I was looking at some text that involves the following equality, where $z \in \mathbb{C}$
$(1+|z|^2+|z|^4+\dots +|z|^{2n-2})=\frac{|z|^{2n}-1}{|z|^2-1}$
I assumed the text used the formula for finite geometric series but shouldn't that give me $\frac{1-|z|^{2n-2}}{1-|z|^2}$ instead of the RHS of the above? I think I'm not realizing some basic properties here, can someone point me out? Thanks!
Geometric series formula says
$$ 1 + x + \dots + x^r = \frac{1 - x^{r + 1}}{1 - x}. $$
In your case we set $x = |z|^2$ and $r = n-1$. So
\begin{align*} 1 + |z|^2 + |z|^4 + \dots + |z|^{2n - 2} &= \frac{1 - (|z|^2)^n }{1 - |z|^2 } \\ &= \frac{|z|^{2n} - 1}{|z|^2 - 1}. \end{align*}