laurent.htm Laurent Series
Laurent Series

Theorem. Let f(z) be analytic in the closed region
DR1, R2 = {0 < R1 ≤ |z| ≤ R2 }.
Then for R1 < |z| < R2,
f(z) = 1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ)
ζ− z
  dζ− 1
2πi
 ⌠
(⎜)
⌡


CR1 
 f(ζ)
ζ− z
  dζ.
 Proof: The proof is similar in spirit to the proof of the Cauchy Integral Formula.
Fix z. For ϵ small, let Cz,ϵ = {ζ| |ζ− z| = ϵ} is in between CR1 and CR2. Define gz(ζ) = [(f(ζ))/(ζ− z)]. Then gz(ζ) is an analytic function of ζ in the region between the two circles CR1 and CR2 and outside Cz,ϵ.

laupic1.gif


2πi  f(z)
= ⌠
(⎜)
⌡


Cz,ϵ 
 gz(ζ)  dζ
= ⌠
(⎜)
⌡


CR2 
 gz(ζ)  dζ − ⌠
(⎜)
⌡


CR1 
 gz(ζ)  dζ
= ⌠
(⎜)
⌡


CR2 
 f(ζ)
ζ− z
  dζ− ⌠
(⎜)
⌡


CR1 
 f(ζ)
ζ− z
  dζ.

Next we write
f(z)
= 1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ)
ζ− z
  dζ− 1
2πi
 ⌠
(⎜)
⌡


CR1 
 f(ζ)
ζ− z
  dζ
=f1(z) + f2(z).
Proceeding as before
f1(z)
= 1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ)
ζ− z
  dζ.
= 1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ) 1
ζ
 ∞
∑
n = 0 
 ⎛
⎝ 		z
ζ
⎞
⎠ 		n

 
  dζ
= 1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ) 1
ζ
 N
∑
n = 0 
 ⎛
⎝ 		z
ζ
⎞
⎠ 		n

 
   dζ
+ 1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ) 1
ζ
⎛
⎝ 		z
ζ
⎞
⎠ 		N+1

 
1 − z
ζ
  dζ
= ∞
∑
n = 0 
 an zn ,
where
an
= 1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ) 1
ζn+1
   dζ.
Note that f1(z) is analytic in the region {z ||z| ≤ R2}.
In the same spirit,
f2(z)
= − 1
2πi
 ⌠
(⎜)
⌡


CR1 
 f(ζ)
ζ− z
  dζ
= 1
2πi
 ⌠
(⎜)
⌡


CR1 
 f(ζ) 1
z
 1
1− ζ
z
  dζ
= 1
2πi
 ⌠
(⎜)
⌡


CR1 
 f(ζ) 1
z
 ∞
∑
m = 0 
 ⎛
⎝ 		ζ
z
⎞
⎠ 		m

 
  dζ
= 1
2πi
 ⌠
(⎜)
⌡


CR1 
 f(ζ) 1
z
 M
∑
m = 0 
 ⎛
⎝ 		ζ
z
⎞
⎠ 		m

 
   dζ
+ 1
2πi
 ⌠
(⎜)
⌡


CR1 
 f(ζ) 1
z
⎛
⎝ 		ζ
z
⎞
⎠ 		M+1

 
1 − ζ
z
  dζ
= ∞
∑
m = 0 
 1
zm+1
 ⎛
⎝ 		1
2πi
 ⌠
(⎜)
⌡


CR2 
 f(ζ) ζm  dζ⎞
⎠ 		.
Note that f2(z) is analytic in the region {z ||z| ≥ R1}.

The Laurent Expansion

Theorem. Let f(z) be analytic in the region {z |R1 < |z| < R2 }. Then for R1 < |z| < R2,
f(z)
= ∞
∑
n = −∞ 
 an zn,
an
= 1
2πi
 ⌠
(⎜)
⌡


Cr 
 f(ζ) ζ−n − 1  dζ.
Here r is any number such that R1 < r < R2.
The series
f1(z) = ∞
∑
n = 0 
 an zn
is analytic in {z | |z| < R2 },
The series
f2(z) = −1
∑
n = −∞ 
 an zn
is analytic in {z | R1 < |z| }.


Consequences and Notes

•
If f(z) be analytic in the region {z ||z| < R2 }, then
an = 1
2πi
 ⌠
(⎜)
⌡


Cr 
 f(ζ) ζ−n − 1   dζ = 0, n = −1,−2, ….
•
If f(z) be analytic in the region {z |0 < |z| < R2 }, then
a−1 = 1
2πi
 ⌠
(⎜)
⌡


Cr 
 f(ζ)   dζ
is called the residue of f(z) at z=0.


Zeroes, Poles, and Essential Singularities
For the moment, we shall consider a function f(z) analytic in the punctured disk
⋅
D
 
R 
 = {z |0 < |z| ≤ R}.
Then
f(z)
= ∞
∑
n = −∞ 
 an zn,
an
= 1
2πi
 ⌠
(⎜)
⌡


Cr 
 f(ζ) ζ−n − 1  dζ.
•
If f(z) = ∑n = 0∞ an zn, f(z) may be extended by defining f(0) = a0, and the resulting function is analytic in |z| ≤ R.

•
If f(z) = ∑n = N∞ an zn, N ≥ 0, aN ≠ 0, f(z) is said to have a zero of order N at z=0. Near z=0,
f(z) = zN ·g(z)
, where g(z) is analytic in |z| ≤ R, g(0) ≠ 0.

•
If f(z) = ∑n = − M∞ an zn, M ≥ 0, a−M ≠ 0, f(z) is said to have a pole of order M at z=0. Near z=0,
f(z) = z−M ·g(z)
, where g(z) is analytic in |z| ≤ R, g(0) ≠ 0.

•
If f(z) = ∑n = − ∞∞ an zn, an ≠ 0 for infinitely many negative n, then f(z) is said to have an essential singularity at z=0.

•
The coefficient of z−1 is called the residue of f(z) at z=0, and is written
Res(f,z=0) = Resf(z)|z=0 = 1
2πi
 ⌠
(⎜)
⌡


Cr 
 f(ζ)  dζ.
Exercises

1.
Let f(z) be analytic in the punctured disk
⋅
D
 
R 
 = {z |0 < |z| ≤ R}.
Then for r small and positive,
⌠
(⎜)
⌡


Cr 
 f(ζ)  dζ = 2 πi Resf(z)|z=0.

2.
Let f(z) be analytic in the punctured disk
⋅
D
 
R 
 = {z |0 < |z| ≤ R}.
Suppose that f(z) has a zero of order N > 0, at z=0.
Then for r small and positive,
⌠
(⎜)
⌡


Cr 
 f′(ζ)
f(ζ)
  dζ = 2 πi ·N.

3.
Let f(z) be analytic in the punctured disk
⋅
D
 
R 
 = {z |0 < |z| ≤ R}.
Suppose that f(z) has a pole of order M > 0, at z=0.
Then for r small and positive,
⌠
(⎜)
⌡


Cr 
 f′(ζ)
f(ζ)
  dζ = − 2 πi ·M.

4.
Let f(z) be analytic in the punctured disk
⋅
D
 
R 
 = {z |0 < |z| ≤ R}.
Suppose that f(z) is bounded as z → 0. Show that
•
limz → 0 f(z) exists.
•
f(z) may be extended to be an analytic function in
DR = {z | |z| ≤ R}.
As a consequence, the singularity of f(z) at z=0 is removable .

5.
Let f(z) be analytic in the punctured disk
⋅
D
 
R 
 = {z |0 < |z| ≤ R}.
Suppose that
f(z) = O(|z|M)  as   z → 0.
Show that for n < M, an = 0.


File translated from TEX by TTH, version 4.03.
On 02 Mar 2012, 09:45.