Been struggling with this for a while now. Just looking for a nudge in the right direction :
Find the Laurent series for $f(z) = \frac{z^2-2z+3}{z-2}$ about $z=2$.
The only thing I can think to do is
$$ \frac{z^2-2z+3}{z-2} = z + \frac{3}{z-2}$$
but I don't think this is correct since the first term is not "about $z=2$," I tried using the infinite geometric series but it didn't seem to get me anywhere...
If the first term is not "about $z=2$", then you can re-write your expression so that it will be : $$z+\dfrac{3}{z-2}=(z-2)+2+\dfrac{3}{z-2}.$$
More generally, looking for the Laurent series of $f$ at $z=a$ is the same as finding the Laurent series of $g:w\mapsto f(w+a)$ at $w=0$ and replace $w$ by $z-a$.