Principal part of Laurent Series for $\frac{1}{(1-\cosh(z))^2}$

Question

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

Answer 1

I would do it quite differently than @Nosrati, taking a very down-to-earth and unimaginative route.

First, $\cosh x=1+x^2/2! + x^4/4! +\cdots$, so that $1-\cosh x=-x^2[1/2!+x^2/4!+\cdots]$. You want to square this and take the reciprocal: \begin{align} \left[x^2\left(\frac12+\frac{x^2}{24}+\frac{x^4}{720}+\cdots\right)\right]^{-2} &=x^{-4}\left[\frac14+\frac{x^2}{24}+\frac{x^4}{320}+\cdots\right]^{-1}\\ &=x^{-4}\left[4-\frac{2x^2}3+\frac{11x^4}{180}+\cdots\right]\\ &=4x^{-4}-\frac23x^{-2}+\frac{11}{180}+\cdots\,, \end{align} where the only thing that slows your hand computation down is doing the long division of $1/4+x^2/14+x^4/320+\cdots$ into $1$.

Answer 2

$$(1-\cosh z)^2=3-4\cosh z+\cosh2z$$ 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.