= Solution
The denominator vanishes at $z_k=1/(2\pi i k)$ for every nonzero <integer> $k$. These points approach zero, the denominator has a <simple zero> at each, and $\sin z_k\ne0$ because a nonzero purely imaginary number is not a zero of sine. Hence there are <poles> arbitrarily close to zero. The origin is a \b[nonisolated singularity], so it has no <residue> defined by a <Laurent series> on a punctured disc. In particular it must not be classified as an isolated <essential singularity> merely because $e^{1/z}$ has one.
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