Binomial expansion with negative exponent

Question

I have the following map

$$\mathbb{C}\ni z\mapsto \dfrac{w}{1-wz^{-1}},\ w\in\mathbb{R}.$$

The book I'm going trough states:

Using the binomial expansion for $(1-wz^{-1})^{-1}$, we may rewrite the exxpression as: $$ \frac{w}{1-wz^{-1}} = w \sum_{l=0}^{\infty}w^{l}z^{-l} $$

I can't figure out how this was done. Where did the $l$ come from? Any help will be appreciated.

Answer 1

This comes from the geometric series

$$S_N:=\sum_{l=0}^N r^l,\ r\in\mathbb{C}.$$

The limit $S:=\lim_{N\to\infty}S_N$ exists if and only if $|r|<1$ and, in such a case, we have that

$$S=\dfrac{1}{1-r}.$$

Now, if we consider the function $$\dfrac{w}{1-wz^{-1}},$$ we can recognize the same shape as $S$ where $r=wz^{-1}$. Therefore, we have that

$$\dfrac{w}{1-wz^{-1}}=w\sum_{l=0}^\infty w^lz^{-l}.$$

Note, however, that this inequality is only valid for all $z\in\mathbb{C}$ such that $|z|>|w|$ while the domain of the original function is $\mathbb{C}-\{w\}$. In fact, the original function is the analytic continuation of the series expression.