# Document

```Chap 5. Series
Series representations of analytic functions
43. Convergence of Sequences and Series
An infinite sequence 數列
z1 , z 2 , ...........,z n ,..........
of complex numbers has a limit z if, for each positive ,
there exists a positive integer n0 such that
zn  z  
w henever n  n0 .
zn
z3
z1
For sufficieutly large n , the points
z n lie in any given  neighborhood of z.
z2

z
n0 , in general, depends on  .
tch-prob
1
The limit z is unique if it exists. (Exercise 6).
When the limit exists, the sequence is said to converge to z.
lim z  z
n  n
Otherwise, it diverges.
Thm 1.
S uppose z n  x n  iy n ,
lim
n 
zn  z
iff
z  x  iy then
lim x n  x, and
n 
tch-prob
lim y n  y .
n 
2
An infinite series

 z n  z1  z 2  .........  z n  ...............
(6)
n 1
of com plex num bers converges to the sum S if the sequence
N
S N   z n  z1  z 2  ..........  z N
( N  1, 2 , ......)
n 1
of partial sum s con verges to S .
W e then w rite

 zn  S
n 1
T hm 2. S uppose z n  x n  iy n

T hen  z n  S
iff
n 1
tch-prob
, S  x  iy


n 1
n 1
 x n  x and  y n  y
3
A necessary condition for the convergence of series
lim z n  0
is that
(6)
n 
The terms of a convergent series of complex numbers are,
therefore, bounded, i.e.,
z M.
n
Absolute convergence:


n 1
n 1
 zn  
S ince
xn 
xn  yn
2
2
xn 2  y n 2 ,
converges
yn 
xn 2  y n 2


  x n and  y n also converges
n 1
n 1


  xn , and  y n converge
n 1
n 1

 From T hm . 2 ,
 z n converge
n 1
Absolute convergence of a series of complex numbers implies
convergence of that series.
tch-prob
4
44. Taylor Series
Thm. Suppose that a function f is analytic throughout an open disk
z  z 0  R0 .
Then at each point z in that disk, f(z) has the series
representation

f ( z )   an ( z - z0 )n
n0
w h ere
f (n ) ( z0 )
an 
n  0 , 1, 2 , .....
n!
z
R0
z0
That is, the power series here converges to f(z)
w hen
z - z 0  R0 .
tch-prob
5
This is the expansion of f(z) into a Taylor series about the point z0
Any function that is known to be analytic at a point z0 must have a
(For, if f is analytic at z0, it is analytic in some neighborhood
z  z 0  ε of z 0
(sec . 20 , p 55) .
 may serve as R0 is the
statement of Taylor’s Theorem)
pf . (a) w hen z 0  0

f ( n ) ( 0) n
prove f ( z )  
z
n
!
n0
Let C 0 : C ircle z  r0
Positively oriented within
( z  R0 )
~ Maclaurin series.
z0=0的case
z  R0
s
z
r
and z is interior to it.
r0
R0
0
C0
tch-prob
6
The Cauchy integral formula applies:
f ( s ) ds
f ( z )  1 c
2 πi 0 s  z
1
now 1  1
s z s
1- z
( z  1),
s
z 1
s
s
N 1
N
n
1
z
and
  z 
1- z n 0
1 z
( z  1)
(Exeicise
13, sec. 6).
z )N
(
s
 1  1 (  ( z )n 
)
s  z s n0 s
z
1
s
N 1
N 1
z
n
N
z
  n 1 
(s  z)s N
n0 s
tch-prob
7
f ( s ) ds N 1 f ( s ) ds n
f ( s ) ds
N

z

z


c s  z
c ( s  z ) s N
c
n 1
0
0
0
s
n0
N 1
  2 i
n0
f
(n)
(0)
n!
z n  z N c
f ( s ) ds
N
0 (s  z)s
(n)
N 1 f
(0) n
f
(
s
)
ds
 f ( z )  1 c
 
z  ρN ( z )
s

z
2 πi 0
n!
n0
N
f ( s ) ds
w her e ρ N ( z )  z
2 πi c0 ( s  z ) s N
S uppose that
z r
then if s is a point on C 0
s  z  s  z  r0  r
tch-prob
8
If
f (s)  M
N
M  2 π r0 M r0 r N
ρN ( z)  r

( )
N
r

r
r0
2 π ( r0  r ) r0
0
B ut r  1,
 lim ρ N ( z )  0
r0

N 
1
s
(n)

f (0) n

 f (z)  
z
n!
n0
(b) For arbitrary z0
Suppose f is analytic when
z  z 0  R0
and note that the
composite function f ( z  z 0 )
must be analytic when
( z  z 0 )  z 0  R0
tch-prob
9
Let g ( z )  f ( z  z 0 ).
The analyticity of g(z) in the disk z  R0 ensures
the existence of a Maclaurin series representation:
 g ( n ) ( 0) n
g (z)  
z
( z  R0 )
n!
n0
 f (n) ( z ) n
0 z
or f ( z  z 0 )  
n!
n0
R eplace z by z - z 0 ,
 f (n) ( z )
0
f (z)  
( z  z0 )n
n 0 n!
tch-prob
10
45 Examples
Ex1. Since f ( z )  e z is entire
It has a Maclaurin series representation which is valid for all z.
f
(n)
(z)  e z ,
f
(n)
(0)  1,
 zn
z
e  
( z  )
n 0 n!
f ( z )  z 2 e 3 z is also entire.
n

z 2e 3 z   3 z n  2
n 0 n!
n2

3
 
zn
( z  )
(
n
2)!
n2
tch-prob
11
Ex2. Find Maclaurin series representation of
f ( z )  sin z
f
(2 n )
(0)  0
f
(2 n  1)
(0)  (  1) n , n  0 , 1, 2 , ....

n
z 2 n 1
sin z   ( - 1)
(2 n  1)!
n0
( z  )
sinh z  -i sin (iz )
Ex3.
f ( z )  cos z
f
(2 n )
( 0)  (  1)
n

n
cos z   ( - 1)
n0
cosh z  cos (iz )
f
(2 n  1)
z 2n
(2 n )!
tch-prob
(0)  0
( z  )
12
Ex4.
1   zn

1 z n0
sin ce
f
( z  1)
(n)
(z) 
n!
(1  z ) n  1
f ( n ) (0)  n !
su b stitu te - z fo r z
1   (  1) n z n

1 z n0
su b stitu te
( z 1)
1 - z fo r z
1   (  1) n ( z  1) n
( z - 1  1)

z
n0


n
n
n
  (1 - z )   ( - 1) ( z - 1)
n0
n0
tch-prob
13
Ex5.
2
2(1  z 2 )  1
1

2
z
1
f (z) 

3
5
z z
z3
1 z 2
 1 (2 - 1 )
z3
1 z 2
can not find a M aclaurin series fo r f ( z )
since it is not analytic at z  0 .
B ut
1  1  z 2  z 4  z 6  ............( z  1 )
1 z 2
H ence, w hen 0  z  1,
f ( z )  1 (2 - 1  z 2  z 4  z 6 ..............)
z3
 1  1  z  z 3  z 5 ...........
z3 z

tch-prob
14
46. Laurent Series
If a function f fails to be analytic at a point z0, we can not
apply Taylor’s theorem at that point.
However, we can find a series representation for f(z)
involving both positive and negative powers of (z-z0).
Thm. Suppose that a function f is analytic in a domain
R1  z  z 0  R 2 ,
and let C denote any positively oriented
simple closed contour around z0 and lying in that domain.
Then at each z in the domain


bn
n
f ( z )   an ( z - z0 )  
n
n0
n 1 ( z - z 0 )
tch-prob
(1)
15
where
f (z) dz
a n  1 c
2π i
( z - z 0 ) n 1
( n  0 , 1, 2 , ......)
f (z) dz
bn  1 c
2 π i ( z - z )  n 1
0
or
( n  1, 2 , .......)

n
 C n ( z - z0 )
n  -
f (z) dz
w h ere C n  1 c
2π i
( z - z ) n 1
f (z) 
( R1  z  z 0  R 2 )
(4)
( n  0 ,  1,  2 , .....)
(5)
0
(1) (4) are Laureut Series
Pf: see textbook.
Z
f (s)
f ( s ) ds
f ( s ) ds
c s  z ds - c s  z   s  z  0
2
1

2 πi f ( z )
tch-prob
r
R2
R1
Z0
C
C2
C1
16
47. Examples
The coefficients in a Laurent series are generally found by
means other than by appealing directly to their integral
representation.
Ex1.
n

z
ez  
( z  )
n0 n !
replace z by 1 z
1

z
1  1  1  1  1  .........
e  
n
z 2! z 2 3! z 3
n 0 n! z
no positive pow ers of z appear.
(0  z   )
1
1
since b1 
e z dz  1

c
2 πi
 c e
1
z
dz  2 πi
Alterative way to calculate
tch-prob
17
Ex2.
1
is already in the form of a laurent series, w here z 0  i.
( z  i)2

f ( z )   C n ( z  i)n
(0  z - i   )
n  
w here C  2  1, all other coefficients are zero.
f (z) 
S ince
f ( z ) dz
1
C n  1 c
 1 c
dz
2π i
n

1
2

i
n

3
( z - i)
( z  i)
0
dz
 c
 
( z - i ) n  3  2  i
w hen n  -2
w hen n= -2
d1
dz z  i
[ c dz  c 1 dz  2 π i
n!
( z  i)2
( z - i)2
tch-prob
 0]
18
Ex3.
f (z) 
1
 1  1
( z  1)( z  2) z  1 z  2
has two singular points z=1 and z=2, and is analytic in
the domains
D1 : z  1
D2 : 1  z  2
D3 : 2  z
D3
D2
Recall that
D1 1
1   zn

1 z n0
( z  1)
2
０
(a) f(z) in D1
f (z)  
1
1 -z
1
2
1
1- z
2
z 1
in D 1
2


z n   (2 -n -1  1) z n
f (z)    zn  

n

1
2
n0
n0
n0
S in ce
z 1
an d
tch-prob
( z  1)
19
(b) f(z) in D2
1
1
f (z)  1
1
z 1 -( 1 ) 2 1 - z
z
2
1 1 ,
z 1 ,
w h en 1  z  2 ,
z
2


1
1
1
n
f (z) 
( ) 
( z )n


z
z
2 n0 2
n0


n
1
z
 
 
(1  z  2)
n

1
n

1
n0 z
n0 2


1
1
n
 
z  
(1  z  2)
n
n

1
z
2
n0
n 1
S in ce
tch-prob
20
(c) f(z) in D3
1
f (z)  1 . 1  1
z 1- 2
z 1-1
z
z
2  1 w hen 2  z  
since 1  1 ,
z
z


n
2
1
 
f (z)  
1

n
n  0 z n 1
n0 z

n
1-2
(2  z   )
 
n  0 z n 1

n 1
2

1
(2  z   )
 
n
z
n 1
tch-prob
21
48. Absolute and uniform convergence of power series
Thm1.

n
 a n ( z - z 0 ) co n verg es
n 0
w h en z  z1 (z 1  z 0 ), th en it is ab so lu tely
(1)
If a p o w er series
co n verg en t at each p o in t z in th e o p en d isk
z - z 0  R1 , w h ere R1  z1  z 0 .
z1
C o n verg es at z1
 C o n verg es at all p o in ts
in sid e th e circle z - z 0  R1.
tch-prob
z
z0
R1
22
•
The greatest circle centered at z0 such that series (1) converges
at each point inside is called the circle of convergence of series
(1).
•
The series CANNOT converge at any point z2 outside that
circle, according to the theorem; otherwise circle of
convergence is bigger.

n
 a n (z-z 0 )
n 0
has circle of convergence z - z 0  R
S uppose that
Let

S (z )   an (z - z0 ) n
n0
N -1
S N (z )   an (z - z0 )n
n 0
tch-prob
(1)
z  z0  R
23
w rite  N ( z )  S ( z )  S N ( z )
since (1) converges,  N ( z )  0 as
( z - z0  R )
N
or for each  , there is a N  sw ch that
 N ( z )   w henever N  N 
When the choice of N  depends only on the value of 
and is independent of the point z taken in a specified region
within the circle of convergence, the convergence is said to
be uniform in that region.
tch-prob
24
T hm 2. If z1 is in c. of. c.
z - z0  R

of  a n ( z - z 0 ) n
n 0
then that series is uniformly convergent in the closed disk
z  z 0  R1 , w here R1  z1  z 0 .
R
z
z0
R1
z1
Corollary.

A pow er series  a n ( z - z 0 ) n represents a continuous function S ( z )
n0
at each point inside its circle of conv ergence z - z 0  R .
tch-prob
25
49. Integration and Differentiation of power series
Have just seen that a power series

S ( z )   an ( z  z0 )n
n0
(1 )
represents continuous function at each point interior to
its circle of convergence.
We state in this section that the sum S(z) is actually
analytic within the circle.
Thm1. Let C denote any contour interior to the circle of
convergence of the power series (1), and let g(z) be any
function that is continuous on C. The series formed by
multiplying each term of the power series by g(z) can be
integrated term by term over C; that is,

n
g
(
z
)
S
(
z
)
dz

a
g
(
z
)
(
z
z
)
dz

c
n c
0
n0
tch-prob
26
Corollary. The sum S(z) of power series (1) is analytic at each
point z interior to the circle of convergence of that series.
Ex1.






sin z
z
w ant to show f ( z ) 
1
w hen z  0
w hen z  0
is entire

2 n 1
z
n
since sin z   ( - 1)
represent sin z for every z ,
(2
n

1)!
n0

n z 2 n 1
(
1)

(2 n  1)! 
z 2 n  1  z 2  z 4 ............
n0
  (  1) n
z
3! 5!
(2 n  1)!
n0
converges to f ( z ) w hen z  0
(4)
But series (4) clearly converges to f(0) when z=0. Hence f(z) is
an entire function.
tch-prob
27
Thm2. The power series (1) can be differentiated term by term. That is,
at each point z interior to the circle of convergence of that series,

S' ( z )   n a n ( z - z 0 ) n 1
n0
Ex2.
Diff.
1 

z
n0
( - 1) n ( z  1) n
( z - 1  1)

 1   ( - 1) n n ( z -1) n-1
z 2 n 1
1   ( - 1) n ( n  1)( z  1) n

2
z
n0
tch-prob
z -1  1
z -1  1
28
50. Uniqueness of series representation
Thm 1. If a series

n
 an ( z  z0 )
n0
co n verg es to f ( z )
at all points interior to some circle z  z 0  R ,
then it is the Taylor series expansion for f in powers
of z  z 0 .
Thm 2. If a series


n 
C n ( z - z0 )n
converges to f(z) at all points in some annular domain
about z0, then it is the Laurent series expansion for f in
powers of z  z 0 for that domain.
tch-prob
29
51. Multiplication and Division of Power Series

f ( z )   an ( z  z0 )n
n0

and
g ( z )   bn ( z  z 0 ) n
n0
converges w ithin z  z 0  R ,
Suppose
then f(z) and g(z) are analytic functions in z  z 0  R
Their product has a Taylor series expansion

f ( z ) g ( z )   cn ( z  z 0 ) n
n0
n
cn   a b
k  0 k n-k
tch-prob
z  z0  R
C auchy product
30
Ex1.
ez
The Maclaurin series for 1  z is valid in disk z  1
ez
1 z
 (1  z  1 z 2  1 z 3  .....)(1 - z  z 2  z 3  ........)
2
6
 1  1 z 2  1 z 3  .....
z 1
2
3
Ex2.
Zero of the entire function sinh z
are the m um bers z  nπi ( n  0 ,  1,  2 , ... )
1
1
So
 1
z 2 sinh z z 3 ( 1  z 2  z 4  .......)
3!
5!
has laurent series in the punctured disk 0  z  π
1
 1  1 . 1  7 z  .....(0  z  π )
z 2 sinh z z 3 6 z 360
z
z
(si nh z  e  e )
2
tch-prob
31
```

– Cards

– Cards

– Cards

– Cards

– Cards