I am trying to determine the Laurent expansion for the function $f(z)=\frac{z}{(z-2)(z-1)}$, which I have successfully decomposed into its partial fraction form: $f(z)=\frac{2}{z-2}-\frac{1}{z-1}$.
I understand that there are three regions to consider: $D_1=\{|z|<1\}, D_2=\{1<|z|<2\}, D_3=\{|z|>2\}$.
I understand that I am now meant to determine the Laurent expansion for each part of the function. I got stuck here so I peeked at the solution, the first part of which shows:
$$-\frac{1}{z-1} = \begin{cases} \frac{1}{1-z} & |z|<1 \\ \\ (\frac{1}{(-z)(1-\frac{1}{z})}) & |z|>1 \end{cases}$$
These two cases appear identical to me, only in the second case $-z$ is factored out. I would greatly appreciate if someone could explain what is at play here because I am failing to grasp the principle behind factoring out $-z$ like this.
Edit: I now notice that in both cases the denominator is factored in such a way as to avoid division by zero in the respective domain. Is there more behind it than this?
Thank you.
Note that$$\frac2{z-2}=-\frac1{1-z/2}=\begin{cases}\displaystyle-\sum_{n=0}^\infty\frac{z^n}{2^n}&\text{ if }|z|<2\\\displaystyle\sum_{-\infty}^{-1}\frac{z^n}{2^n}&\text{ if }|z|>2\end{cases}$$and that$$-\frac1{z-1}=\frac1{1-z}=\begin{cases}\displaystyle\sum_{n=0}^\infty z^n&\text{ if }|z|<1\\\displaystyle-\sum_{-\infty}^{-1}z^n&\text{ if }|z|>1.\end{cases}$$This is enough to solve your problem.