📂Complex Anaylsis

Residue at a Simple Pole

Theorem ¹

Suppose a function $f$ can be expressed as $\displaystyle f(z) = {{g(z)} \over {h(z)}}$. Here, if $g$ and $h$ are analytic at $\alpha$, and $g(\alpha) \ne 0 , h(\alpha) = 0, h ' (\alpha) \ne 0$, then $\alpha$ is a simple pole of $f$ and $$ \text{Res}_{\alpha} f(z) = {{g(\alpha)} \over {h ' (\alpha)}} $$

Since $h$ does not necessarily have to be a polynomial in the form $\displaystyle f(z) = {{g(z)} \over {h(z)}}$, this cannot merely be called a theorem that restricts the residue at a pole to $m=1$. Because it can cover many more kinds of functions $h$ as long as the conditions are well satisfied, its applications are boundless.

One thing to keep well in mind is that, upon reading the theorem carefully, the fact that $f$ has a simple pole $\alpha$ is not a condition but a result.

Since $\alpha$ is not something we have to show is a simple pole but rather something that is shown to be so, we only need to worry about the conditions on $g$ and $h$.

Proof

Since $h ' (\alpha) \ne 0$ from the assumption, if we set $\displaystyle H(z) = {{ h(z) - h(\alpha) } \over { z - \alpha }}$, then $H(\alpha) = h ' (\alpha) \ne 0$, and since $h(\alpha) = 0$ from the assumption, $$ f(z) = {{g(z)} \over {h(z)}} = {{g(z)} \over {h(z) - h(\alpha) }} = {{g(z)} \over {(z - \alpha) H(z) }} $$ Since $g / H$ is analytic at $\alpha$ and $\displaystyle {{g(\alpha)} \over {H(\alpha)}} \ne 0$, $\alpha$ is a pole of $f$ of order $1$.

Residue at a Pole: If $\alpha$ is a simple pole, then $\displaystyle \text{Res}_{\alpha} f(z) = \lim_{z \to \alpha} (z - \alpha) f(z)$

The residue at the pole is $$ \begin{align*} \text{Res}_{\alpha} f(z) =& \lim_{z \to \alpha} g(z) {{1} \over {H(z)}} \\ =& \lim_{z \to \alpha} g(z) \cdot \lim_{z \to \alpha} {{z - \alpha} \over {h(z) - h(\alpha) }} \\ =& g(\alpha) \cdot {{1} \over {h ' (\alpha) }} \end{align*} $$

■

Pole of Order 2

Meanwhile, although its practicality as a formula is lower, the following theorem is known for $2$. The method of proof is essentially not much different from what was done for the simple pole.

Residue at a Pole of Order $2$: Suppose a function $f$ can be expressed as $\displaystyle f(z) = {{g(z)} \over {h(z)}}$. Here, if $g$ and $h$ are analytic at $\alpha$, and $g(\alpha) \ne 0 , h(\alpha) = h ' (\alpha) = 0, h’’(\alpha) \ne 0$, then $\alpha$ is a pole of $f$ of order $2$, and $$\displaystyle \text{Res}_{\alpha} f(z) = {{2g ' (\alpha)} \over {h’’(\alpha)}} - {{2g(\alpha) h’’’(\alpha) } \over {3 (h’’(\alpha))^2 }}$$