Singular points
Chap 6 Residues and Poles
Cauchy-Goursat Theorem:
c f dz  0
if f analytic.
What if f is not analytic at finite number of points interior
to C
Residues. 殘值
C
53. Residues
z0 is called a singular point of a function f if f fails to be analytic at
z0 but is analytic at some point in every neighborhood of z0.
A singular point z0 is said to be isolated if, in addition, there is a
deleted neighborhood 0  z  z0  ε of z0 throughout which f
is analytic.
除了那點Z0之外的小圈圈(半徑為)之內 f 都是可解析的
1
Ex1.
z 1
z 2(z 2 1 )
has isolated singnlar points z  0,  i
Ex2. The origin is a singular point of Log z, but is not isolated
Ex3.
1
sin( z)
singular points z  0 and z  1n n  1,  2, ....
not isolated
isolated
When z0 is an isolated singular point of a function f, there is a R2
such that f is analytic in 0  z  z0  R2
0
2
0  z  z0  R2
Consequently, f(z) is represented by a Laurent series
R.O.C.

bn
b
b2
f ( z)   an ( z  z0 )n  z 1z 

.....

 .......
(1)
2
n
(
z

z
)
(
z

z
)
0
0
0
n0
0  z  z0  R2
where bn  1 c f ( z)dz
(n 1, 2, ... )
2 i (z  z )n1
0
and C is positively oriented simple closed contour
around z0 and lying in 0  z  z0  R2
When n=1, 2πi b1  c f (z) dz
(2)
1
The complex number b1, which is the coefficient of z  z in
0
expansion (1) , is called the residue of f at the isolated singular
point z0.
A powerful tool for evaluating certain
Re s f ( z)
z  z0
integrals.
3
c
Ex4.
dz
z( z  2)4
C : z- 2 1
1
has singular points at z  0, z  2
z( z  2)4
it has Laurent series representation
in 0  z - 2  2
dz  2πi Re s
1
based on (2)
c
z  2 z ( z  2)4
z( z  2)4
but
1  1 .
1
z( z- 2)4 ( z  2)4 2  ( z  2)
1
 1 .
2( z- 2)4 1 ( z  2)
2
 (-1)n
 
( z- 2)n-4
n0 2n1
b1   1
16
c
0
2
湊出z-2在分母
0 z -2  2
dz  2 i   1   π i
 16  8
z( z- 2)24


4
Ex5. show c exp ( 12 ) dz  0
z
where
C : z 1
1 is analytic everywhere except at the origin
z2
0
2 3
e z 1 z  z  z  ......
z 
1! 2! 3!
1
e z2 1 1  1 1  1 1  ..... 0  z  
1! z 2 2! z 4 3! z6
b1  0
c exp ( 1 ) dz  0
z2
f analytic on and within C  c f dz  0
The reverse is not necessarily true.
5
More on Cauchy Integral Formula (1)
Simply-Connected and Multiply-Connected
C
Simply Connected
c f ( z) dz  0
Multiply Connected
n
c f ( z) dz   ck f ( z)dz  0
k 1
C1
C
C2
6
More on Cauchy Integral Formula (2)
Simply-Connected and Multiply-Connected
to show c dz
z  2π i
con stmct a circle C0,
C
Z  ρeiθ
C0
iθ
iρρ
π
2
dz
 2π i
sin ce c z  0
iθ
0
ρe
and 1 is analytic everywhere except at Z  0
Z
c dz
z  2π i
c Z dz  0
2
c Z dz  0
2π iθ dθ  ρ2i 2π ei2θ dθ  0
0
0 ρe
same.........
7
More on Cauchy Integral Formula (3)
Simply-Connected and Multiply-Connected
dz 
dz  d ( z  z0)  2π i
c Z c Z-Z c Z-Z
0
0
f(z ) dz
0
 2π i f(z )
c Z-Z
0
0
f (s)ds  N 1 f (s)ds z n  z N
f (s)ds
f(z
)
dz


c0 (s  z)s N
f(z) dz   0(n)  2π i f(z ) c0 s  z
c0
n 1
s
g
(
0)
n

0
c Z-Z g ( z) c  Z-Z z n
( z  R00 )
0
0
n
!
N 1
f (n) (0) n N
n

0
1
  2 i
z  z c f (s)dsN
f(s) ds  f(z) (n)
 f ( z0 ) n
n!
c
0 (s  z )s
n 0
z
2π i or fS(zZ z0 )  
N 1 f (n) (0)
Then connection to
f
(
s
)
ds
nTaylor
0 n! Series…….
1
 f ( z) 

z n  ρN ( z)


c
2πi 0 s  z n0 n!
Replace z by z - z0 ,
N
f (s)ds
z
 f (n) ( z0 )
wher
e
ρ
(
z
)

n
f ( z)  
( z  z0 )
N
2πi c0 (s  z)s N
n0 n!
8
Why
f(z) dz  f(z0 ) dz  2π i f(z )
c Z-Z
c Z-Z
0
0
0
(chap4)
f(z) f(z )
0 dz
 c
0 Z  Z0
but f(z )c dz  f(z )02π 1
dz  2π i f(z )
0 0 Z-Z
0
0
Z Z
0
0
f(z) f(z )
f(z)
0 dz
 c
dz - 2π i f(z )  c
(5)
0
Z Z
Z

Z
0
0
0
f(z) f(z )
f(z) f(z )
0 dz  
0 dz
btu c
c
0 Z  Z0
0 Z  Z0
f(z) f(z )
0 dz   2πρ
 c
ρ
0 Z Z
0
 2π 
9
More on Cauchy Integral Formula (4)
Simply-Connected and Multiply-Connected
C
C
c f ( z) dz  0
C0 (  )

C
C1

C0 (  )


C2
C1
C0 (  )



C1 (  )


C2 (  )




C1 (  )

C3 (  )

C2 (  )

C4 (  )
C1 (  )
C1
n
f
(
z
)
dz

 c f ( z)dz  0
c
k 1 k

C
C2
10
 ...
54. Residue Theorems
Thm1. Let C be a positively oriented simple closed contour. If f
is analytic inside and on C except for a finite number of (isolated)
singular points zk inside C, then
n
f ( z)
c f ( z) dz  2πi  Res
z  zk
k 1
pf:
but
n
f
(
z
)
dz

 c f ( z) dz  0
c
k 1 k
c f ( z) dz  2π i Res f ( z)
k
z  zk
Cauchy’s residue theorem
C
Z3
Z1
Z2
11
Ex1.
Evaluate c 5z - 2 dz
where C : z  2
z(z-1 )
Two singularities z  0, z 1
a. When 0  z 1
5z- 2  5z  2 -1  (5  2) (-1- z- z 2.....)
z
z
1- z
z( z-1)
b1  B1  Re s f ( z)  2
z 0
b. when 0  z -1 1
5z - 2  5( z 1)  3 .
1
z 1
z( z -1)
1 ( z 1)
 (5  3 ) [1- ( z -1)  ( z -1)2.......]
z 1
b1  B2  3
c 5z  2 dz  2π i (2  3) 10π i
z( z 1)
12
分解大突破
展開法
Evaluate c 5z- 2 dz where C: z  2
z(z-1 )
Two singularities z  0, z 1
a. When 0  z 1
5z- 2  5z  2 -1  (5  2) (-1-z-z 2.....)
z
z( z-1) z 1-z
b1  B1  Re s f (z)  2
z 0
b. when 0  z-1 1
5z- 2  5( z 1)  3 . 1
z 1 1 ( z 1)
z( z-1)
 (5  3 ) [1- (z-1)  (z-1)2.......]
z 1
b1  B2  3
c 5z  2 dz  2π i (2  3) 10π i
z( z 1)
係數比較法
因式分解法
5z  2
A
B
5z  2
A
B
 
 
z ( z  1) z z  1 z ( z  1) z z  1
A( z  1)  Bz  5 z  2
5z  2
A
2
z  1 z 0
A  2, B  3
B
5z  2
3
z z 1
13
n
f
(
z
)
dz

2
πi
f ( z)
 Res
c
z  zk
k 1
pf:
展開法
but
n
f
(
z
)
dz

 c f ( z) dz  0
c
係數比較法
k 1 k
f ( z)
ck f ( z) dz  2π i Res
zz
因式分解法
k
分別找k個singular
points 的c-1(Residue)
g(z)
1個z=0的residue
Thm2: If a function f is analytic everywhere in the finite plane except
for a finite number of singular points interior to a positively
oriented simple closed contour C, then
2
1
1
f
(
z
)
dz

2
πi
Re
s
[
f
(
)]
c
2
z 0
z
z
Z-1 0
Z1
14
Thm2: If a function f is analytic everywhere in the finite plane except
for a finite number of singular points interior to a positively
oriented simple closed contour C, then
C0
1
1
s [ 2 f ( )]
c f ( z) dz  2πi Re
z 0
z
z
Pf:
From Laurent Theorem

(R1  z  ) (3)
f ( z)   cn z n
n
where cn  1 c f ( z) dz
2πi 0 z n1
c0 f ( z) dz  2πi c1 (c1 is not the residue of f at z=0)
C
R1 R0
Replace z by 1 in (3),
z
 c

cn
1
1 1
n2
)

z

(0


)
(
f


n
n2
2
R1
z
z n 
z
n z
now c1 is the residue
1
1
c1  Re s [ 2 f ( )]
z 0
z
z
of 12 f (1z ) at z  0 15
z
Ex2.
f ( z)  5z  2
z  ( z 1)
1 f (1 )  5  2 z  5  2 z . 1
z 1 z
z 2 z z(1 z)
 (5z  2)(1 z  z 2  ....)
 5z  3  3z  ......
Re s 1 f (1)  5
z
z 0 z 2
c f ( z)dz 10πi
(0  z 1)
16
55. Three Types of Isolated Singular points
If f has an isolated singular point z0, then f(z) can be represented by
a Laurent series

bn
b1
b2
n
f ( z)   an ( z  z0 )  z  z 
 .....
 ....
2
n
( z  z0 )
0 ( z  z0 )
n0
in a punctured disk 0  z  z0  R2
bn
b1
b2


.....

 ....
n
z  z0 ( z  z )2
( z  z0 )
0
is called the principal part of f at z0.
The portion
17
(i) Type 1.
bm  0 and bm1  bm 2  .......  0

bm
b
f ( z)   an ( z  z0 )n  1  ..............
m
(
z

z
)
(
z

z
)
0
n0
0
0  z  z0  R2
The isolated singular point z0 is called a pole of order m.
m 1,  simple pole
Ex1.
z 2  2 z  3  z  3  2  ( z  2)  3
z 2
z 2
z 2
(0  z- 2  )
Simple pole m 1 at z0  2, b1  3.
18
Ex2.
sinh z  1 ( z  z3  z5  ....)
3! 5!
z4
z4
3
1
1
1
z
z
 3  z    ...........0  z  
5! 7!
z 3!
has pole of order m  3 at z0  0, b1  1
6
(ii) Type 2
bn=0, n=1, 2, 3,……

f ( z)   an ( z  z0 )n  a0  a1 ( z  z0 )  a2 ( z  z0 )2  ......
n0
0  z  z0  R2
z0
is known as a removable singular point.
* Residue at a removable singular point is always zero.
19
* If we redefine f at z0 so that f(z0)=a0
define
Above expansion becomes valid throughout the entire disk
z  z0  R2
* Since a power series always represents an analytic function
Interior to its circle of convergence (sec. 49), f is analytic
at z0 when it is assigned the value a0 there. The singularity
at z0 is therefore removed.
Ex3.
2 4 6
f ( z)  1 cos z  1 [1 (1 z  z  z  ....)]
2! 4! 6!
z2
z2
2 4
 1  z  z  .......
(0  z )
2! 4! 6!
when the value f (0)  1 is assigned, f become entire,
2
the point z0  0 is a removable singular point.
* another example f ( z)  sinz z .
20
(iii) Type 3:
Infinite number of bn is nonzero.
z0 is said to be an essential singular point of f.
In each neighborhood of an essential singular point, a
function assumes every finite value, with one possible
exception, an infinite number of times. ~ Picard’s theorem.
21
Ex4.

exp(1z )   1 1n 1 1 1z  1 12  ...... 0  z 
1! 2! z
n0 n! z
has an essential singular point at z0  0
where the residue b1 1
* Note that exp z  -1 when z  (2n 1)π i
(n  0, 1,  2, ...)
1
i
 exp (1z )  1 when z 
(2n 1)πi (2n 1)π
(n  0, 1,  2, ...)
an infinite number of these points clearly lie in any given
neighborhood of the origin.
* Since exp(1z )  0 for any value of z, zero is the exceptional
value in Picard's theorem.
22
* exp z 1 when z  2nπ i
(n  0, 1,  2, ...)
 exp (1z ) 1 when z  1  - i
2nπi 2nπ
(n  0, 1,  2, ...)
* exp z  i when z  (2n 1/ 2)π i
(n  0, 1,  2, ...)
1
i
 exp (1z )  i when z 
(2n 1/ 2)πi (2n 1/ 2)π
(n  0, 1,  2, ...)
an infinite number of these points clearly lie in any given
neighborhood of the origin.
23
56. Residues at Poles
identify poles and find its corresponding residues.
Thm. An isolated singular point z0 of a function f is a pole
of order m iff f(z) can be written as
f ( z)   ( z) m
( z  z0 )
where  ( z) is analytic and nonzero at z0 .
Moreover, Res f ( z)   ( z0 ) if m 1
z  z0
and
Res f ( z) 
z  z0
 (m-1) ( z0 )
(m 1)!
if m  2
24
Pf: “<=“
Suppose f ( z)   ( z) m .
( z - z0 )
Since  ( z) is analytic at z0 , it has a Taylor series representation
 ( z)   ( z0 ) 
 '( z0 )
( z  z0 ) 
 ''( z0 )
( z  z0 )2  .......
1!
2!
(m 1)
  ( n) ( z )

( z0 )
0

( z  z0 )m1  
( z  z0 ) n
n!
(m 1)!
nm
 ( z0 )   '( z0 ) /1!   ''( z0 ) / 2!  .......
f ( z) 
( z  z0 ) m ( z  z0 ) m1 ( z  z0 ) m 2
(m 1)

( z0 ) /(m 1)!




 ( n ) ( z0 )
( z  z0 ) n  m
n!
( z  z0 )
nm
Since  ( z0 )  0, z0 is a pole of order m of f ( z) and
z  z0  
0  z  z0  
(m 1)

( z0 )
Res f ( z) 
.
z  z0
(m 1)!
25
“=>”
If z0 is a pole of order m of f , or f ( z) has a Laurent series representation

b
b2
bm
f ( z)   an ( z  z0 )n  z 1z 

.....

(bm  0)
2
m
(
z

z
)
(
z

z
)
0
n0
0
0
in a punctured disk 0  z  z0  R2
The function defined by
( z  z0 ) m f ( z ) when z  z0
 ( z)  
bm
when z  z0

has the power series representation
 (z)  bm  bm1(z  z0 )   b2 (z  z
(z  z0
)m2  b1
0
)m1 

 an ( z  z0 )
m n
n 0
throughout z  z0  R2.
Consequently,  ( z) is analytic in that disk (sec.49)
and, in particular at z0.
Also  (z0 )  bm  0.
26
Ex1.
f ( z)  z 1 has an isolated singular point at z  3 i
z2  9
f ( z)   ( z)
z 3 i
where  ( z)  z 1
z 3 i
 ( z) is analytic at z  3 i
 (3 i)  3 i 1  0 a simple pole
6i
Re s  3  i
6
z 3i
another simple pole
z  -3 i
residue 3  i
6
27
Ex3.
f ( z)  sinh z
z4
To find residue at z0  0,
can not write f ( z)   ( z) ,  ( z)  sinh z
z4
since  ( z0 )  0
Need to write out the Laurent series for f(z) as in Ex 2.
Sec. 55.
sinh z  1  1 1  1 z  ..........
z4
z3 3! z 5!
z0  0 is a pole of the third order, its residue  1
6
28
Ex4.
Since z(e z 1) is entire and its zeros are
z  2n i (n  0, 1,  2, ..... )
z  0 is an isolated singular point of
f ( z)  z1
z(e 1)
2 3
e z 1 z  z  z  ....
z 
1! 2! 3!
2
z
z
z
2
z(e 1)  z (1   .....) z  
2! 3!
1
Thus f ( z)   ( z)
 ( z) 
2
z2
1 z  z  .......
2! 3!
Since  ( z) is analytic at z  0, and  (0) 1  0
z  0 is a pole of the second order
( 1  2 z  .....)
2! 3!
b1  ' (0) 
2
(1 z  z  .....)2
z 0
2! 3!
-1
2
29
57. Zeros and Poles of order m
Consider a function f that is analytic at a point z0.
(From Sec. 40). f (n)( z) (n 1, 2, ....) exist at z
0
If f ( z0 )  0 ,
f '( z0 )  0
:
f (m-1) ( z )  0
0
f (m) ( z0 )  0
Then f is said to have a zero of order m at z0.
Lemma:
f ( z)  ( z  z0 )m g ( z)
analytic and non-zero at z0.
30
Ex1.
f ( z)  z(e z 1)
2
 z 2(1 z  z  ......)
2! 3!
has a zero of order m  2 at z0  0
(e z 1) / z when z 0
g ( z)  
when z 0
 1
is analytic at z  0.
Thm. Functions p and q are analytic at z0, and p( z0 )  0.
If q has a zero of order m at z0, then
p( z) has a pole of order m there.
q( z)
q( z)  ( z  z )m g ( z)
0
analytic and non zero
p( z)  p( z) /g ( z)
q( z) ( z  z0 )m
31
Ex2.
f ( z) 
1
has a pole of order 2 at z0  0
z(ez 1)
Corollary: Let two functions p and q be analytic at a point z0.
If p( z0 )  0 , q( z0 )  0 , and q' (z0 )  0
then z0 is a simple pole of p(z) and
q(z)
p( z0 )
Re s p( z) 
z  z0
q( z) q' ( z0 )
Pf:
q( z)  ( z  z0 ) g ( z),
g ( z) is analytic ard non zero at z0
p( z)  p( z)/ g ( z)
z  z0
q( z)
p(z)  p( z0 )
Res
Form Theorem in sec 56,
zz
q(z) g ( z )
0
But g ( z0 )  q' ( z0 )
0

p( z0 )
q' ( z0 )
32
58. Conditions under which f ( z)  0
連續周圍2D, 3D
Lemma : If f(z)=0 at each point z of a domain or arc
containing a point z0, then f ( z)  0 in any
neighborhood N0 of z0 throughout which f is
analytic. That is, f(z)=0 at each point z in N0.
Pf:
f(z)
Z0
Under the stated condition,
arc
For some neighborhood N of z0
N
f(z)=0
N0
Otherwise from (Ex13, sec. 57)
There would be a deleted neighborhood of z0
throughout which f ( z)  0
0
 inconsistent with f ( z)  0 in a domain or arc
containing z0.
34
f(z0)
w0  f ( z0 )  0
z0
0
w0  f ( z0 )  0
w
f(z0)
z0
n 
 an ( z  z 0 ) n
0
z
n 0
w  w0  
w
n 
 a (z  z )
n 0
n
n
0
f(z0)
z0
0
z
w
n 
n
a
(
z

z
)
 00
 n
0
n 0
35
Since f ( z)  0 in N, an in the Taylor series for f(z) about z0
must be zero.
Thus f ( z)  0 in neighborhood N0 since that Taylor series
also represents f(z) in N0.
圖解
Z0 Z
若有一點 f ( z )  0
則全不為 0
全為 0
Z0
Z
若在 arc or domain為 0 ,則
Ex13, sec 57
Theorem. If a function f is analytic throughout a domain D
and f(z)=0 at each point z of a domain or arc
contained in D, then f ( z)  0 in D.
P
Z0 Z1 Z 2 Z3
Zn
36
Corollary: A function that is analytic in a domain D is uniquely
determined over D by its values over a domain, or
along an arc, contained in D.
D
f ( z), g ( z) analytic in D
f ( z)  g ( z) in some domain or arc contained in D
h( z)  f ( z) - g ( z)  0 in a domain or acr
h( z)  0 in D
f ( z)  g ( z) in D
domain
arc
Example:
Since sin2 x  cos2 x 1
along real x-axis
(an arc)
f ( z)  sin 2 z  cos2 z 1
is zero along the real axis
 f ( z)  0 throughout the complex plane
 sin 2 z  cos2 z 1 for all z
37
Sect. 59
• Reference, study for strengthen your theory
background
• Is not covered in final exam
38
Cauchy Integral 補充
1. Cauchy-Goursat Theorem: c f ( z) dz  0if f analytic.
2 1







f ( z)
2. c
zz
0







z
dz  2i  f ( z )
0
3
f(z)
3.
c
f ( z)
z  z0 n 1
dz  2i 
f
z
X
z0







f ( z)
zz
0
(n)
(z )
0
n!
43







c
Ex4.
dz
z( z  2)4
C : z- 2 1
1
has singular points at z  0, z  2
z( z  2)4
it has Laurent series representation
in 0  z - 2  2
dz  2πi Re s
1
based on (2)
c
z  2 z ( z  2)4
z( z  2)4
but
1  1 .
1
z( z- 2)4 ( z  2)4 2  ( z  2)
1
 1 .
2( z- 2)4 1 ( z  2)
2
 (-1)n
 
( z- 2)n-4
n0 2n1
b1   1
16
c
0
2
湊出z-2在分母
0 z -2  2
dz  2 i   1   π i
 16  8
z( z- 2)24


44
Ans:2πi·3
Ans:0
z 2
2z
3e
C z dz
for C:
z 1
z 1
3e 2 z
C ( z  5)( z  1) 2 dz
1  e2z
C ( z  3) 2 dz
3e 2 z
C ( z  3) 4 dz
for C:
z 2
1  e2z
C z 2 dz
for C:
for C:
for C:
z 2
z 1
z 3  1
45