Complex Analysis/Laurent Expansion

Laurent Expansion Around a Point

Let $G\subseteq \mathbb {C}$ be a domain, $z_{0}\in G$ , and $f\colon G\setminus {z_{0}}\to \mathbb {C}$ a holomorphic function. A Laurent expansion of $f$ around $z_{0}$ is a representation of $f$ as a Laurent Series:

f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}

where $a_{n}\in \mathbb {C}$ , 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 $0\leq r_{1}<r_{2}$ be two radii (the expansion around a point corresponds to $r_{1}=0$ ), and let $A_{r_{1},r_{2}}:={z\in \mathbb {C} :r_{1}<|z-z_{0}|<r_{2}}$ be an annular region around $z_{0}$ , and let $f\colon A_{r_{1},r_{2}}\to \mathbb {C}$ be a holomorphic function. Then the Laurent Series

f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}

with $a_{n}\in \mathbb {C}$ 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}}\int _{|z-z_{0}|=r}{\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}}\int _{|z-z_{0}|=r}{\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}}\int _{C_{1}}{\frac {f(w)}{w-z}}\,dw

and

0={\frac {1}{2\pi i}}\int _{C_{2}}{\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}}\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{w-z}}\,dw-{\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{1}}{\frac {f(w)}{w-z}}\,dw

For $|w-z_{0}|=R_{2}$ , we have:

{\begin{array}{rl}\displaystyle {\frac {1}{w-z}}&=\displaystyle {\frac {1}{(w-z_{0})-(z-z_{0})}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\cdot {\frac {1}{1-{\frac {z-z_{0}}{w-z_{0}}}}}\\&=\displaystyle {\frac {1}{w-z_{0}}}\sum _{n=0}^{\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}{rl}{\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{w-z}}\,dw&={\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\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}^{\infty }\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}\\&={\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{|w-z_{0}|=r}{\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}{rl}\displaystyle {\frac {1}{w-z}}&=\displaystyle {\frac {1}{(w-z_{0})-(z-z_{0})}}\\&=\displaystyle {\frac {-1}{z-z_{0}}}\cdot {\frac {1}{1-{\frac {w-z_{0}}{z-z_{0}}}}}\\&=\displaystyle {\frac {-1}{z-z_{0}}}\sum _{n=0}^{\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}{rl}-{\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{1}}{\frac {f(w)}{w-z}}\,dw&={\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{1}}{\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}^{\infty }\int _{|w-z_{0}|=R_{1}}{\frac {f(w)}{(w-z_{0})^{-n}}}\,dw\cdot (z-z_{0})^{-n-1}\\&={\frac {1}{2\pi i}}\sum _{n=0}^{\infty }\int _{|w-z_{0}|=r}{\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=-\infty }^{\infty }\int _{|w-z_{0}|=r}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\cdot (z-z_{0})^{n}

which proves the existence of the claimed Laurent expansion.

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Date: 11/26/2024

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