Laurent expansion of $\frac{z}{z^2+1}$ valid for $|z-3|>2$ how to find the Laurent expansion of $\frac{z}{z^2+1}$ valid for $|z-3|>2$.
$$\frac{z}{z^2+1} = \frac{1}{2}\left( \frac{1}{z-3 + (3 + i)} + \frac{1}{z-3 + (3 - i)}\right)$$
Given $|z-3| < \sqrt{10}$ or $|z-3| > \sqrt{10}$, I think we have two different expansions, how do I proceed? Do I need to consider both cases and thus have two answers?

A related problem. First convert to the partial fraction form
$$ \frac{z}{z^2+1}=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{z+i}. $$
Then, we consider the first term, since the second is the same
$$ \frac{1}{2}\frac{1}{z-i}=\frac{1}{2}\frac{1}{(z-3)+(3-i)}= \frac{1}{2}\frac{1}{z-3}\frac{1}{1+\frac{3-i}{z-3}}$$
$$ = \frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k\,(3-i)^k}{(z-3)^{k+1}}\, $$
where 
$$ \Big|\frac{3-i}{z-3}\Big| <1 \implies |z-3| > |3-i| \implies |z-3| > {\sqrt{10}}. $$
The same can be done with the other term which has the form
$$ = \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k(3+i)^k}{(z-3)^{k+1}}\quad |z-3|>\sqrt{10}. $$
So, we have
$$ \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k(3-i)^k}{(z-3)^{k+1}}+\frac{1}{2}\frac{1}{z-3}\sum_{k=0}^{\infty} \frac{(-1)^k(3+i)^k}{(z-3)^{k+1}}  $$
$$ =\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k((3-i)^k+(3+i)^k)}{(z-3)^{k+1}}\,,\quad |z-3|> \sqrt{10}. $$  
