Find poles and residues of \begin{equation} f(z)=z^2/(\cosh z-1) \end{equation}

Question

I understand that the point
\begin{equation}
z=0
\end{equation}

is an obvious singularity, but since it's also a root of multiplicity 2 of the nominator the residue of the given function at 0 will be 0.

As for the other singularities we have

\begin{equation}
\cosh z=1
\end{equation}
or

\begin{equation}
\cos(iz)=1
\end{equation}

and finally we find the singularities

\begin{equation}
iz=2n\pi
\end{equation}
or
\begin{equation}
z_n=-i2n\pi
\end{equation}
for any integer.

As I have said, the residue at 0 is 0, but the other singularites are all simple poles, and therefore the residue at those points will be
\begin{equation}
\text{Res} f(z_n)=\lim_{z\to z_n} \frac{(z-z_n)z^2}{\cosh z-1}
\end{equation}
This is a 0/0 case and we can use L'Hospital's rule, and we get
\begin{equation}
\lim_{z\to z_n} \frac{3z^2-2z_nz}{\sinh z}
\end{equation}
Now this is no longer in indeterminate form, but we have \begin{equation}
a/0
\end{equation}

Clearly I must have done something wrong, but I cannot seem to grasp what that something is. Any help will be greatly appreciated.

Answer 1

$z_n$ is a pole of order $2$ since $e^{z}+e^{-z}-2=(e^{z/2}-e^{-z/2})^{2}$. So you have to multiply by $(z-z_n)^{2}$ and compute the derivative at $z=z_n$.