📂Complex Anaylsis

Residue at Poles

Theorem ¹

Let $\alpha$ be represented as the pole of order $m$, that is, $\displaystyle f(z) = {{g(z)} \over { (z - \alpha)^m }}$ of the function $f: A \subset \mathbb{C} \to \mathbb{C}$. Here, if $g$ is analytic within $\alpha$ and we denote $g(\alpha) \ne 0$ as $$ \text{Res}_{\alpha} f(z) = {{g^{(m-1)} (\alpha)} \over {(m-1)!} } $$

Although we can transform the integration problem into a problem of finding residues through the Residue Theorem, it’s futile if finding the residues is as difficult as the integration itself. Each time we apply the Residue Theorem, we would have to perform the Laurent expansion according to its definition and find the residues, which is realistically too complex for computations. Therefore, scholars have developed methods to easily find residues, and fortunately, they have discovered formulas that allow us to quickly and easily find the residues for poles. However, in this theorem’s case, since it requires differentiation of $g$ by $(m-1)$ times, it might be easier to find the residues using their definition once it reaches around $m \ge 3$.

Proof

Since $g$ is analytic within $\alpha$, it can be expressed as $\displaystyle g(z) = \sum_{n=0}^{\infty} {{g^{(n)} ( \alpha ) ( z - \alpha )^{n} } \over {n!}}$ through Taylor expansion. According to the 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 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 $\alpha$ at $f$.

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Especially when $m=1$, that is, if $\alpha$ is a simple pole, it becomes $\displaystyle \text{Res}_{\alpha} f(z) = g(\alpha) = \lim_{z \to \alpha} (z - \alpha) f(z)$, which allows us to solve the problem very conveniently.