Complex Analysis/Laurent Expansion

Laurent Expansion Around a Point

Let $G\subseteq ℂ$ be a domain, $z_0\in G$, and $f:G\setminus z_0\to ℂ$ a holomorphic function. A Laurent expansion of $f$ around $z_0$ is a representation of $f$ as a Laurent Series:

$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$

where

$a_n\in ℂ$

, and the series converges on an annular region around

$z_0$

(i.e., excluding the point

$z_0$

).

Laurent Expansion on an Annulus

A slightly more general form of the expansion above is the following: Let $0\le r_1<r_2$ be two radii (the expansion around a point corresponds to $r_1=0$), and let $A_{r_1,r_2}:=z\in ℂ:r_1<|z-z_0|<r_2$ be an annular region around $z_0$, and let $f:A_{r_1,r_2}\to ℂ$ be a holomorphic function. Then the Laurent Series

$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$

with

$a_n\in ℂ$

is a Laurent expansion of

$f$

on

$A_{r_1,r_2}$

, provided the series converges for all

$z\in A_{r_1,r_2}$

.

Existence

Every holomorphic function on $A_{r_1,r_2}$ has a Laurent expansion around $z_0$, and the coefficients $a_n$ in the expansion are given by:

$a_n=\frac{1}{2\pi i}\underset{|z-z_0|=r}{\int }\frac{f(z)}{(z-z_0{)}^{n+1}}dz$

for a radius

$r$

with

$r_1<r<r_2$

.

Uniqueness

The coefficients are uniquely determined by:

$a_n=\frac{1}{2\pi i}\underset{|z-z_0|=r}{\int }\frac{f(z)}{(z-z_0{)}^{n+1}}dz$

Proof of Existence and Uniqueness of the Laurent Representation

Uniqueness follows from the Identity Theorem for Laurent Series. To prove existence, choose a radius $r$ such that $r_1<r<r_2$ and choose $R_1,R_2$ so that $r_1<R_1<r<R_2<r_2$. Let $z\in A_{R_1,R_2}$ be arbitrary. "Cut" the annular region $A_{R_1,R_2}$ at two points using radii $D_1$ and $D_2$ such that the cycle $\partial K_{R_2}-\partial K_{R_1}$ is represented as the sum of two closed curves $C_1$ and $C_2$ in $A$ that are null-homotopic. Choose $D_1$ and $D_2$ so that $z$ is encircled by $C_1$. By the Cauchy Integral Theorem, we have:

$f(z)=\frac{1}{2\pi i}\underset{C_1}{\int }\frac{f(w)}{w-z}dw$

and

$0=\frac{1}{2\pi i}\underset{C_2}{\int }\frac{f(w)}{w-z}dw$

since

$C_2$

does not encircle

$z$

. Thus, because

$C_1+C_2=\partial K_{R_2}-\partial K_{R_1}$

, we have:

$f(z)=\frac{1}{2\pi i}\underset{|w-z_0|=R_2}{\int }\frac{f(w)}{w-z}dw-\frac{1}{2\pi i}\underset{|w-z_0|=R_1}{\int }\frac{f(w)}{w-z}dw$

For

$|w-z_0|=R_2$

, we have:

$\begin{array}{cc}\frac{1}{w-z}& =\frac{1}{(w-z_0)-(z-z_0)}\\ & =\frac{1}{w-z_0}\cdot \frac{1}{1-\frac{z-z_0}{w-z_0}}\\ & =\frac{1}{w-z_0}\sum _{n=0}^{\mathrm{\infty }}\frac{(z-z_0{)}^n}{(w-z_0{)}^n}\end{array}$

The series converges absolutely because

$|z-z_0|<|w-z_0|$

, and we obtain:

$\begin{array}{cc}\frac{1}{2\pi i}\underset{|w-z_0|=R_2}{\int }\frac{f(w)}{w-z}dw& =\frac{1}{2\pi i}\underset{|w-z_0|=R_2}{\int }\frac{1}{w-z_0}\cdot \frac{f(w)(z-z_0{)}^n}{(w-z_0{)}^n}dw\\ & =\frac{1}{2\pi i}\sum _{n=0}^{\mathrm{\infty }}\underset{|w-z_0|=R_2}{\int }\frac{f(w)}{(w-z_0{)}^{n+1}}dw\cdot (z-z_0{)}^n\\ & =\frac{1}{2\pi i}\sum _{n=0}^{\mathrm{\infty }}\underset{|w-z_0|=r}{\int }\frac{f(w)}{(w-z_0{)}^{n+1}}dw\cdot (z-z_0{)}^n\end{array}$

Now, consider the integral over the inner circle, which is analogous to the above for

$|w-z_0|=R_1$

:

$\begin{array}{cc}\frac{1}{w-z}& =\frac{1}{(w-z_0)-(z-z_0)}\\ & =\frac{-1}{z-z_0}\cdot \frac{1}{1-\frac{w-z_0}{z-z_0}}\\ & =\frac{-1}{z-z_0}\sum _{n=0}^{\mathrm{\infty }}\frac{(w-z_0{)}^n}{(z-z_0{)}^n}\end{array}$

Thus, due to

$R_1=|w-z_0|<|z-z_0|$

, the series converges, and we obtain:

$\begin{array}{cc}-\frac{1}{2\pi i}\underset{|w-z_0|=R_1}{\int }\frac{f(w)}{w-z}dw& =\frac{1}{2\pi i}\underset{|w-z_0|=R_1}{\int }\frac{-1}{z-z_0}\cdot \frac{f(w)(w-z_0{)}^n}{(z-z_0{)}^n}dw\\ & =\frac{1}{2\pi i}\sum _{n=0}^{\mathrm{\infty }}\underset{|w-z_0|=R_1}{\int }\frac{f(w)}{(w-z_0{)}^{-n}}dw\cdot (z-z_0{)}^{-n-1}\\ & =\frac{1}{2\pi i}\sum _{n=0}^{\mathrm{\infty }}\underset{|w-z_0|=r}{\int }\frac{f(w)}{(w-z_0{)}^{-n}}dw\cdot (z-z_0{)}^{-n-1}\end{array}$

Thus, it follows that for

$z\in A_{R_1,R_2}$

:

$f(z)=\frac{1}{2\pi i}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\underset{|w-z_0|=r}{\int }\frac{f(w)}{(w-z_0{)}^{n+1}}dw\cdot (z-z_0{)}^n$

which proves the existence of the claimed Laurent expansion.

See Also

• Examples of Laurent Series Expansion
• Laurent Series

Page information

Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Kurs:Funktionentheorie/Laurententwicklung - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Laurententwicklung

• Date: 11/26/2024