Laurent series outside region of convergence of $\frac{1}{z^2+1}$

Question

I have already computed the Laurent series of $\frac{1}{z^2+1}$ at $z_0=i$ in $R=\{z\in \mathbb{C} : 0<|z-i|<2\}$. I have to compute the Laurent series in $|z-i|>2$ right now, how can I do that?

Answer 1

Since $\dfrac1{z^2+1}=\dfrac1{(z+i)(z-i)}$, you can just compute the Laurent series of $\dfrac1{z+i}$ and then divide by $z-i$. And$$\begin{align}\frac1{z+i}&=\frac1{z-i+2i}\\&=\frac{-i}{2-i(z-i)}\\&=-\frac12\frac i{1-i(z-i)/2}\\&=\frac12\sum_{n=-\infty}^{-1}\left(\frac{i(z-i)}2\right)^n\\&=\sum_{n=-\infty}^{-1}\frac{i^n}{2^{n+1}}(z-i)^n\end{align}.$$