I need to find the residue of $$ f(z) = \frac{ e^{\pi z} }{ (z-i)^2 } \; \text{at} \; z_0 = i $$
I tried using Laurent series expansions, but I can't seem to find expansions that make the coefficients $a_k$ not be functions of $z$. Also I tried the equality $$\text{Res}\left(z_0, \frac{f}{g} \right) = \frac{f(z_0)}{ g\prime(z_0)} $$ but $g\prime(z) = 2(z-i) \implies g\prime(z_0) = 0 $, so this doesn't work either.
Building the limit $\lim_{z\to 0} f(z) $ doesn't get me further either as it leads to $\frac{-1}{0}$.
I'm at a loss at what to try next, and feel like missing something obvious. Can someone give me a hint? Thanks in advance!
Let $c$ be a constant. Then $$e^{cz} = e^{c(z-i)+ci} = e^{ci} e^{c(z-i)} = e^{ci} \sum_{k=0}^\infty \frac{(c(z-i))^k}{k!}.$$ This furnishes the series expansion of $e^{cz}$ at $z = i$. Then choosing $c = \pi$, we have $$\begin{align} f(z) &= \frac{e^{\pi z}}{(z-i)^2} \\ &= e^{\pi i} \sum_{k=0}^\infty \frac{\pi^k}{k!} (z-i)^{k-2} \\ &= - \sum_{k=0}^\infty \frac{\pi^k}{k!} (z-i)^{k-2} \\ &= -(z-i)^{-2} - \pi(z-i)^{-1} - \frac{\pi^2}{2} - \frac{\pi^3}{6}(z-i) - \cdots \end{align}$$ is the Laurent expansion of $f$ about $z = i$, from which we easily read off the desired residue.