Find the Laurent expansion on $0<|z-i|<2$ of:
$$f(z)=\frac{1}{(z^2-1)^2}$$
I would like to try this exercise by imitating another that is similar:
$$f(z)=\frac{1}{(z^2+i)^2}$$
where the Laurent expansion on $0<|z-i|<2$, is:
$$\frac{-i}{4(z-i)}-\frac{1}{4(z-i)^2}+\sum_{n=0}^{\infty}\frac{(n+3)i^n(z-i)^n}{2^{n+4}}$$
but I can't get to that expansion. I have used that
$$\frac{1}{4(z-i)^2}=-\frac{1}{4}\frac{d}{dz} \Big (\frac{1}{z+i} \Big)$$