Laurent Series Difficulty Hello all at StackExchange,
I'm having some trouble understanding computing the Laurent series for different domains.
Here's my approach to finding the Laurent series for $\dfrac{3}{(z+1)(z-2)}$ for the following domains.
1) $|z| < 1$
2) $ 1 < |z| < 2$
3) $|z| > 2$
So, using partial fraction decomposition, I obtain $\dfrac{1}{z-2} - \dfrac{1}{z+1}$.
The idea is to obtain series for both fractions, which is understandable (Taylor series and power series of complex functions is understood), but I am not understanding $\bf{why}$ the fractions become modified as follows (the algebra is understood, not the idea).
1) We change $\dfrac{1}{z-2} = -\dfrac{1}{2}\left[\dfrac{1}{1-\frac{z}{2}}\right]$ and leave $\dfrac{1}{z+1} = \dfrac{1}{z-(-1)}$.
2) We still have $\dfrac{1}{z-2} = -\dfrac{1}{2}\left[\dfrac{1}{1-\frac{z}{2}}\right]$ but instead $\dfrac{1}{z+1} = \dfrac{1}{z}\left[\dfrac{1}{1+\frac{1}{z}}\right]$.
3) We change $\dfrac{1}{z-2} = \dfrac{1}{z}\left[\dfrac{1}{1-\frac{2}{z}}\right]$ but keep  $\dfrac{1}{z+1} = \dfrac{1}{z}\left[\dfrac{1}{1+\frac{1}{z}}\right]$.
Do the fractions change because of analyticity in the domains, and in particular, why do we factor a term of $\dfrac{1}{z}$ for some of them? Much help would be appreciated, and regards.
 A: I think that the reason is that you need to use the identity $$\frac{1}{1-w} = \sum_{n=0}^\infty w^n$$ which holds only for $|w| < 1$. Notice that if $|z|<1$, then $\frac{1}{z+1} = \sum_{n=0}^\infty (-z)^n$ is valid, but not for $|z| > 1$. That's why in the other two cases, you need to write it as $$\frac{1}{z} \cdot \frac{1}{1+\frac 1 z} = \frac{1}{z} \sum_{n=0}^\infty \left(-\frac 1 z\right)^{n}$$
in order to use the identity, since now $\left|\frac 1z\right| < 1$.
I might be mistaken, so please correct me if you find a mistake in what I have said.
A: I think blf gave the answer you probably want. But I'll add some extra reasoning:
For 1) with the factored fractions you've given, the corresponding geometric series you find for $-\frac{1}{2}\left[\frac{1}{1-\frac{z}{2}}\right]$ will be valid in $|z|<2$ and the series for $\frac{1}{1-(-z)}$ will be valid in $|z|<1$ so together they are both valid in the smaller of the two circles (which is your domain $|z|<1$).
For 2) you have your series representation for $\frac{1}{z-2}$ is already valid inside $|z|<2$ but the series for $\frac{1}{z+1}$ that you found in part 1) is only valid for $|z|<1$ and you need it valid in $|z|>1$ so that is the series that you will need to alter.
For 3) your series for $\frac{1}{z+1}$ from part 2) will be valid in $|z|>1$ so it can be left alone. But the series representation for $\frac{1}{z-2}$ is still only valid inside the circle $|z|<2$ so you will need to alter it to make it valid in $|z|>2$ so that they will be both be valid in the larger of the two circles which is your domain $|z|>2$.
