= Solution
{c}
For $z\ne0$, the denominator vanishes exactly when $1/z=(2k+1)\pi i$. Thus set
$$
z_k=\frac1{(2k+1)\pi i},\qquad k\in\mathbb Z.
$$
Each denominator zero is simple, since
$$
\left(e^{1/z}+1\right)'\big|_{z=z_k}=-\frac{e^{1/z_k}}{z_k^2}=\frac1{z_k^2}\ne0.
$$
Every $z_k$ is purely imaginary and nonzero. The identity $\sin(iy)=i\sinh y$ for real $y$ shows $\sin z_k\ne0$, 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. Therefore
$$
\boxed{z_k\text{ are simple poles, and }0\text{ is a non-isolated singularity}.}
$$
In 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>.
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