Singularities and residues of $f(z) = \frac{\sin(z)}{(1-e^{-z})^3}$ We are given the meromorphic function $f(z) = \frac{\sin(z)}{(1-e^{-z})^3}$. I know the singularities lie at $z = 2\pi \mathrm{i} k$ for $k \in \mathbb{Z}$. But how do I show that the singularities are poles of order $3$ and how do I calculate the residues? I only end up with really messy calculations, so I assume there must be some sort of trick...

$\sin z = \frac {1}{2i}(e^z - e^{-z})$
$\frac {\sin z}{(1-e^{-z})^3} = \frac {(e^z - e^{-z})}{2i(1-e^{-z})^3}$
multiply top and bottom by $e^{3z}$
$\frac {(e^{4z} - e^{2z})}{2i(e^z-1)^3}$
$\frac {(e^{4z} - e^{2z})}{2i(e^z-1)^3}$
$u= e^z$
$\frac {(u^4 - u^2)}{2i(u-1)^3} = \frac {u^3+u)}{2i(u-1)^2} = \frac 1{2i}( u^2 + 2u+ \frac {4}{u-1} + \frac {2}{(u-1)^2})\\
\frac 1{2i}( e^{2z} + 2e^z + \frac {4}{e^z-1} + \frac {2}{(e^z-1)^2})$
Check my work on that partial fraction decomposition.
