# Residue at Multiple Pole

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## Theorem

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

Let $f$ have a single pole in $D$, of order $N$, at $a$.

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

$\displaystyle \Res f a = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac { \d^{N - 1} } { \d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$

## Proof

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

$\displaystyle \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$

convergent on $D \setminus \set a$.

As $f$ has a pole of order $N$ at $a$, we have $c_n = 0$ for $n < -N$.

So:

$\displaystyle \paren {z - a}^N \map f z = \sum_{n \mathop = -N}^\infty c_n \paren {z - a}^{n + N}$

Which can be rewritten:

$\displaystyle \paren {z - a}^N \map f z = \sum_{n \mathop = 0}^\infty c_{n - N} \paren {z - a}^n$

Note that this is a Taylor series with centre $a$.

By the definition of a residue:

$\displaystyle \Res f a = c_{-1}$

This corresponds to the $\paren {N - 1}$th in the Taylor series of $\paren {z - a}^N \map f z$ about $a$.

We therefore have by Taylor Series of Holomorphic Function:

$\displaystyle c_{-1} = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac { \d^{N - 1} } { \d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$

Hence the result.

$\blacksquare$