Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 1 / 2D / i / Solution

The zeroes of $\cosh z$ occur at $z=i\pi (n+\frac{1}{2})$, $n\in \mathbb{Z}$. They are simple because $\sinh z\ne 0$ there. Squaring the denominator therefore givesThere are no other finite singularities; this follows from the local zero-order criterion for a pole.