Laurent Series expansion about the point $z_0 = i$ of $\frac{z}{z^2+1}$ I am trying to construct the Laurent series expansion of $f(z) = \frac{z}{z^2+1}$ about $z_0 = i$ in the region $\{z \in \mathbb{C}: 0 < |z - i| < 2\}$ but I am stuck.
We can re-write $f(z) = \dfrac{z}{z^2+1} = \dfrac{z}{(z-i)(z+i)}$
This allows us to see more easily that there are first order poles at $\pm i$. We can use partial fractions to obtain:
$$\frac{1}{(z+i)(z-i)} = \frac{i}{2(z+i)} - \frac{i}{2(z-i)}$$
We want to re-write everything in terms of $z-i$:
$$\frac{1}{z+i} = \frac{1}{2i + (z - i)} = \frac{1}{2i}\left(\frac{1}{1+\frac{1}{2i}(z-i)}\right)$$
Using the geometric series this gives:
$$ = \frac{-i}{2} \left(\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right)$$
Then we have that:
$$f(z) = z\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)$$
But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here? 
 A: \begin{align}&f(z)=\frac{z}{z^2+1}=\frac{A}{z-i}+\frac{B}{z+i}\\&=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{z+i}\end{align}
\begin{align}&A=Res[f,i]=\lim\limits_{z\rightarrow i}\frac{z}{z+i}=\frac{1}{2}
\\&B=Res[f,-i]=\lim\limits_{z\rightarrow -i}\frac{z}{z-i}=\frac{1}{2}\end{align}\begin{align}f(z)&= \frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{2i\left(1+\frac{z-i}{2i}\right)}\\&=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n (z-i)^n}{(2i)^{n+1}}\end{align}\begin{align}=&\frac{1}{2}\frac{1}{z-i}-\frac{i}{4}+\frac{1}{8}(z-i)\\&+\frac{i}{16}(z-i)^2+\cdots\end{align}
in the region $\{z\in\mathbb{C}:0<|z-i|<2\}.$
A: Following your argument, 
$$f(z) = z\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right) =  (z-i+i)\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)\\ = \left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^{n+1}}{(2i)^n}\right) - 1\right) -\left(\frac{1}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)$$
