📂Complex Anaylsis

Residue at a Pole

Theorem ¹

Suppose $\alpha$ is a pole of order $m$ of a function $f: A \subset \mathbb{C} \to \mathbb{C}$, that is, $f$ can be written as $\displaystyle f(z) = {{g(z)} \over { (z - \alpha)^m }}$. Here, if $g$ is analytic at $\alpha$ and $g(\alpha) \ne 0$, then $$ \text{Res}_{\alpha} f(z) = {{g^{(m-1)} (\alpha)} \over {(m-1)!} } $$

Being able to turn an integration problem into a problem of finding a residue via the residue theorem is all well and good, but it is useless if finding the residue is as hard as the integration. If we had to perform a Laurent expansion according to the definition and extract the residue every time we use the residue theorem, the computation would be far too difficult in practice. Thus scholars set out to find easy ways to compute residues, and at least for poles they found a formula that computes the residue extremely easily and quickly. However, even with this theorem, since one must differentiate $g$ a total of $(m-1)$ times, once $m \ge 3$ or so it may be more convenient to just use the definition of the residue.

Proof

Since $g$ is analytic at $\alpha$, it can be expressed via a Taylor expansion as $\displaystyle g(z) = \sum_{n=0}^{\infty} {{g^{(n)} ( \alpha ) ( z - \alpha )^{n} } \over {n!}}$. By assumption, $$ \begin{align*} f(z) =& \sum_{n=0}^{\infty} {{g^{(n)} ( \alpha ) ( z - \alpha )^{n-m} } \over {n!}} \\ =& \sum_{n=0}^{\infty} {{g^{(n)} ( \alpha ) } \over {n! ( z - \alpha )^{m-n}} } \end{align*} $$ By the definition of the residue, the coefficient $\displaystyle {{g^{(m-1)} (\alpha)} \over {(m-1)!} }$ of $\displaystyle {{1} \over {z - \alpha}}$ becomes the residue $\text{Res}_{\alpha} f(z)$ of $f$ at $\alpha$.

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In particular, when $m=1$, that is, when $\alpha$ is a simple pole, we have $\displaystyle \text{Res}_{\alpha} f(z) = g(\alpha) = \lim_{z \to \alpha} (z - \alpha) f(z)$, allowing us to solve problems very conveniently.