# Residue at Simple Pole

## Theorem

Let $f: \C \to \C$ be a function meromorphic on some region, $D$, containing $a$.

Let $f$ have a simple pole at $a$.

Then the residue of $f$ at $a$ is given by:

$\displaystyle \operatorname{Res} \left({f, a}\right) = \lim_{z \mathop \to a} \left({z - a}\right) f \left({z}\right)$

## Proof

By Existence of Laurent Series, there exists a Laurent series:

$\displaystyle f\left({z}\right) = \sum_{n \mathop = -\infty}^\infty c_n \left({z - a}\right)^n$

which is convergent in $D \setminus \left\{a\right\}$, where $\left({c_n}\right)$ is a doubly infinite sequence of complex coefficients.

We are given that $f$ has only a simple pole at $a$.

Thus $c_n = 0$ for $n < -1$.

So we can write:

$\displaystyle f\left({z}\right) = \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^n + \frac {c_{-1} } {z - a}$

Then:

 $\displaystyle \lim_{z \mathop \to a} \left({z - a}\right) f\left({z}\right)$ $=$ $\displaystyle \lim_{z \mathop \to a} \left({z - a}\right) \left({ \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^n + \frac {c_{-1} } {z - a} } \right)$ $\displaystyle$ $=$ $\displaystyle \lim_{z \mathop \to a} \left({ \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^{n + 1} + c_{-1} }\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty c_n \left({a - a}\right)^{n+1} + c_{-1}$ $\displaystyle$ $=$ $\displaystyle 0 \sum_{n \mathop = 0}^\infty c_n + c_{-1}$ $\displaystyle$ $=$ $\displaystyle c_{-1}$ $\displaystyle$ $=$ $\displaystyle \operatorname{Res} \left({f, a}\right)$ Definition of Residue

$\blacksquare$