Laurent series of $\sin z/(1 - \cos z)$ I have trouble solving this exercise: find the first three terms of the Laurent series of $\sin z/(1 - \cos z)$ centered at $z=0$. 
I have found the first two. I proved that at $z=0$ we have a first order pole and the first one I calculated the residue. I also thought that the second term is zero because this function is odd. Now I have problems with the third. Someone can help me?

You know that laurent series of $sin(z) = z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})$.
Then, the laurent series of $ 1-cos(z)= \frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})$
Overall you have $\frac{z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7})}{\frac{z^2}{2}-\frac{z^4}{24}+O(z^{6})}$.
Now look at the denominator: $(\frac{z^2}{2}-\frac{z^4}{24}+\frac{z^6}{720}+O(z^8))^{-1}=(\frac{z^2}{2})^{-1}(1-\frac{z^2}{12}+\frac{z^4}{360}+O(z^{6}))^{-1}=\frac{2}{z^2}(1-(-\frac{z^2}{12}+\frac{z^4}{360})+\frac{z^4}{144}+O(z^6))
=\frac{2}{z^2}(1+\frac{z^2}{12}+\frac{z^4}{240})=\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120}$
Multiply everything together to get: $(z - \frac{z^3}{6} + \frac{z^5}{120} + O (z^{7}))$ $(\frac{2}{z^2}+\frac{1}{6}+\frac{z^2}{120} )$ $ =\frac{2}{z}-\frac{z}{6}-\frac{z^3}{360} +O(z^5).$
