$f(z)=\frac{1}{z^4-1}, z_0=3+2i$
My idea:
$f(z)=\frac{1}{z^4-1}=\frac{1}{4(z-1)}-\frac{1}{4(z+1)}-\frac{2}{4(z^2+1)}=-\frac{1}{4}\sum\limits_{n=0}^{\infty } z^n - \frac{1}{4}\sum\limits_{n=0}^{\infty } (-z)^n-\frac{1}{2}\sum\limits_{n=0}^{\infty } (-z^2)^n $
But I am not sure how to get all convergence ranges from it?