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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.

2 Answers 2

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

$|z^n|\gt(1+|z|+\cdots+|z^{n-1}|)(|z|-1)$

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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    Oh!! You are right!!! I edit the answer immediately! (and the tex as well!)2011-12-16
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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*} $