Let $\Omega$ be a domain (open subset), and $z_0 \in \Omega$ a point. We say a holomorphic function $f: \Omega \RM \{z_0\} \to \C$ has a pole of order $m$ at $z_0$, if $h(z) = (z-z_0)^m f(z)$ can be extended to the entire $\Omega$, and $h(z_0) \neq 0$.

Suppose $f$ has an order $m$ pole at $z_0$, then near $z_0$ (more precisely, for $z \in \C$ with $0 < |z-z_0| < r$ for some small positive $r$), we can write Laurent expansion $f(z) = \frac{a_{-m}}{ (z-z_0)^m} + \cdots + \frac{a_{-1}}{ (z-z_0)^1} + a_0 + a_1 (z - z_0)^1 + \cdots$

The coefficient $a_{-1}$ is called the residue, and denoted as $Res_{z_0} f := a_{-1}$. ($:=$ read 'is defined as')

1. If the pole is order $1$, then $Res_{z_0}f(z) = \lim_{z \to z_0} f(z) (z-z_0)$.

Example: residue of $f(z) = (z-2) / (z-1)$ at $z=1$.

2. If the pole at $z_0$ is of order $m>1$, then just Laurent expand. More precisely, do Taylor expansion of $(z-z_0)^m f(z) = a_{-m} + \cdots + a_{-1} (z-z_0)^{m-1} + \cdots$ and then you get $a_{-1}$.

Example: residue of $f(z) = [(z-2) / (z-1)]^2$ at $z=1$.

Thm: let $f$ has a pole (of some order) at $z_0$, and let $C$ be a small circle of radius $\epsilon$ around $z_0$ positively oriented (i.e in counter-clockwise direction) so that $f$ is holomorphic on $C$, and $C$ encloses only one pole. Then $\oint_{C} f(z) dz = (2\pi i) Res_{z_0} f.$

Proof sketch: write $f$ as a Laurent expansion, then integrate term by term.