Expand the function $f(z)=\frac{1}{z-1}$ as as a series around $z_{0}$ in two regions
a) $$|z-z_{0}| < |1-z_{0}|$$ b) $$|z-z_{0}| > |1-z_{0}|$$ and find coefficient $a_{n}$ is each case.
I found radius of convergence to be $|1-z_{0}|$ , so for $|z-z_{0}| < |1-z_{0}|$ I found $$f(z)= -\sum \frac{(z-z_{0})^n}{(1-z_{0})^{n+1}}$$ But what happens for case b) $|z-z_{0}| > |1-z_{0}|$, doesn't it diverge? Can we still find series expansion for it? Will that be a Laurent or Taylor?
The two cases are very similar. In case (a), you write $${1\over z-1}={1\over (z-z_0)-(1-z_0)}={-1\over 1-z_0}\cdot {1\over 1-{z-z_0\over 1-z_0}}$$ and expand the last factor as a geometric series $\sum_{n=0}^\infty w^n$ with $w={z-z_0\over 1-z_0}$. Note that $|w|<1$ in case (a).
In case (b), you need to pull out the larger term $z-z_0$ instead, obtaining $${1\over z-1}={1\over (z-z_0)-(1-z_0)}={1\over z-z_0}\cdot {1\over 1-{1-z_0 \over z-z_0}}$$ Similarly as above, let $w={1-z_0\over z-z_0}$, and observe that $|w|<1$ in case (b), so we can use the same trick again. Since $z-z_0$ now appears in the denominator of $w$, the Taylor series in $w$ will give rise to a Laurent series in $z-z_0$.