first 3 terms of $1/(z\sin z)$ in laurent series How could I calculate the first $3$ terms of Laurent series on $1/ (z \sin z)$ around $z=0$?
For $z=0$ it is a pole of $n= 2$.
$C_n$ are $0$ for $n < -2$, therefore the first $3$ terms are $C_{-2}, C_{-1}, C_0$. I failed to integrate to get the coefficients.
Thanks for your kick.

Since $\sin z = z - z^3/6 + o(z^4) $, you have 
$$\frac 1{z\sin z} = \frac{1}{z^2}\frac 1{ 1 - z^2/6 + o(z^3)} = \frac{1}{z^2} \sum_{n=0}^\infty (-z^2/6 + o(z^3))^n$$
Expanding a little you find $$\frac{1}{z^2} (1 - z^2/6 + o(z^3))$$ hence
$$c_{-2} = 1, c_{-1}=0, c_0 = -1/6$$
(Note: the fact that the residue $c_{-1}=0$ could have been seen from the beginning, because $\frac 1{z \sin z}$ is an even function. So you can say that all coefficients with odd powers are $0$ ) 
