# express代写 | 作业 R – Main

### Main

express代写 | 作业 R – 这个题目属于一个 R 的代写任务, 是比较有代表性的express/ R 等代写方向 Problem 1.Using Cauchys integral formula calculate

``````
``````
``````dz
``````
``````z
``````
``````2
2 z+ 10
``````

#### ,

where is a counter-clockwise oriented simple contour, not passing through

eiter of 1 3 iin the following cases

(a) The point 1 + 3iis inside and 1 3 iis outside it;

(b) The point 1 3 iis inside and 1 + 3iis outside it;

(c) Both points 1 3 iare inside .

Solution. Observing that

``````1
``````
``````z
``````
``````2
2 z+
``````

#### =

``````1
``````
``````(z 1  3 i)(z1+3i)
``````

(a) We apply Cauchys formula forf(z) =

``````1
``````
``````(z1+3i)
``````
``````resulting in 2if(1 + 3i) =
``````
``````
``````
``````3
``````

#### .

(b) We apply Cauchys formula forf(z) =

``````1
``````
``````(z 1  3 i)
``````
``````resulting in 2if(1 3 i) =
``````
``````
``````
``````3
``````

#### .

(c) Using additivity we conclude that in this case we need just sum results

of (a) and (b) resulting in 0.

Problem 2. (a) Find the decomposition into power series atz= 0 off(z) =

(1z)

``````
``````
``````1
``````
``````2
``````

. What is the radius of convergence?

(b) Plugging inz

``````2
instead ofzand integrating, obtain a decomposition at
``````

z= 0 of arcsin(z).

Solution. (a) Obviously f

``````(n)
(z) =
``````
``````1
``````
``````2
``````
``````3
``````
``````2
``````
``````2 n 1
``````
``````2
``````
``````(1z)
``````
``````
``````
``````n+
``````

(^2) ; then f (n) (0) = (2n1)!! 2 n forn1 and (1z) 1 2 = 1 + n= (2n1)!!

#### 2

``````n
n!
``````
``````z
``````
``````n
.
``````

The radius of convergence is 1 sincef(z) is analytic in {z:|z|< r}if and

only ifr1.

(b) Pluggingz

``````2
``````
``````(1z
``````
``````2
)
``````
``````
``````
``````1
``````
``````2
= 1 +
``````
``````

``````
``````n=
``````
``````(2n1)!!
``````

#### 2

``````n
n!
``````
``````z
``````
``````2 n
``````

and integrating we get

``````arcsin(z) =z+
``````
``````

``````
``````n=
``````
``````(2n1)!!
``````

#### 2

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

Problem 3.Find all singular points of

``````f(z) =z
``````
``````2
(z
``````
``````2

``````
``````2
) cot
``````
``````2
(z)
``````

and determine their types (removable, pole (in which case what is its order),

essential singularity, not isolated singularity, branching point).

In particular, determine singularity at (what kind of singularity we get

atw= 0 forg(w) =f(1/w)?).

Solution. Since cot(z) =

``````cos(z)
``````
``````sin(z)
``````
``````we need to consider points where sin(z) = 0,
``````

which arez n

``````=n withnZ.
``````

Since at each such point cot(z) has a simple pole cos(z n

#### ) = (1)

``````n
6 = 0,
``````

sin(z n

``````) = 0, but (sin(z))
``````
``````
= cos(z) = (1)
``````
``````n
atz=z
n
``````

#### .

Therefore cot

``````2
(z) has double poles (of order 2) atz
n
``````

. Howeverz

``````2
(z
``````
``````2

``````
``````2
) has
``````

a double zero point atz= 0 and simple zero points atz=. Therefore

``````z
n
=n is
``````
``````removable singularity as n= 0,
``````
``````simple pole as n= 1 ,
``````
``````double pole as n= 2 , 3 ,....
``````

Meanwhile, infinity is not an isolated singularity.

Problem 4.Calculate an improper integral

#### I =

``````
``````
``````0
``````
``````dx
``````
``````x(x
``````
``````2
+ 1)
``````

Hint:

(a) Calculate

#### J

``````R,
``````

#### =

``````
R,
``````
``````f(z)dz, f(z):=
``````

#### 1

``````z(z
``````
``````2
+ 1)
``````

where we have chosen the branch of

``````zsuch that it is analytic on the upper
``````

half-plane{z: Imz > 0 }and is real-valued forz=x >0. R, is the contour

on the figure below:

``````

``````
``````R

``````
``````
R
``````

(b) Prove that

``````
R
``````
``````dz

``````
``````z(z
``````
``````2
+1)
``````
``````0 asR and
``````
``````

``````
``````dz

``````
``````z(z
``````
``````2
+1)
``````
``````0 as 0
``````
``````+
``````

where R and are large and small semi-circles on the picture. This will

give you a value of

``````
0
``````
``````
``````
``````f(z)dz+
``````
``````
``````
``````0
``````
``````f(z)dz. (4.1)
``````

(c) express both integrals usingI.

Solution. (a) AsR > 1 there is just one singularity inside R, , namely a

simple pole atz=i. The residue is Res(

``````1

``````
``````z(z
``````
``````2
+1)
``````
``````,i) =
``````
``````1

``````
``````z 2 z
``````

#### |

``````z=i
``````

#### =

``````1
``````
``````2 i
``````
``````e
``````
``````i/ 4
due
``````

to the branch selection and thereforeJ= 2i

``````1
``````
``````2 i
``````
``````e
``````
``````i/ 4
=e
``````
``````i/ 4
.
``````

(b)

#### |

``````
R
``````
``````dz
``````
``````z(z
``````
``````2
+ 1)
``````

#### |

``````R
``````

#### R(R1)

``````2
``````
`````` 0 as R
``````

and

#### |

``````

``````
``````dz
``````
``````z(z
``````
``````2
+ 1)
``````

#### |

``````
``````
``````(1)
``````
``````2
``````
`````` 0 as  0
``````

(c) In (4.1) the second integral isI and the first one is

``````0
``````
``````
``````
``````dx
``````
``````e
``````
``````i/ 2
``````
``````|x|(x
``````
``````2
+ 1)
``````
``````=e
``````
``````i/ 2
``````
``````0
``````
``````
``````
``````dx
``````
``````|x|(x
``````
``````2
+ 1)
``````
``````=e
``````
``````i/ 2
I
``````

after change of variables. Thus

``````(e
``````
``````i/ 2
+ 1)I =e
``````
``````i/ 4
=
``````

#### I=

``````e
``````
``````i/ 4
``````
``````e
``````
``````i/ 2
+ 1
``````

#### =

``````
``````
``````e
``````
``````i/ 4
+e
``````
``````i/ 4
``````

#### =

``````
``````
``````2 cos(/4)
``````

#### =

``````
``````

#### .

Problem 5. Consider f(z) =

``````5
``````
``````(z2)(z+3)
``````
``````and decompose it into Laurents
``````

series converging

(a) As|z|<2;

(b) As 2<|z|<3;

(c) As|z|>3.

Solution.

#### 5

``````(z2)(z+ 3)
``````

#### 1

``````(z2)
``````

#### 1

``````(z+ 3)
``````

with

#### 1

``````z 2
``````

#### =

``````

``````
``````n=
``````

#### 2

``````n 1
z
``````
``````n
|z|< 2 ,
``````
`````` 1

``````
``````n=
``````

#### 2

``````n 1
z
``````
``````n
|z|> 2
``````

and

#### 1

``````z+ 3
``````

#### =

``````

``````
``````n=
``````

#### 3

``````n 1
(1)
``````
``````n
z
``````
``````n
|z|< 3 ,
``````
`````` 1

``````
``````n=
``````

#### 3

``````n 1
(1)
``````
``````n 1
z
``````
``````n
|z|> 3.
``````

Then

#### 5

``````(z2)(z+ 3)
``````

#### =

``````

``````
``````n=
``````

#### 2

``````n 1
3
``````
``````n 1
(1)
``````
``````n
``````

#### )

``````z
``````
``````n
|z|< 2
``````
`````` 1

``````
``````n=
``````

#### 2

``````n 1
z
``````
``````n

``````
``````

``````
``````n=
``````

#### 3

``````n 1
(1)
``````
``````n
z
``````
``````n
2 <|z|< 3
``````
`````` 1

``````
``````n=
``````

#### 2

``````n 1
+ 3
``````
``````n 1
(1)
``````
``````n
``````

#### )

``````z
``````
``````n
|z|> 3 ,
``````

where in fact for |z| > 3 summation is from to 2 (since term with

n=1 is 0).

### Day

Problem 1.Using Cauchys integral formula calculate

``````
``````
``````z dz
``````
``````z
``````
``````2
4 z+ 5
``````

#### ,

where is a counter-clockwise oriented simple contour, not passing through

eiter of 2iin the following cases

(a) The point 2 +iis inside and 2iis outside it;

(b) The point 2iis inside and 2 +iis outside it;

(c) Both points 2iare inside .

Solution. Observing that

``````z
``````
``````z
``````
``````2
4 z+
``````

#### =

``````1
``````
``````(z 2 i)(z2+i)
``````

(a) We apply Cauchys formula forf(z) =

``````z
``````
``````(z2+i)
``````
``````resulting in 2if(2 +i) =(2 +i).
``````

(b) We apply Cauchys formula forf(z) =

``````z
``````
``````(z 2 i)
``````
``````resulting in 2if(2i) =
``````

(2 +i).

(c) Using additivity we conclude that in this case we need just sum results

of (a) and (b) resulting in 2i.

Problem 2. (a) Find the decomposition into power series atz= 0 off(z) =

(1z)

`````` 1 / 3
``````

. What is the radius of convergence?

(b) Plugging inz

``````2
instead of z, integrating and multi[lying byz
``````
`````` 1
, obtain
``````

a decomposition atz= 0 ofF(z) =

``````1
``````
``````z
``````
``````z
``````
``````0
``````
``````(1z
``````
``````2
)
``````
`````` 1 / 3
dz(which is the special
``````

case of the generalized hypergeometric function).

Solution. (a) Obviously f

``````(n)
(z) =
``````
``````1
``````
``````3
``````
``````4
``````
``````3
``````
``````3 n 1
``````
``````2
``````
``````(1z)
``````
``````
``````
``````3 n+
``````
``````3
; then f
``````
``````(n)
(0) =
``````
``````(3n1)!!!
``````
``````3
``````
``````n
forn 1
``````
``````1
and
``````
``````(1z)
``````
``````
``````
``````1
``````
``````3
= 1 +
``````
``````

``````
``````n=
``````
``````(3n1)!!!
``````

#### 3

``````n
n!
``````
``````z
``````
``````n
.
``````

The radius of convergence is 1 sincef(z) is analytic in {z:|z|< r}if and

only ifr1.

(b) Pluggingz

``````2
``````
``````(1z
``````
``````2
)
``````
``````
``````
``````1
``````
``````3
= 1 +
``````
``````

``````
``````n=
``````
``````(3n1)!!!
``````

#### 3

``````n
n!
``````
``````z
``````
``````2 n
``````

and integrating we get

``````F(z) = 1 +
``````
``````

``````
``````n=
``````
``````(3n1)!!!
``````

#### 3

``````n
(2n+ 1)n!
``````
``````z
``````
``````2 n
.
``````
``````1
withk!!!:=k(k3)(k 3 bk/ 3 c) so called multifactorial.
``````

Problem 3.Find all singular points of

``````f(z) = (z
``````
``````2
1) cot(z
``````
``````2
)
``````

and determine their types (removable, pole (in which case what is its order),

essential singularity, not isolated singularity, branching point).

In particular, determine singularity at (what kind of singularity we get

atw= 0 forg(w) =f(1/w)?).

Solution. Since cot(z

``````2
) =
``````
``````cos(z
``````
``````2
)
``````
``````sin(z
``````
``````2
)
``````
``````we need to consider points where sin(z
``````
``````2
) = 0,
``````

which are 0,

``````n,
``````
``````ni,n= 1, 2 ,....
``````

At each such point = n with n = 1, 2 ,… we see that cot() has a

simple pole, we conclude that cot(z

``````2
) has also simple poles at
``````
``````n and
``````
``````ni. However (z
``````
``````2
1) has zero points atz= 1. Therefore z=1 is
``````

a removable singularity. On the other hand, sin(z

``````2
) has a double zero point
``````

atz= 0 and thus

`````` 1 removable singularity,
``````
``````n, simple pole as n= 2, 3 ,...,
``````
``````ni, simple pole as n= 1, 2 , 3 ,...,
``````
``````0 double pole.
``````

Meanwhile, infinity is not an isolated singularity.

Problem 4.Calculate an improper integral

#### I =

``````
``````
``````0
``````
``````xdx
``````
``````(x
``````
``````2
+ 2x+ 2)
``````

Hint:

(a) Calculate

#### J

``````R,
``````

#### =

``````
R,
``````
``````f(z)dz, f(z):=
``````
``````z
``````
``````(z
``````
``````2
+ 2z+ 2)
``````

where we have chosen the branch of

``````zsuch that it is analytic inside  and
``````

is real-valued forz =x+i0 withx >0. R, is the contour on the figure

below:

``````

``````
`````` R
``````
``````
R
``````

(b) Prove that

``````R
``````
``````
``````
``````z dz
``````
``````(z
``````
``````2
+1)
``````
``````0 asRand
``````
``````
``````
``````
``````
``````z dz
``````
``````(z
``````
``````2
+1)
``````
``````0 as 0
``````
``````+
where
``````

R and are large and small circles on the picture. This will give you a

value of 0

``````
``````
``````f(xi0)dx+
``````
``````
``````
``````0
``````
``````f(x+i0)dx (4.1)
``````

wheref(xi0) = lim 0

``````+f(x+i).
``````

(c) Express both integrals usingI.

Solution. (a) AsR >1 there are two singularities inside R,

``````, namely, sim-
``````

ple poles atz=1 +i=

``````2 e
``````
``````3 i/ 4
andz=1 +i=
``````
``````2 e
``````
``````5 i/ 4
due to branch
``````

selection. The residues are

``````Res(
``````
``````z
``````
``````(z
``````
``````2
+ 2z+ 2)
``````
``````,1 +i) =
``````
``````z
``````
``````2 z+ 2
``````
``````1+i
``````

#### 1

``````2 i
``````

#### 2

``````1 / 4
e
``````
``````3 i/ 8
``````

and

``````Res(
``````
``````z
``````
``````(z
``````
``````2
+ 2z+ 2)
``````
``````, 1 i) =
``````
``````z
``````
``````2 z+ 2
``````
``````z= 1 i
``````

#### 1

``````2 i
``````

#### 2

``````1 / 4
e
``````
``````5 i/ 8
=
``````

#### 1

``````2 i
``````

#### 2

``````1 / 4
e
``````
`````` 3 i/ 8
``````

due to the branch selection and therefore

``````J= 2i 2
``````
``````1 / 4
``````

#### 1

``````2 i
``````

#### [

``````e
``````
``````3 i/ 8
+e
``````
`````` 3 i/ 8
``````

#### = 2

``````5 / 4
cos(3/8).
``````

(b)

#### |

``````
R
``````
``````z dz
``````
``````(z
``````
``````2
+ 1)
``````

#### |

``````2 R
``````

#### (R1)

``````2
``````
`````` 0 as R
``````

and

#### |

``````
``````
``````z dz
``````
``````(z
``````
``````2
+ 1)
``````

#### |

``````2
``````
``````
``````
``````(1)
``````
``````2
``````
`````` 0 as  0
``````

(c) In (4.1) the second integral isI and the first one is

``````0
``````
``````
``````
``````xdx
``````
``````(x
``````
``````2
+ 1)
``````

#### =I

after change of variables. Thus

#### 2 I= 2

``````5 / 4
sin(3/8) = I= 2
``````
``````1 / 4
cos(3/8).
``````

Problem 5. Consider f(z) =

``````3 z
``````
``````(z2)(z+1)
``````
``````and decompose it into Laurents
``````

series converging

(a) As|z|<1;

(b) As 1<|z|<2;

(c) As|z|>2.

Solution.

``````3 z
``````
``````(z2)(z+ 1)
``````

#### 2

``````(z2)
``````

#### 1

``````(z+ 1)
``````

with

#### 1

``````z 2
``````

#### =

``````

``````
``````n=
``````

#### 2

``````n 1
z
``````
``````n
|z|< 2 ,
``````
`````` 1

``````
``````n=
``````

#### 2

``````n 1
z
``````
``````n
|z|> 2
``````

and

#### 1

``````z+ 1
``````

#### =

``````

``````
``````n=
``````

#### (1)

``````n
z
``````
``````n
|z|< 1 ,
``````
`````` 1

``````
``````n=
``````

#### (1)

``````n 1
z
``````
``````n
|z|> 1.
``````

Then

``````3 z
``````
``````(z2)(z+ 3)
``````

#### =

``````

``````
``````n=
``````

#### 2

``````n
+ (1)
``````
``````n
``````

#### )

``````z
``````
``````n
|z|< 1
``````
``````

``````
``````n=
``````

#### 2

``````n
z
``````
``````n
+
``````
`````` 1

``````
``````n=
``````

#### (1)

``````n 1
z
``````
``````n
1 <|z|< 2
``````
`````` 1

``````
``````n=
``````

#### 2

``````n
+ (1)
``````
``````n 1
``````

#### )

``````z
``````
``````n
|z|> 2.
``````

### Morning

Problem 1.Using Cauchys integral formula calculate

``````
``````
``````dz
``````
``````z
``````
``````2
6 z+ 25
``````

#### ,

where is a counter-clockwise oriented simple contour, not passing through

eiter of 1 3 iin the following cases

(a) The point 3 + 4iis inside and 3 4 iis outside it;

(b) The point 3 4 iis inside and 3 + 4 is outside it;

(c) Both points 3 4 iare inside .

Solution. Observing that

``````1
``````
``````z
``````
``````2
6 z+
``````

#### =

``````1
``````
``````(z 3  4 i)(z3+4i)
``````

(a) We apply Cauchys formula forf(z) =

``````1
``````
``````(z3+4i)
``````
``````resulting in 2if(3 + 4i) =
``````
``````
``````
``````4
``````

#### .

(b) We apply Cauchys formula forf(z) =

``````1
``````
``````(z 3  4 i)
``````
``````resulting in 2if(1 3 i) =
``````
``````
``````
``````4
``````

#### .

(c) Using additivity we conclude that in this case we need just sum results

of (a) and (b) resulting in 0.

Problem 2. (a) Find the decomposition into power series atz= 0 off(z) =

(1z)

`````` 1
``````

. What is the radius of convergence?

(b) Plugging inz

``````2
instead ofz and integrating, obtain a decomposition
``````

atz= 0 of arctan(z).

Solution. (a) Obviouslyf

``````(n)
(z) =n!(1z)
``````
``````n 1
; thenf
``````
``````(n)
(0) =n! forn 0
``````

and

``````(1z)
``````
`````` 1
=
``````
``````

``````
``````n=
``````
``````z
``````
``````n
.
``````

The radius of convergence is 1 sincef(z) is analytic in {z:|z|< r}if and

only ifr1.

(b) Pluggingz

``````2
``````
``````(1 +z
``````
``````2
)
``````
`````` 1
=
``````
``````

``````
``````n=
``````

#### (1)

``````n
z
``````
``````2 n
``````

and integrating we get

``````arctan(z) =
``````
``````

``````
``````n=
``````

#### 1

``````(2n+ 1)
``````

#### (1)

``````n
z
``````
``````2 n+
.
``````

Problem 3.Find all singular points of

``````f(z) =
``````
``````sin(z)
``````
``````sin(z
``````
``````3
)
``````

and determine their types (removable, pole (in which case what is its order),

essential singularity, not isolated singularity, branching point).

In particular, determine singularity at (what kind of singularity we get

atw= 0 forg(w) =f(1/w)?).

Solution. The singularities are obviously atz=

``````3
``````
``````nwithnZand ifn 6 = 0
``````

denominators have simple zeroes and numerator is not zero unless n is a

perfect cube. On the other hand, atz= 0 numerator has a simple zero and

denominator has a triple zero.

z n

#### =

``````3
``````
``````n is
``````
``````removable singularity ifnZ,n 6 = 0 is a perfect cube,
``````
``````simple pole ifnZ,n 6 = 0 is not a perfect cube,
``````
``````double pole if n= 0.
``````

Meanwhile, infinity is not an isolated singularity.

Problem 4.Calculate an improper integral

#### I=

``````
``````
``````0
``````
``````ln(x)
``````
``````xdx
``````
``````(x
``````
``````2
+ 1)
``````

Hint:

(a) Calculate

#### J

``````R,
``````

#### =

``````
R,
``````
``````f(z)dz, f(z):=
``````
``````zlog(z)
``````
``````(z
``````
``````2
+ 1)
``````

where we have chosen the branches of log(z) and

``````z such that they are
``````

analytic on the upper half-plane {z: Imz > 0 } and is real-valued for z=

x >0. R, is the contour on the figure below:

``````

``````
``````R

``````
``````
R
``````

(b) Prove that

``````R
``````
``````
``````
``````zlog(z)dz
``````
``````(z
``````
``````2
+1)
``````
`````` 0 as  R   and
``````
``````
``````
``````
``````
``````zlog(z)dz
``````
``````(z
``````
``````2
+1)
``````
`````` 0 as
``````

0

``````+
0 where
R
``````
``````and

``````
``````are large and small semi-circles on the picture.
``````

This will give you a value of

``````0
``````
``````
``````
``````f(z)dz+
``````
``````
``````
``````0
``````
``````f(z)dz. (4.1)
``````

(c) Express both integrals usingI.

Solution. (a) AsR > 1 there is just one singularity inside R,

``````, namely a
``````

simple pole atz=i. The residue is Res(

``````
``````
``````zlog(z)
``````
``````(z
``````
``````2
+1)
``````
``````,i) =
``````
``````
``````
``````zlog(i)
``````
``````2 z
``````

#### |

``````z=i
``````

#### =

``````i/ 2
``````
``````2 i
``````
``````e
``````
``````i/ 4
``````

due to the branch selection and thereforeJ = 2i

``````i/ 2
``````
``````2 i
``````
``````e
``````
``````i/ 4
=
``````
``````
``````
``````2
``````
``````2
``````
``````e
``````
``````3 i/ 4
=
``````
``````
``````
``````2
``````
``````2
``````
``````
``````
``````2
``````
``````( 1 i).
``````

(b)

``````R
``````
``````zlog(z)dz
``````
``````(z
``````
``````2
+ 1)
``````

#### |

``````R
``````
``````R(ln(R) +)
``````

#### (R1)

``````2
``````
`````` 0 as R
``````

and

``````
``````
``````

``````
``````zlog(z)dz
``````
``````(z
``````
``````2
+ 1)
``````

#### |

``````
``````
``````(|log|+)
``````
``````(1)
``````
``````2
``````
`````` 0 as  0
``````

(c) In (4.1) the second integral isI and the first one is

``````0
``````
``````
``````
``````e
``````
``````i/ 2
``````
``````|x|(ln|x|+i)dx
``````
``````(x
``````
``````2
+ 1)
``````
``````=i
``````
``````0
``````
``````
``````
``````|x|(ln|x|+i)dx
``````
``````(x
``````
``````2
+ 1)
``````
``````=iIK
``````

with

#### K=

``````
``````
``````0
``````
``````|x|dx
``````
``````(x
``````
``````2
+ 1)
``````

after change of variables. Thus

``````(i+ 1)IK =
``````
``````
``````
``````2
``````

#### 2

``````(1 +i).
``````

since bothI,Kare real

#### I =

``````
``````
``````2
``````

#### .

As an added value we getIK=

``````
``````
``````2
``````
``````2
``````
``````
``````
``````2
``````
``````andK=
``````
``````

``````
``````2
``````

#### .

Problem 5. Consider f(z) =

``````8
``````
``````(z3)(z+5)
``````
``````and decompose it into Laurents
``````

series converging

(a) As|z|<3;

(b) As 3<|z|<5;

(c) As|z|>5.

Solution.

#### 8

``````(z3)(z+ 5)
``````

#### 1

``````(z3)
``````

#### 1

``````(z+ 5)
``````

with

#### 1

``````z 3
``````

#### =

``````

``````
``````n=
``````

#### 3

``````n 1
z
``````
``````n
|z|< 3 ,
``````
`````` 1

``````
``````n=
``````

#### 3

``````n 1
z
``````
``````n
|z|> 3
``````

and

#### 1

``````z+ 5
``````

#### =

``````

``````
``````n=
``````

#### 5

``````n 1
(1)
``````
``````n
z
``````
``````n
|z|< 5 ,
``````
`````` 1

``````
``````n=
``````

#### 5

``````n 1
(1)
``````
``````n 1
z
``````
``````n
|z|> 5.
``````

Then

#### 8

``````(z3)(z+ 5)
``````

#### =

``````

``````
``````n=
``````

#### 3

``````n 1
5
``````
``````n 1
(1)
``````
``````n
``````

#### )

``````z
``````
``````n
|z|< 3
``````
`````` 1

``````
``````n=
``````

#### 3

``````n 1
z
``````
``````n

``````
``````

``````
``````n=
``````

#### 5

``````n 1
(1)
``````
``````n
z
``````
``````n
3 <|z|< 5
``````
`````` 1

``````
``````n=
``````

#### 3

``````n 1
+ 5
``````
``````n 1
(1)
``````
``````n
``````

#### )

``````z
``````
``````n
|z|> 5 ,
``````

where in fact for |z| > 5 summation is from to 2 (since term with

n=1 is 0).