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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!

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    $\begingroup$ $a/b = (-a)/(-b)$. Then fix the index. $\endgroup$ Nov 20, 2022 at 1:33
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    $\begingroup$ To elaborate on @EthanBolker's comment, it may be that your expression is for $n=1,\dots,N$ while the quoted formula is for $n=0,\dots,N-1$ $\endgroup$ Nov 20, 2022 at 1:39
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    $\begingroup$ To be exact, this is formula for geometrical progression with ratio $|z|^2$, first member $1$ and sum is taken for first $n$ members $ar^0+ar^1+\cdots +ar^{n-1} = \frac{a(1-r^n)}{1-r} $. BTW, here is nothing from complex analysis. $\endgroup$
    – zkutch
    Nov 20, 2022 at 2:01

1 Answer 1

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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*}

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