Laurent series of a trigonometric function

Question

Find the Laurent Serie(and residue) aroud $z_0=0$ of the function

$f(z) = \frac{1}{1-\cos z}$.

Progress:

It looks very trivial but it seems to get complicated so I'll only try with 3-4 terms:

We can use that $\cos z = 1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0}^{\infty}\frac{(-1)^k z^{2k}}{(2k)!}$ and hence

$f(z) = \frac{1}{1-\cos z} = \frac{1}{1-(1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0}^{\infty}\frac{(-1)^k z^{2k}}{(2k)!})} \Rightarrow f(z)= \frac{1}{\frac{z^2}{2!}-\frac{z^4}{4!}+\frac{z^6}{6!}} $

This is not correct apparently so I tried to factor out $z^2$ and$\frac{z^2}{2!}$ but I was unable to proceed. I suspect that I'll arrive to a nestled sum.

Answer 1

Since we're interested only around $\,z=0\,$ we can try the following using what you've already done:

$$\frac1{1-\cos z}=\frac1{\frac{z^2}2\left(1-\frac{z^2}{12}+\mathcal O(z^4)\right)}=\frac2{z^2}\left(1+\frac{z^2}{12}+\frac{z^4}{12^2}+\ldots\right)=$$

$$\frac2{z^2}+\frac16+\frac{2z^2}{12^2}+\ldots$$

We used, of course, the development

$$\frac1{1-z}=1+z+z^2+\ldots\;\;,\;\;|z|<1$$

Answer 2

Factor out a $z^2/2!$ from the denominator to get

$$\begin{align}\frac{1}{1-\cos{z}} &= \frac{2}{z^2} \frac{1}{\displaystyle1-2 \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!}}\\ &= \frac{2}{z^2} + \frac{4}{z^2} \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!} + \frac{2}{z^2} \left ( \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!}\right )^2+\cdots \\ &\approx \frac{2}{z^2} +\frac{1}{6} +\cdots\end{align}$$

The residue at $z=0$ is the coefficient of $1/z$ and is thus zero.