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In continuation to my previous post : Inequality in Complex Plane I'm still having a small problem with a similar inequality :

For $z$ such that: $|z|> 1$ I wish to prove:

$$1+|z|+\dots+|z^{n-1}| < \frac{|z^n|}{|z|-1}$$

This reminds me of $\frac{1}{1-z} = 1+z+z^2+\dots$ but this is true only for $z$ such that $|z| < 1$ so I'm not sure what to do.

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up vote 2 down vote accepted

Your inequality is clearly equivalent, if $|z|\gt 1$, to:


Now observe that $|z^l|=|z|^l$, rewriting everything in this form gives:

$$|z|^n\gt|z|+|z|^2+\cdots+|z|^n-1-|z|-\cdots-|z|^{n-1}=|z|^n -1.$$

Which is thus true for any $z \in \mathbb{C}$ with $|z| \gt 1$.

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I´m sorry for the double answer, I was typing... – Giovanni De Gaetano Dec 16 '11 at 13:52
hmm...take z such that |z| < 1 than 1+|z|+...+|z^(n-1)| > 0 but |z|-1 < 0 hence |z^n|/(|z|-1) < 0 – Belgi Dec 16 '11 at 14:01
I deleted my answer because it is basically the same as yours. I made a small correction: you multiply the inequality by $(|z|-1)$ so you have to ensure that this is $\gt 0$ (not just $\neq 1$). Note also that \neq in TeX gives you $\neq$. This is better than typing on the keyboard. – t.b. Dec 16 '11 at 14:04
Oh!! You are right!!! I edit the answer immediately! (and the tex as well!) – Giovanni De Gaetano Dec 16 '11 at 14:06
thank you both! – Belgi Dec 16 '11 at 14:07

Here's an alternate approach:

$$ \begin{align*} 1 + |z| + |z|^2 + \cdots + |z|^{n-1} &= |z|^{n-1} + \cdots + |z| + 1 \\ &= |z|^n \cdot \left( \frac{1}{|z|} + \frac{1}{|z|^2} + \cdots + \frac{1}{|z|^{n}} \right), \end{align*} $$ which is an infinite geometric series whose successive terms have the ratio $\frac{1}{|z|} < 1$. Therefore, $$ \begin{align*} 1 + |z| + |z|^2 + \cdots + |z|^{n-1} &= |z|^n \cdot \sum_{i=1}^{n} \frac{1}{|z|^i} \\ &\lt |z|^n \cdot\sum_{i=1}^{\infty} \frac{1}{|z|^i} \\ &= |z|^n \cdot \frac{\frac{1}{|z|}}{1 - \frac{1}{|z|}} \\ &= \frac{|z|^n}{|z| - 1}. \end{align*} $$

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