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.

Say $$\frac{1}{z \sin z}=c_{-2}z^{-2}+c_{-1}z^{-1}+c_0+\dots. $$
It follows that
$$1=(c_{-2}z^{-2}+c_{-1}z^{-1}+c_0+\dots)z \sin z=(c_{-2}z^{-1}+c_{-1}+c_0z+\dots)(z-\frac{z^3}{3!}+- \dots). $$
Try to expand the RHS, collect the coefficients and compare with the LHS. This should give you a sufficient amount of equations to find $c_{-2},c_{-1},c_0$.
