$f(z)=\frac{1}{z^2+1}$ Laurent Series

Question

Determine the Laurent series of the function $f(z)=\frac{1}{z^2+1}$ on the set $A=\{z|\:0<|z-i|<2\}$.

I know I should use the expansion $\frac{1}{z-1}=\sum_\limits{n=0}^{\infty}z^n$ or $\frac{1}{z+1}=\sum_\limits{n=0}^{\infty}(-1)^n z^n$.

The solution to the problem provided by the book is $\frac{-i}{2(z-i)}+\frac{1}{4}\sum_\limits{n=0}^{\infty}(\frac{-1}{2i})^n (z-i)^n$.

I have solve lots of different problems but the ring $A$ is always centred around $|z|$ not $|z-i|$.

Question:

How should I solve the exercise?

Thanks in advance!

Answer 1

You have$$\frac1{z^2+1}=\frac1{(z+i)(z-i)}.$$But\begin{align}\frac1{z+i}&=\frac1{2i+(z-i)}\\&=\frac1{2i}\frac1{1+\frac{z-i}{2i}}\\&=\frac1{2i}\sum_{n=0}^\infty(-1)^n\left(\frac{z-i}{2i}\right)^n\\&=\sum_{n=0}^\infty\frac{(-1)^n}{(2i)^{n+1}}(z-i)^n\end{align}and therefore\begin{align}\frac1{z^2+1}&=\frac1{(z+i)(z-i)}\\&=\sum_{n=0}^\infty\frac{(-1)^n}{(2i)^{n+1}}(z-i)^{n-1}\\&=\sum_{n=-1}^\infty\frac{(-1)^{n+1}}{(2i)^{n+2}}(z-i)^n.\end{align}