I must find the Laurent series of the function $\displaystyle g(z)=\frac{1}{z(z^2+1)}$ for 3 different regions: $\displaystyle 0<|z-i|<1$, $\displaystyle 1<|z-i|<2$ and $\displaystyle 2<|z-i|<\infty$.

My attempt: I've found out that g has 3 singularities (at $\displaystyle z=\pm i$ and $\displaystyle z=0$) and is I think otherwise analytic.

So for the first region, there is 1 singularity inside it (just at the center of the region).

So I think I should write the Laurent series centered at the singularity $\displaystyle z=i$.

I've tried many things like rewriting $\displaystyle g(z)=\frac{1}{z(z+i)} \cdot \frac{1}{z-i}$ but I realized it was better to keep it as it was. I was thinking about writing $\displaystyle \frac{1}{z^2+1}$ as an infinite series (now I realize that as it should be centered in i, I should have factored out by (z-i)...) and then divide by z the whole series. I'm going into circles and I don't click yet on how to get it. I'd like a tip.