Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 12F / c

Since $\tan z=\sin z/\cos z$, its singularities occur at the simple zeros of the denominator,At each one, $\sin a_k\ne 0$ and $(\cos z{)}^{\prime }{|}_{a_k}=-\sin a_k\ne 0$, so all are simple poles. There are no other finite singularities:

PutNear $a_k$, writing $\delta =1/z-a_k$ givesButand hence, exactly,Since $\delta =O(w)$, the first two Laurent terms areEquivalently, substitute $z_k=2/((2k+1)\pi )$ in this expression.

At every $z_k$, the function $\tan (1/z)$ has a simple pole. The exponential of a pole is an essential singularity, soare essential singularities of $\exp (\tan (1/z))$.

These points accumulate at $0$. Therefore no punctured neighbourhood of zero is a domain of holomorphy for the function: zero is a non-isolated singularity, specifically an accumulation point of essential singularities, rather than an isolated essential singularity. There are no other singularities in $\mathbb{C}$.