Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7 / c

The zeros of $\sin z$ are $z=n\pi$ for $n\in \mathbb{Z}$, all simple, and $\cos (n\pi )=(-1{)}^n\ne 0$. Hence the finite singularities of $\cot z$ are exactlyLocally $\cot z\sim 1/(z-n\pi )$, so each pole has residue $1$.

For $z\ne 0$, singularities occur when $1/z=n\pi$, with $n\ne 0$. Each is a simple pole, since the substitution has nonzero derivative there:More explicitly, $1/z-n\pi \sim -(z-z_n)/z_n^2$, so $\cot (1/z)\sim -z_n^2/(z-z_n)$ and its residue is $-1/(n^2{\pi }^2)$.

These poles accumulate at zero. Consequently $\boxed{z=0\text{ is a non-isolated singularity}}$. The removable/pole/essential classification for isolated singularities cannot be applied at that accumulation point; it is not an isolated essential singularity.

At every $z=n\pi$, the inner function has a simple pole, with $\cot z=1/(z-n\pi )+h(z)$ for analytic $h$. ThusThe first factor is analytic and nonzero, while the second has infinitely many negative powers in its Laurent series. Therefore, by the exponential of a pole is an essential singularity result,At all other finite points the composition is analytic, so there are no additional finite singularities.