# Complex Analysis/development in Laurent series
## 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:
$f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n}$
with $a_{n}\in \mathbb {C}$, 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 $0\leq r_{1}<r_{2}$ 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 annulus around $z_{0}$. If $f\colon A_{r_{1},r_{2}}\to \mathbb {C}$ is 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 $A_{r_{1},r_{2}}$ possesses a Laurent expansion around $z_{0}$. Coefficients $a_{n}$ with $n\in \mathbb {Z}$ exist and can be calculated with
$a_{n}={\frac {1}{2\pi i}}\int _{|z-z_{0}|=r}{\frac {f(z)}{(z-z_{0})^{n+1}}}\,dz$
for any radius $r$ with $r_{1}<r<r_{2}$.

### Uniqueness

The coefficients are uniquely determined by:
$a_{n}={\frac {1}{2\pi i}}\int _{|z-z_{0}|=r}{\frac {f(z)}{(z-z_{0})^{n+1}}}\,dz$

### 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:
$f(z)={\frac {1}{2\pi i}}\int _{C_{1}}{\frac {f(w)}{w-z}}\,dw$
and
$0={\frac {1}{2\pi i}}\int _{C_{2}}{\frac {f(w)}{w-z}}\,dw$
since $C_{2}$ does not enclose $z$. Hence, due to $C_{1}+C_{2}=\partial K_{R_{2}}-\partial K_{R_{1}}$, we obtain:
$f(z)={\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{w-z}}\,dw-{\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{1}}{\frac {f(w)}{w-z}}\,dw$
We now expand ${\frac {1}{w-z}}$ for $|w-z_{0}|=R_{2}$ using:
${\begin{array}{rl}{\frac {1}{w-z}}&={\frac {1}{(w-z_{0})-(z-z_{0})}}\\&={\frac {1}{w-z_{0}}}\cdot {\frac {1}{1-{\frac {z-z_{0}}{w-z_{0}}}}}\\&={\frac {1}{w-z_{0}}}\sum _{n=0}^{\infty }{\frac {(z-z_{0})^{n}}{(w-z_{0})^{n}}}\end{array}}$
This series converges absolutely for $|z-z_{0}|<|w-z_{0}|$, yielding:
${\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{w-z}}\,dw=\sum _{n=0}^{\infty }\left({\frac {1}{2\pi i}}\int _{|w-z_{0}|=R_{2}}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\right)(z-z_{0})^{n}.$
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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