<s>[INST] Principal part of Laurent Series for $\frac{1}{(1-\cosh(z))^2}$ in this exercise I am asked to provide the principal part of the Laurent series of
$$\frac{1}{(1-\cosh(z))^2}$$
And i am kinda struggling with fonding a solution or even a pattern towards one
Thanks in advance to everyone keen to help
 [/INST] $$(1-\cosh z)^2=3-4\cosh z+\cosh2z$$
then
\begin{align}
\dfrac{1}{(1-\cosh z)^2}
&= \dfrac{1}{3-4\cosh z+\cosh2z} \\
&= \dfrac{1}{3-4 \left(1+\frac{1}{2!}z^2+\frac{1}{4!}z^4+\frac{1}{6!}z^6+\cdots\right) + \left(1+\frac{4}{2!}z^2+\frac{16}{4!}z^4+\frac{64}{6!}z^6+\cdots\right)} \\
&= \dfrac{1}{\frac{1}{4}z^4+\frac{1}{24}z^6+\frac{1}{320}z^8+\cdots} \\
&= \frac{4}{z^4}-\frac{2}{3z^2}+\dfrac{11}{180}+\sum_{n\geq1}a_{2n}z^{2n}\cdots
\end{align}
and you find the principal part.
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