Find the Laurent series of the complex function $\frac{z}{1-\cos(z)}$ around $z=0$ ?
My goal is to find the residue of this function at $z=0$, in order to calculate the integral : $$\int_{|z-1|=\pi} \frac z{1-\cos z}$$
because $z=0$ is the only singular point at the integral domain. (at least that's what I figured)
I tried to say that when we get close to $z=0$, we can say that $|\cos(z)|\le 1$, so we can expand $\frac{1}{1-\cos(z)}$ into power series and from there to find the coefficient of $\frac{1}{z}$ but couldn't get such power.
What am I missing here?
To computer the residue at $z=0$ you only need to find $$ \begin{align} \lim_{z\to 0} z\frac{z}{1-\cos z} =& \frac{z^2}{1-\cos z} \\ &= \frac{z^2(1+\cos z)}{(1-\cos z)(1+\cos z)} \\ &= \frac{z^2(1+\cos z)}{1-\cos^2 z} \\ &= \frac{z^2(1+\cos z)}{\sin^2 z} \\ &= 2 \end{align}$$