I am having some difficulty wrapping my head around the techniques used to perform the Laurent/Taylor series expansion, especially the factoring for a specific domain. Here is an example problem, and my thought process going forward:
Find the Laurent series expansion centered at $z=0$ of $f(z)=\frac{z}{(z+1)(z-2)}$ in each of the following domains.
(a) $|z|<1$
(b) $1<|z|<2$
(c) $|z|>2$
My first step for solving this problem is to use partial fraction decomposition of $f(z)$. $$f(z)=\frac{z}{(z+1)(z-2)}=\frac{1}{3(z+1)}+\frac{2}{3(z-2)}$$
Next I attempt to expand both fractions on the right side. To do this, I need to factor/reorganize the denominator to meet my needs for a given domain. First let's attempt the first fraction in the domain $|z|<1$.
First, I think my goal is to put it in a form where I can use my knowledge of the geometric series. In addition to this, I want to account for my domain, I think this means I want the denominator in the form $(1-z)$:
$$\frac{1}{3(z+1)}=\frac{1}{3(1-(-z))}=\frac{1}{3}\sum_{n=0}^\infty(-1)^nz^n$$
Next for the second fraction I have trouble factring for it to be in the same domain:
$$\frac{2}{3(z-2)}=\frac{2}{-6(1-\frac{z}{2})}\Rightarrow|\frac{z}{2}|<1\Rightarrow|z|<2$$
I can only get it to work for the domain $|z|<2$. I am also having trouble with the domain $1<|z|<2$ since I am not sure how to ensure it is between two points. Hopefully this question makes sense, and I hope I am correct in my understanding of the restrictions of the domain. Any help or advice is very appreciated.
You will have to do your geometric series differently depending on the domain. For example, take the second of your partial fractions. If $|z|<2$ then we have $$\frac23\frac1{z-2}=-\frac13\frac1{1-\frac z2}=-\frac13\sum_{n=0}^\infty\Bigl(\frac z2\Bigr)^n\ .$$ If $|z|<1$ then $|z|<2$ and so this still works - this is the answer to the question at the beginning of your last paragraph.
If $|z|>2$ then the above does not work as the series diverges. So do it this way: $$\frac23\frac1{z-2}=\frac2{3z}\frac1{1-\frac2z}=\frac2{3z}\sum_{n=0}^\infty\Bigl(\frac2z\Bigr)^n=\frac23\sum_{n=0}^\infty\frac{2^n}{z^{n+1}}\ .$$ The series converges because $|\frac2z|<1$.
For the other fraction you will need different series depending whether $|z|<1$ or $|z|>1$. Notice that the latter includes $|z|>2$ as in your third region.
Hope this helps.