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:

$\ds \Res f a = \lim_{z \mathop \to a} \paren {z - a} \map f z$

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

which is convergent in $D \setminus \set a$, where $\sequence {c_n}$ 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:

$\ds \map f z = \sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n + \frac {c_{-1} } {z - a}$

Then:

$\ds \lim_{z \mathop \to a} \paren {z - a} \map f z$

$=$

$\ds \lim_{z \mathop \to a} \paren {z - a} \paren {\sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n + \frac {c_{-1} } {z - a} }$

$\ds $

$=$

$\ds \lim_{z \mathop \to a} \paren {\sum_{n \mathop = 0}^\infty c_n \paren {z - a}^{n + 1} + c_{-1} }$

$\ds $

$=$

$\ds \sum_{n \mathop = 0}^\infty c_n \paren {a - a}^{n + 1} + c_{-1}$

$\ds $

$=$

$\ds 0 \sum_{n \mathop = 0}^\infty c_n + c_{-1}$

$\ds $

$=$

$\ds c_{-1}$

$\ds $

$=$

$\ds \Res f a$

Definition of Residue

$\blacksquare$