📂Complex Anaylsis

Flow in Simple Extremes

Theorem ¹

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

It’s not just that $h$ in the form of $\displaystyle f(z) = {{g(z)} \over {h(z)}}$ has to be a polynomial, so it can’t be said to be simply a theorem limiting the residue to $m=1$. As long as the conditions are met well, it’s possible to cover much more types of functions, $h$, which makes its application infinite.

One thing to note is that if you read through the theorem properly, it is not a condition but a result that $f$ has a simple pole $\alpha$.

$\alpha$ being a simple pole is not something to show but something shown, so all that is needed is to pay attention to the conditions regarding $g$ and $h$.

Proof

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

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

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Order-2 Pole

While less practical as a formula, the following theorem is known for $2$. The method of proof is fundamentally no different from that used for simple poles.

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