Most of the time I deal with questions which asked me to find poles, e.g., $$ \dfrac{\sin(x)}{x+1} \quad\text{and}\quad \dfrac{x}{(x-i)(x-1)} \;, $$ which I can find the poles and residue easily.
But this time round, if the question asks me to find poles of $$ \frac{\mathrm{e}^{z}}{\sin(2\mathrm{i}z)} $$ at first glance I know it's gonna be $z=n\mathrm{i}\pi$ where $n$ is an integer.
So, for this case is the residue just $\mathrm{e}^{n\mathrm{i}\pi}$? I doubt my answer because when I check using the Wolfram residue calculator, it gives $\dfrac{-\mathrm{i}(-1)^n}{2}$ which I have no idea how it did it.
Since the poles are all of order one, the residues at the poles $z = \frac{1}{2}n\pi i$ (note the $1/2$) are given by $$ \lim_{z\to \frac{1}{2}n\pi i}(z-\frac{1}{2}n\pi i)\frac{e^z}{\sin(2iz)}.$$ To evaluate this limit you just need to Taylor expand $\sin(2iz)$ at $z = \frac{1}{2}n\pi i.$ We have $$ \left.\frac{d}{dz} \sin(2iz)\right|_{z=\frac{1}{2}n\pi i} = 2i\cos(2iz)\Big|_{z=\frac{1}{2}n\pi i} = 2i\cos(n\pi)$$ so the residue is $$ \frac{e^{\frac{1}{2}n\pi i}}{2i\cos(n\pi)}.$$ For $n=2m$ even (which correspond to the poles you didn't forget about) this reduces to $$ \frac{e^{m\pi i}}{2i\cos(2m\pi)} = -\frac{i}{2}e^{m\pi i} = -\frac{i}{2}(-1)^m$$ which agrees with what Wolfram gave you.