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
2pi
 ó
(ç)
õ


CR2 
 f(z)
z- z
  dz- 1
2pi
 ó
(ç)
õ


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

laupic1.gif


2pi  f(z)
= ó
(ç)
õ


Cz,e 
 gz(z)  dz
= ó
(ç)
õ


CR2 
 gz(z)  dz - ó
(ç)
õ


CR1 
 gz(z)  dz
= ó
(ç)
õ


CR2 
 f(z)
z- z
  dz- ó
(ç)
õ


CR1 
 f(z)
z- z
  dz.

Next we write
f(z)
= 1
2pi
 ó
(ç)
õ


CR2 
 f(z)
z- z
  dz- 1
2pi
 ó
(ç)
õ


CR1 
 f(z)
z- z
  dz
=f1(z) + f2(z).
Proceeding as before
f1(z)
= 1
2pi
 ó
(ç)
õ


CR2 
 f(z)
z- z
  dz.
= 1
2pi
 ó
(ç)
õ


CR2 
 f(z) 1
z
 ¥
å
n = 0 
 æ
è 		z
z
 ö
ø 		n

 
  dz
= 1
2pi
 ó
(ç)
õ


CR2 
 f(z) 1
z
 N
å
n = 0 
 æ
è 		z
z
 ö
ø 		n

 
   dz
+ 1
2pi
 ó
(ç)
õ


CR2 
 f(z) 1
z
æ
è 		z
z
 ö
ø 		N+1

 
1 - z
z
  dz
= ¥
å
n = 0 
 an zn ,
where
an
= 1
2pi
 ó
(ç)
õ


CR2 
 f(z) 1
zn+1
   dz.
Note that f1(z) is analytic in the region {z ||z| £ R2}.
In the same spirit,
f2(z)
= - 1
2pi
 ó
(ç)
õ


CR1 
 f(z)
z- z
  dz
= 1
2pi
 ó
(ç)
õ


CR1 
 f(z) 1
z
 1
1- z
z
  dz
= 1
2pi
 ó
(ç)
õ


CR1 
 f(z) 1
z
 ¥
å
m = 0 
 æ
è 		z
z
 ö
ø 		m

 
  dz
= 1
2pi
 ó
(ç)
õ


CR1 
 f(z) 1
z
 M
å
m = 0 
 æ
è 		z
z
 ö
ø 		m

 
   dz
+ 1
2pi
 ó
(ç)
õ


CR1 
 f(z) 1
z
æ
è 		z
z
 ö
ø 		M+1

 
1 - z
z
  dz
= ¥
å
m = 0 
 1
zm+1
 æ
è 		1
2pi
 ó
(ç)
õ


CR2 
 f(z) zm  dz ö
ø 		.
Note that f2(z) is analytic in the region {z ||z| ³ R1}.

The Laurent Expansion

Theorem. Let f(z) be analytic in the region {z |R1 < |z| < R2 }. Then for R1 < |z| < R2,
f(z)
= ¥
å
n = -¥ 
 an zn,
an
= 1
2pi
 ó
(ç)
õ


Cr 
 f(z) z-n - 1  dz.
Here r is any number such that R1 < r < R2.
The series
f1(z) = ¥
å
n = 0 
 an zn
is analytic in {z | |z| < R2 },
The series
f2(z) = -1
å
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
2pi
 ó
(ç)
õ


Cr 
 f(z) z-n - 1   dz = 0, n = -1,-2, ¼.
·
If f(z) be analytic in the region {z |0 < |z| < R2 }, then
a-1 = 1
2pi
 ó
(ç)
õ


Cr 
 f(z)   dz
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)
= ¥
å
n = -¥ 
 an zn,
an
= 1
2pi
 ó
(ç)
õ


Cr 
 f(z) z-n - 1  dz.
·
If f(z) = ån = 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) = ån = N¥ an zn, N ³ 0, aN ¹ 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) ¹ 0.

·
If f(z) = ån = - M¥ an zn, M ³ 0, a-M ¹ 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) ¹ 0.

·
If f(z) = ån = - ¥¥ an zn, an ¹ 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
2pi
 ó
(ç)
õ


Cr 
 f(z)  dz.
Exercises

1.
Let f(z) be analytic in the punctured disk
×
D
 
R 
 = {z |0 < |z| £ R}.
Then for r small and positive,
ó
(ç)
õ


Cr 
 f(z)  dz = 2 pi 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,
ó
(ç)
õ


Cr 
 f¢(z)
f(z)
  dz = 2 pi ·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,
ó
(ç)
õ


Cr 
 f¢(z)
f(z)
  dz = - 2 pi ·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 3.76.
On 26 Apr 2007, 16:03.