Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 4 / b

The expression is $F(z)=e^{z^2+1/z}=e^{z^2}{e}^{1/z}$, which is holomorphic for every $z\ne 0$. At zero,has infinitely many nonzero negative-power terms, so it has an essential singularity. Multiplication by the nonvanishing entire function $e^{z^2}$ preserves that classification: if $F$ had a removable singularity or a pole, multiplying by $e^{-z^2}$ would give the same type for $e^{1/z}$, a contradiction. Hence

The only possible finite singularities are $z_{\pm }=\pm i\pi /2$, the zeros of $4z^2+{\pi }^2$. They are simple zeros of the denominator, because its derivative $8z_{\pm }$ is nonzero. They are also simple zeros of the numerator: $\cosh z_{\pm }=0$ and $\sinh z_{\pm }=\pm i\ne 0$.

The common simple factors cancel, so both points are removable singularities, with extension valuesThere are no other finite singularities; after inserting these two values the function is entire.

For $z\ne 0$, the denominator vanishes exactly when $1/z=(2k+1)\pi i$. Thus setEach denominator zero is simple, sinceEvery $z_k$ is purely imaginary and nonzero. The identity $\sin (iy)=i\sinh y$ for real $y$ shows $\sin z_k\ne 0$, so there is no cancellation. Each $z_k$ is consequently a simple pole, with residue $z_k^2\sin z_k$.

These poles accumulate at zero. ThereforeIn particular, zero has no punctured neighbourhood on which the function is holomorphic, so it cannot be classified as a removable singularity, a pole, or an isolated essential singularity. There are no other finite singularities.