## Laurent Series (Example)

The Laurent series is something like the Taylor series, but with terms with negative exponents, e.g. $z^{-1}$. The below Laurent Series formula may not be the most practical way to compute the coefficients, usually we will use known formulas, as the example below shows.

## Laurent Series

The Laurent series for a complex function $f(z)$ about a point $c$ is given by: $\displaystyle f(z)=\sum_{n=-\infty}^\infty a_n(z-c)^n$ where $\displaystyle a_n=\frac{1}{2\pi i}\oint_\gamma\frac{f(z)\, dz}{(z-c)^{n+1}}.$

The path of integration $\gamma$ is anticlockwise around a closed, rectifiable path containing no self-intersections, enclosing $c$ and lying in an annulus $A$ in which $f(z)$ is holomorphic. The expansion for $f(z)$ will then be valid anywhere inside the annulus.

## Example

Consider $f(z)=\frac{e^z}{z}+e^\frac{1}{z}$. This function is holomorphic everywhere except at $z=0$. Using the Taylor series of the exponential function $\displaystyle e^z=\sum_{k=0}^\infty\frac{z^k}{k!},$ we get
\begin{aligned} \frac{e^z}{z}&=z^{-1}+1+\frac{z}{2!}+\frac{z^2}{3!}+\dots\\ e^\frac{1}{z}&=1+z^{-1}+\frac{1}{2!}z^{-2}+\frac{1}{3!}z^{-3}+\dots\\ \therefore f(z)&=\dots+(\frac{1}{3!})z^{-3}+(\frac{1}{2!})z^{-2}+2z^{-1}+2+(\frac{1}{2!})z+(\frac{1}{3!})z^2+\dots \end{aligned}
Note that the residue (coefficient of $z^{-1}$) is 2.