3.7 Residues at poles
Theorem
3.35
Suppose \(f\) has an isolated singularity at \(z_0\). Then \(f\) has a pole of order \(m {\lt}\infty \) if and only if there exist an analtyic function \(g\in A(D(z_0,r))\) for some \(r {\gt} 0\) such that \(g(z_0)\neq 0\) and
\begin{equation} \label{poles1} f(z) = \frac{g(z)}{(z-z_0)^m} \end{equation}
3.36
for \(z\in D^*(z_0,r)\).
Moreover, in this case,
\begin{equation} \label{poles2} \operatorname*{\mathrm{Res}}_{z=z_0} f = \frac{g^{(m-1)}(z_0)}{(m-1)!}. \end{equation}
3.37
Proof
The first assertion is pretty obvious from looking at the Laurent series.
To prove the second assertion, just apply Cauchy’s formula for \(g^{(m-1}(z_0)\): If \(C\) is a circle in \(D^* = D^*(z_0,r)\), then
\begin{align*} g^{(m-1)}(z_0) & = \frac{(m-1)!}{2\pi i} \int _C \frac{g(z)}{(z-z_0)^m} \\ & = \frac{(m-1)!}{2\pi i} 2\pi i \operatorname*{\mathrm{Res}}_{z=z_0} f(z)\\ & = (m-1)! \operatorname*{\mathrm{Res}}_{z=z_0} f(z)\\ \end{align*}
as desired.
Example
3.38
Let’s compute the residue \(\displaystyle \operatorname*{\mathrm{Res}}_{z=2} \frac{z}{z^2 - 4}\). Here can take \(g(z) = z/(z+2)\). Then \(\displaystyle \operatorname*{\mathrm{Res}}_{z=2} \frac{z}{z^2 -4} = g(2) = 2/4 = 1/2\).
Example
3.39
We have
\begin{align*} \operatorname*{\mathrm{Res}}_{z=0} \frac{e^z}{z^5} & = \frac{\frac{d^4}{dz^4}e^z|_{z=0}}{4!}\\ & = 1/4! = 1/24. \end{align*}