Complex Analysis/development in Laurent series

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:

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

with $a_n\in ℂ$, which converges on a punctured disk (i.e., excluding the center $z_0$) around $z_0$.

Laurent Expansion on an Annulus

A more general case than the 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 annulus around $z_0$. If $f:A_{r_1,r_2}\to ℂ$ is 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 $A_{r_1,r_2}$ possesses a Laurent expansion around $z_0$. Coefficients $a_n$ with $n\in \mathbb{Z}$ exist and can be calculated with

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

for any 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 a Laurent Expansion

The uniqueness follows from the identity theorem for Laurent Series. For existence, choose $r$ with $r_1<r<r_2$ and $R_1,R_2$ such that $r_1<R_1<r<R_2<r_2$. Let $z\in A_{R_1,R_2}$ be arbitrary. "Cut" the annulus $A_{R_1,R_2}$ at two points using radii $D_1$ and $D_2$ such that the cycle $\partial K_{R_2}-\partial K_{R_1}$ can be expressed as the sum of two closed, null-homotopic curves $C_1$ and $C_2$. Choose $D_1$ and $D_2$ such that $z$ is enclosed 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 enclose $z$. Hence, due to $C_1+C_2=\partial K_{R_2}-\partial K_{R_1}$, we obtain:

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

We now expand

$\frac{1}{w-z}$

for

$|w-z_0|=R_2$

using:

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

This series converges absolutely for $|z-z_0|<|w-z_0|$, yielding:

$\frac{1}{2\pi i}\underset{|w-z_0|=R_2}{\int }\frac{f(w)}{w-z}dw=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{2\pi i}\underset{|w-z_0|=R_2}{\int }\frac{f(w)}{(w-z_0{)}^{n+1}}dw\right)(z-z_0{)}^n.$

Similarly, for the inner circle $|w-z_0|=R_1$, we expand and calculate analogously. The final result shows that for $z\in A_{R_1,R_2}$, the Laurent series converges, proving the existence of the Laurent expansion.

See Also

• Example Computation with Laurent Series
• Laurent Series
• Holomorphic function

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• Source: Kurs:Funktionentheorie/Laurententwicklung - URL:

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• Date: 1/1/2025

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