I am to find regions where $f(z) = \frac{1}{z-2}$ around $i$ has Laurent series. As Jose pointed out, the natural regions are rings $A(i, 0, \sqrt{5})$ and $A(i, \sqrt{5}, \infty)$ because $\sqrt5$ is the distance between the center and singularity.
Starting with $|z-i|<\sqrt5$. $$ \frac{1}{z-i+i-2} = \frac{1}{i-2}\frac{1}{1+\frac{z-i}{i-2}} = \frac{1}{i-2}\frac{1}{1-\frac{-(z-i)}{i-2}} = \frac{1}{i-2}\sum_\limits{n=0}^{\infty}\frac{(-1)^n(z-i)^n}{(i-2)^{n}}= \sum_\limits{n=0}^{\infty}\frac{(-1)^n(z-i)^n}{(i-2)^{n+1}} $$
If $|z-i|>\sqrt5$ we do as follows: $$ \frac{1}{z-i+i-2} = \frac{1}{z-i}\frac{1}{1+\frac{i-2}{z-i}} = \frac{1}{z-i}\frac{1}{1-\frac{-(i-2)}{z-i}} = \frac{1}{z-i}\sum_\limits{n=0}^{\infty}\frac{(-1)^n(i-2)^n}{(z-i)^{n}} = \\ = \sum_\limits{n=0}^{\infty}\frac{(-1)^n(i-2)^n}{(z-i)^{n+1}} = \sum_\limits{k=-\infty}^{-1}(-1)^{-k-1}(i-2)^{-k-1}(z-i)^{k} $$
So for ring $A(i, 0, \sqrt{5})$ Laurent series is $\sum_\limits{n=0}^{\infty}\frac{(-1)^n(z-i)^n}{(i-2)^{n+1}}$ and for $A(i, \sqrt{5}, \infty)$ it equals to $\sum_\limits{k=-\infty}^{-1}(-1)^{-k-1}(i-2)^{-k-1}(z-i)^{k}$.
Is this correct?
No, it is not correct. The natural regions are $\left\{z\in\Bbb C\,\middle|\,|z-i|<\sqrt5\right\}$ nad $\left\{z\in\Bbb C\,\middle|\,|z-i|>\sqrt5\right\}$, since the distance from $i$ to $2$ is $\sqrt5$.