determining a residue by taking a limit

Question

To determine a residue, I need to take this limit: $$\lim_{z\to 2\pi ik} \frac{d}{dz}\frac{(z-2\pi ik)^2}{z(e^z-1)^2}$$ with $k$ any integer number (like -1, -1, 0, 3, 7) I have tried l'Hopital's rule to take the limit but it gets a mess real quickly. Is there an easy way to take this limit?

Answer 1

Start taking $w=z-2\pi ik$. Then, $$ \begin{align} \lim_{z\to 2\pi ik}\frac{d}{dz}\frac{(z-2\pi ik)^2}{z(e^z-1)^2} & = \lim_{w\to 0}\frac{d}{dw}\frac{w^2}{(w+2\pi ik)(e^{(w+2\pi ik)}-1)^2}\\ & = \lim_{w\to 0}\frac{d}{dw}\frac{w^2}{(w+2\pi ik)(e^w-1)^2}\\ & = \lim_{w\to 0}\left(\frac{d}{dw}\frac1{(w+2\pi ik)}\right)\frac{w^2}{(e^w-1)^2}+ \lim_{w\to 0}\frac1{(w+2\pi ik)}\frac{d}{dw}\frac{w^2}{(e^w-1)^2}\\ & = -\frac1{(2k\pi i)^2}\cdot 1 + \frac1{2k\pi i}\cdot\lim_{w\to 0}{{-2\,w\,\left(w\,e^{w}-e^{w}+1\right)}\over{\left(e^{w}-1\right)^3 }}\\ & = \frac1{4k^2\pi^2}+ \frac{i}{2k\pi}\lim_{w\to 0}\frac{w^3+\cdots}{w^3+\cdots}=\frac1{4k^2\pi^2}+ \frac{i}{2k\pi} \end{align} $$

Alternate solution: find the first (negative) terms of the Laurent series like in First four terms of the power series of $f(z) = \frac{z}{e^z-1}$?. The residue is by definition the coefficient of the $(-1)$-th term.