= Solution
The exponential numerator is entire and never zero. The denominator vanishes precisely when $e^{2z}=-1$, giving
$$
z_n=i\pi\left(n+\tfrac12\right),\qquad n\in\mathbb Z.
$$
At these points $\sinh z_n=i(-1)^n\ne0$, so every zero of the denominator is simple and hence every singularity is a <simple pole>. Using the <residue> formula in (b),
$$
\boxed{\operatorname{Res}(f,z_n)=\frac{e^{-ikz_n}}{\sinh z_n}=\frac{(-1)^n}{i}\exp\left[k\pi\left(n+\tfrac12\right)\right].}
$$
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