How to compute the Laurent series for $f(z) = \frac{z+i}{z-i} $

Question

How to compute the series of $$ f(z) = \frac{z+i}{z-i} $$ centered at $z_0 = 1$ if $|z-z_0|< \sqrt 2$ and if $|z-z_0|> \sqrt 2\mbox{ ?}$

Answer 1

$$\frac{z+i}{z-i}=1+\frac{2i}{z-i}=1+\frac{2i}{z-1+(1-i)}=1+\frac{2i}{1-i}\frac1{1+\frac{z-1}{1-i}}=$$

$$1+\frac{2i}{1-i}\left(1-\frac{z-1}{1-i}+\frac{(z-1)^2}{(1-i)^2}-\ldots+(-1)^n\frac{(z-1)^n}{(1-i)^n}+\ldots\right)$$

The above is valid for

$$\left|\frac{z-1}{1-i}\right|<1\iff |z-1|<\sqrt2\;\ldots$$

Now you try something similar for $\;|z-1|>\sqrt2\;$, but this time factor out $\;z-1\;$ instead of $\;1-i\;$!