= Solution
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 $\cosh(n\pi i)=(-1)^n\ne0$. The numerator is $e^{2n\pi i}=1$ there, so no cancellation occurs. Therefore
$$
\boxed{\text{simple poles at }z=n\pi i\ (n\in\mathbb Z);\qquad\text{no finite zeros}.}
$$
For 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.
Back to article page
