Laurent series in two different regions... Compute the Laurent series of $f(z)={(z+2)\over{(z-2)(z+1)}}$ for:
1) $2<|z|<\infty$
2) $1<|z|<2$ 
I have started my breaking it into partial fractions, and am totally lost on where to go from there. How to I deal with the "(z+1)" fraction after writing the "(z-2)" fraction as a laurent series?
 A: If you know how to write $1/(z-2)$ as a Laurent series in both cases, you should know how to write $1/(z+1)$ as a Laurent series.  Hint: whether you use positive or negative powers of $z$ is determined by the fact that you want to take powers of something small, not something large...
A: Note that for $z\neq 0,1$ we can write $$\frac1{z+1}=\frac1z\cdot\cfrac1{1-\left(-\frac1z\right)}.$$
Now, we can expand this as a multiple of a geometric series in the annulus $1<|z|,$ using the fact that $$\frac1{1-w}=\sum_{k=0}^\infty w^k$$ whenever $|w|<1$. That will give you the Laurent expansion for $\frac1{z+1}$ that is valid in both of the annuli in question.
We can do something similar for $\frac1{z-2}$, noting that for $z\neq0,2$ we can write $$\frac1{z-2}=\frac1z\cdot\cfrac1{1-\left(-\frac2z\right)}$$ and $$\frac1{z-2}=-\frac12\cdot\cfrac1{1-\frac{z}2}.$$
Now, one of these can be expanded as a multiple of a geometric series in the annulus $|z|>2$ and the other can be expanded as a multiple of a geometric series in the annulus $0<|z-2|<2$. You should figure out which annulus works for which rewritten version, and find the respective expansions in both cases. That will give you two different Laurent expansions for $\frac1{z-2}$, one valid in each of the two annuli in question.
Use these expansions, and your partial fraction decomposition, to get the Laurent expansion for $f(z)$ in each annulus.
