Laurent series expansion for $\lvert z\rvert >1$.

Question

I have a simple complex function like this:

$$\frac{z+1}{z-1}$$

When I expand it by its Maclaurin series:

$$\frac{z+1}{z-1} = \frac{z-1+2}{z-1} = 1 - \frac{2}{1-z} = 1 - 2\sum_{k=0}^{\infty}z^{k} $$

subject to $\mathbf{|z|<1}$.

I am supposed to represent this function by its Laurent series for the domain $ |z| > 1 $ but for $ |z| > 1 $, the function hasn't any singularity, also there is no descending term.

How to approach this problem?

Answer 1

It is asking you to consider $\lvert z\rvert > 1\iff 1/\lvert z\rvert < 1$. Therefore, when you do the expansion, you want to have something of the form $$ \frac{A}{1-1/\lvert z\rvert} $$

Answer 2

Note that $$\frac{z+1}{z-1}=\frac{(z+1)/z}{(z-1)/z}=\frac{1+\frac{1}{z}}{1-\frac{1}{z}},$$and the right hand expression is the sum of a geometric series with quotient $=1/z$.