Find the Laurent series representation of $(z^2+1)^{-1}$ on $0 < |z-i| < 2$ . I want to see if I understand correctly this kind of problems. So we need to represent this function as a series that is convergent on $0 < |z-i| < 2$. Then $$\dfrac{1}{z^2 + 1} = \dfrac{1}{(z-i)(z+i)} = \dfrac{1}{z-i} \dfrac{1}{z-i + 2i} = \dfrac{1}{z-i}\dfrac{1}{2i}\dfrac{1}{1 - \dfrac{-(z-i)}{2i}} ^{(*)}= \dfrac{1}{z-i}\dfrac{1}{2i} \sum_{n=0}^\infty{\dfrac{i^{2n}(z-i)^n}{2^ni^n}}=\sum_{n=0}^\infty {\dfrac{i^{n-1} (z-i)^{n-1}}{2^{n+1}}}$$ $(*)$ here $0 < \big| \dfrac{-(z-i)}{2i}\big| < 1$.
It seems alright to me, but I am new to these problems and this problem isn't from the textbook so I don't know if my solution is correct.