Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 7 / d

The denominator factors as $z^2+iz+6=(z-2i)(z+3i)$, with no cancellation against the numerator. Thus there areTo inspect infinity, put $w=1/z$:Hence infinity is a removable singularity at infinity, with limiting value zero. After this natural extension to the Riemann sphere it is also a simple zero. There are no other zeros or singularities.

The complex exponential never vanishes. The zeros of $\sinh z$ occur at $z=n\pi i$, $n\in \mathbb{Z}$, and they are simple because $\cosh (n\pi i)=(-1{)}^n\ne 0$. The numerator is $e^{2n\pi i}=1$ there, so no cancellation occurs. ThereforeFor example their residues are $1/\cosh (n\pi i)=(-1{)}^n$.

Infinity requires care: the poles themselves tend to infinity. Equivalently $f(1/w)$ has poles at $w=1/(n\pi i)$ accumulating at zero. Thus $\boxed{\infty \text{ is a non-isolated singularity}}$, and it is not a zero. In the usual classification of isolated singularities, it cannot be called a removable singularity, pole or isolated essential singularity. The non-isolated singularity distinction is necessary even though the function is meromorphic at every finite point.