Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 3 / a / iii / Solution

The denominator vanishes exactly when $e^z=-1$, namely at $z=(2k+1)\pi i$, $k\in \mathbb{Z}$. Its derivative there is $e^z=-1\ne 0$, so each such zero is simple and the numerator does not cancel it. Henceand every omitted point is a simple pole.