Suppose
$$z\frac{\cos z}{\sin z}= \sum_{-\infty}^{n} a_nz^n $$ the laurent series of $f(z)= z\frac{cosz}{sinz} $ on the ring π<|z|<2π.Find the $a_n$.
Now i know $a_n= \frac{1}{2πi} \int\frac{f(z)}{z^{n+1}}dz$ so for $$n=0$$i plug in the $f$ and i try to use the residues theorem but i dont know the order of the anomalies at 0,π,-π and i dont know what formula to use to calculate the residues.I know that to find out what order the anomalies are i have to calculate the limit $\frac{cosz}{sinz}$ but it is a complex limit 1/0 is infinite in complex numbers. so that means i have to expand the laurent series for that function and cant use a residues formula which looks like the initial task. im confused.
You seem to want the series around $\;z=\pi\;$ , which is a pole of the function:
$$z\frac{\cos z}{\sin z}=z\frac{\cos(z-\pi)}{\sin(z-\pi)}=\left[(z-\pi)+\pi\right]\frac{1-\frac{(z-\pi)^2}{2}+\mathcal O((z-\pi)^4)}{(z-\pi)\left(1-\frac{(z-\pi)^2}{6}+\mathcal O((z-\pi)^4)\right)}=$$$${}$$
$$=\frac{\left[(z-\pi)+\pi\right]}{z-\pi}\left(1-\frac{(z-\pi)^2}{2}+\mathcal O((z-\pi)^4)\right)\left(1+\frac{(z-\pi)^2}6+\mathcal O((z-\pi)^4)\right)=$$$${}$$
$$=\frac\pi{z-\pi}+\ldots$$
and there's your residue.