Solution
= Solution
Put $w=z-i\pi$. Since $\sinh(w+i\pi)=-\sinh w$ and $\sinh w=w(1+w^2/6+O(w^4))$, the local <Laurent series> is
$$
\frac1{\sinh^3z}=-w^{-3}(1+w^2/6+O(w^4))^{-3}=-w^{-3}+\frac12w^{-1}+O(w).
$$
Thus all singular terms are
$$
\boxed{-\frac1{(z-i\pi)^3}+\frac1{2(z-i\pi)}.}
$$
There is a third-order <pole>, with no second-order term.