.
no hint
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
adds the following residue
res(f(z), 0) =1The singularity
z=Iadds the following residue
res(f(z),I) = eThe 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 sumres(f(z), z) = 0
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Info.
not givenComment.
no comment
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2.17 Problem.
Check the Zero Sum theorem for the following function
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
adds the following residue
res(f(z), 1) =1
5eI
The singularity
z= 2 I
adds the following residue
res(f(z), 2 I) = ( 110
+ 1
20I) e(2)
The singularity
z =2 Iadds the following residue
res(f(z),2 I) = ( 110
120
I) e2
At infinity we get the residue
res(f(z),) =15eI + ( 110 120I) e(2) + ( 110+ 120I) e2and finally we obtain the sum
res(f(z), z) = 0
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Info.
not givenComment.
no comment
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2.18 Problem.
Check the Zero Sum theorem for the following function
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
adds the following residue
res(f(z), 0) = 1
The singularity
z=I
adds the following residue
res(f(z),I) =12
e
The singularity
z = I
adds the following residue
res(f(z), I) =12
e(1)
At infinity we get the residue
res(f(z),) =1 +12e +12e(1)and finally we obtain the sum
res(f(z), z) = 0
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Info.
not given
Comment.
no comment
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2.19 Problem.
Check the Zero Sum theorem for the following function
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
z =3 Iadds the following residue
res(f(z),3 I) = 1324
e3
The singularity
z= 0
adds the following residue
res(f(z), 0) = 1
81
The singularity
z= 3 I
adds the following residue
res(f(z), 3 I) = 5324
e(3)
At infinity we get the residue
res(f(z),) = 1324e3 181+ 5324e(3)and finally we obtain the sum
res(f(z), z) = 0
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Info.
not givenComment.
no comment
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2.20 Problem.
Check the Zero Sum theorem for the following function
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
adds the following residue
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
adds the following residue
res(f(z), 1 + I) = (14 1
4I) eI e(1)
At infinity we get the residue
res(f(z),) =12+ ( 14 14I) eI e + ( 14+ 14I) eI e(1)and finally we obtain the sum
res(f(z), z) = 0
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Info.
not givenComment.
no comment
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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=Iadds the following residue
res(f(z),I) = 12
I e
The singularity
z = I
adds the following residue
res(f(z), I) =12
I e(1)
At infinity we get the residue
res(f(z),) =12
I e +1
2I e(1)
and finally we obtain the sumres(f(z), z) = 0
Info.not given
Comment.
no comment
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2.22 Problem.
Check the Zero Sum theorem for the following function
e(I z)
z2 + 4 z+ 20
Hint.
no hint
Solution.We denote
f(z) = e(I z)
z2 + 4 z+ 20
We find singularities
[{z=2 4 I},{z=2 + 4 I}]The singularity
z=2 4 Iadds the following residue
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
At infinity we get the residue
res(f(z),) =18
I e4
(eI)2+
1
8
I e(4)
(eI)2
and finally we obtain the sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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2.23 Problem.
Check the Zero Sum theorem for the following function
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
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 sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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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 = 12 1
2I
3},{z= 1
2+
1
2I
3}]
The singularity
z =1adds the following residue
res(f(z),1) = 13
1
eI
The singularity
z=
1
21
2I
3
adds the following residue
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
adds the following residue
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)
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and finally we obtain the sum
res(f(z), z) = 0
Info.
not given
Comment.
no comment
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2.25 Problem.
Check the Zero Sum theorem for the following function
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
adds the following residue
res(f(z), 1) =1
4eI
The singularity
z =
1
adds the following residue
res(f(z),1) =14
1
eI
The singularity
z=Iadds the following residue
res(f(z),I) =14
I e
The singularity
z = I
adds the following residue
res(f(z), I) =1
4I e(1)
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At infinity we get the residue
res(f(z),) =14
eI +14
1eI
+14
I e 14
I e(1)
and finally we obtain the sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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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 sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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2.27 Problem.
Check the Zero Sum theorem for the following function
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
adds the following residue
res(f(z), 1) =14
eI
The singularity
z =
1
adds the following residue
res(f(z),1) =14
1
eI
The singularity
z=Iadds the following residue
res(f(z),I) = 14
e
The singularity
z = I
adds the following residue
res(f(z), I) =1
4e(1)
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At infinity we get the residue
res(f(z),) = 14
eI +14
1eI
14
e 14
e(1)
and finally we obtain the sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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2.28 Problem.
Check the Zero Sum theorem for the following function
z e(I z)
z2 + 4 z+ 20
Hint.
no hint
Solution.We denote
f(z) = z e(I z)
z2 + 4 z+ 20
We find singularities
[{z=2 4 I},{z=2 + 4 I}]The singularity
z=2 4 Iadds the following residue
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
At infinity we get the residue
res(f(z),) = 18
I(4 I e4 + 2 e4)
(eI)2 +
1
8
I(4 I e(4) 2 e(4))(eI)2
and finally we obtain the sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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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 Iadds the following residue
res(f(z),3 I) = 12
e3
The singularity
z= 3 I
adds the following residue
res(f(z), 3 I) =1
2e(3)
At infinity we get the residue
res(f(z),) =12
e3 12
e(3)
and finally we obtain the sumres(f(z), z) = 0
Info.not given
Comment.
no comment
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2.30 Problem.
Check the Zero Sum theorem for the following function
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
adds the following residue
res(f(z), 1 + 3 I) =16
I(3 I eI e(3) + eI e(3))
At infinity we get the residue
res(f(z),) = 16
I(3 I eI e3 eI e3) +16
I(3 I eI e(3) + eI e(3))
and finally we obtain the sumres(f(z), z) = 0
Info.not given
Comment.
no comment
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2.31 Problem.
Check the Zero Sum theorem for the following function
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
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 sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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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 sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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2.33 Problem.
Check the Zero Sum theorem for the following function
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
z=Iadds the following residue
res(f(z),I) =16
e
The singularity
z= 2 I
adds the following residue
res(f(z), 2 I) =2
3e(2)
The singularity
z =2 Iadds the following residue
res(f(z),2 I) = 23
e2
The singularity
z = I
adds the following residue
res(f(z), I) =16
e(1)
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At infinity we get the residue
res(f(z),) = 16
e 23
e(2) 23
e2 +16
e(1)
and finally we obtain the sumres(f(z), z) = 0
Info.
not given
Comment.
no comment
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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
We find singularities
[{z=}]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
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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)
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),) =1and finally we obtain the sum
res(f(z), z) = 0
Info.
not given
Comment.
no comment
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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)
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) =IThe singularity
z = I
adds the following residue
res(f(z), I) = I
At infinity we get the residue
res(f(z),) =1and finally we obtain the sum
res(f(z), z) = 0
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Info.
not givenComment.
no comment
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2.37 Problem.
Check the Zero Sum theorem for the following function1
(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 =1adds the following residue
res(f(z),1) =12
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
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2.38 Problem.
Check the Zero Sum theorem for the following function1
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
adds the following residue
res(f(z), 1) =1
2
The singularity
z =1adds the following residue
res(f(z),1) =12
The singularity
z= 0
adds the following residue
res(f(z), 0) = 1
At infinity we get the residue
res(f(z),) = 0and finally we obtain the sum
res(f(z), z) = 0
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Info.
not givenComment.
no comment
134
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2.39 Problem.
Check the Zero Sum theorem for the following function1
z (z2 + 4)2
Hint.
no hint
Solution.We denote
f(z) = 1
z (z2 + 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= 2 I
adds the following residue
res(f(z), 2 I) =1
32
The singularity
z =2 Iadds the following residue
res(f(z),2 I) =132
At infinity we get the residue
res(f(z),) = 0and finally we obtain the sum
res(f(z), z) = 0
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Info.
not given
Comment.
no comment
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2.40 Problem.
Check the Zero Sum theorem for the following function1
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) =1At 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
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2.41 Problem.
Check the Zero Sum theorem for the following function1
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) =1At 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
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2.42 Problem.
Check the Zero Sum theorem for the following function1
(z2 + 1)2
Hint.
no hint
Solution.We denote
f(z) = 1
(z2 + 1)2
We find singularities
[{z=I},{z = I}]The singularity
z=Iadds the following residue
res(f(z),I) = 14
I
The singularity
z = I
adds the following residue
res(f(z), I) =14
I
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
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2.43 Problem.
Check the Zero Sum theorem for the following function1
z3 z5Hint.
no hint
Solution.We denote
f(z) = 1
z3 z5We 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) =12
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 sumres(f(z), z) = 0
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Info.
not givenComment.
no comment
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2.44 Problem.
Check the Zero Sum theorem for the following function1
z4 + 1
Hint.
no hint
Solution.We denote
f(z) = 1
z4 + 1
We find singularities
[{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
2 I
2 2 2The singularity
z =12
2 12
I2
adds the following residue
res(f(z),12
2 1
2I
2) = 1
2 I
2 2 2The singularity
z=1
2
2 1
2I
2
adds the following residue
res(f(z), 1
2
2
1
2I
2) =
1
2 I2 + 2 2The singularity
z =12
2 +
1
2I
2
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adds the following residue
res(f(z),12
2 +12
I2) = 12 I
2 + 2
2
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
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2.45 Problem.
Check the Zero Sum theorem for the following function1
z z3Hint.
no hint
Solution.We denote
f(z) = 1
z z3We 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) =12
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 sumres(f(z), z) = 0
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Info.
not givenComment.
no comment
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2.46 Problem.
Check the Zero Sum theorem for the following function1
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),) =1and finally we obtain the sum
res(f(z), z) = 0
Info.
not given
Comment.
no comment
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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
We find singularities
[{z =},{z =3 I},{z= 0},{z = 3 I}]The singularity
z =3 Iadds the following residue
res(f(z),3 I) =154
I
(eI)3
The singularity
z= 0
adds the following residue
res(f(z), 0) =1
9
The singularity
z= 3 I
adds the following residue
res(f(z