= Solution
{c}
The zeros of the <hyperbolic sine> are exactly $z=k\pi i$, $k\in\mathbb Z$, since $\sinh z=0$ is equivalent to $e^{2z}=1$. Each zero is simple because $\cosh(k\pi i)=(-1)^k\ne0$. The quotient is therefore <holomorphic> precisely on
$$
\boxed{\mathbb C\setminus\{k\pi i:k\in\mathbb Z\}.}
$$
At zero, the <Taylor expansion> $\sinh z=z+O(z^3)$ gives $z/\sinh^2z=1/z+O(z)$, so zero is a <simple pole>. At $k\pi i\ne0$, the denominator has a double zero and the numerator is nonzero, giving a <pole> of order two. Thus none of the omitted points admits a <holomorphic extension>.
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