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? 

Following your argument, 
$$f(z) = z\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right) =  (z-i+i)\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)\\ = \left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^{n+1}}{(2i)^n}\right) - 1\right) -\left(\frac{1}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)$$
