Integral $\int_{C(0,3/2)} \frac{z^3}{1-\cos(\pi iz^2)}\,dz$ using residue theorem. I'm stuck with this exam problem. It says : Resolve the following integral
$$\int_{C(0,3/2)} \frac{z^3}{1-\cos(\pi iz^2)}\,dz$$
Through the circumference $C(0,3/2)$
I asume we have to use the residue theorem. The poles are the $z=\sqrt{2ki}$ for any integer $k$. The ones that matters,i.e. the $z_j$ with $\operatorname{ind}\left(z_j\right)=1$ are these: 
$$z_1=0,z_2=1+i,z_3=1-i,z_4=-1+i,z_5=-1-i$$
Thank you

The calculations are very nasty, indeed. Alternative idea: Laurent series. In $z_0=1+i$:
$$
-2+2\,i+6\,i\,\left(z-z_0\right)+\left(3\,i+3\right)\,(z-z_0)^2+(z-z_0)^3=
$$
$$
= z^3 = f(z)(1-\cos(\pi i z^2))=
$$
$$=\left({a_{-2}\over(z-z_0)^2}+{a_{-1}\over z-z_0}+a_0+a_1(z-z_0)+\cdots\right)
\times
$$
$$
(-4\,\pi^2\,i\,(z-z_0)^2+(-2\,\pi^2\,i-2\,\pi^2)\,(z-z_0)^3+\cdots).
$$
