Residue of $\frac{z}{1-\cos(z)}$ at $z= 0$

Question

I've managed so far to find the Laurent Series Expansion at $0$ and I have this: $\frac{2}{z} - \frac{z}{6} + \frac{z^3}{120} + O(z^5)$

We know that the residue is the $a_{-1}$ term of the series $\sum_{ n = -m}{a_nz^n}$ so I guess it should be in $\frac{2}{z} - \frac{z}{6}$ but I do not know how to proceed.

Answer 1

Yes, the Laurent series of your function centered at $0$ has that form. Therefore, the residue of the function at $0$ is the coefficient of $z^{-1}$, which is $2$.