hard complex inequality

jason32 · **Joined:** Wed, 28 Sep 2011 15:35:23 UTC **Posts:** 8

Hello.
Can you please help me to solve this very difficult exercice (for me).
n is an integer (n>0) ; z is a complex such |z|=1
I have to show that
$n|1+z| + |1+z^2| + |1+z^3| +...+ |1+z^{2n}|+ |1+z^{2n+1}| \ge 2n$

I try to write $z=e^{i\theta}$ et $1+z^k=1+e^{ik\theta}=2e^{ik\frac{\theta}2}\cos k\frac{\theta}2$
so $\vert 1+z^k\vert=2\vert\cos k\frac{\theta}2\vert$
and the sum is equal to $2n\vert\cos\frac{\theta}2\vert+2\sum_1^{2n}\vert\cos(k+1)\frac{\theta}2\vert$
But it doesn't work

Thanks for any help.

Shadow · **Posted:** Wed, 5 Oct 2011 16:17:48 UTC

jason32 wrote:

Hello.
Can you please help me to solve this very difficult exercice (for me).
n is an integer (n>0) ; z is a complex such |z|=1
I have to show that
$n|1+z| + |1+z^2| + |1+z^3| +...+ |1+z^{2n}|+ |1+z^{2n+1}| \ge 2n$

I try to write $z=e^{i\theta}$ et $1+z^k=1+e^{ik\theta}=2e^{ik\frac{\theta}2}\cos k\frac{\theta}2$
so $\vert 1+z^k\vert=2\vert\cos k\frac{\theta}2\vert$
and the sum is equal to $2n\vert\cos\frac{\theta}2\vert+2\sum_1^{2n}\vert\cos(k+1)\frac{\theta}2\vert$
But it doesn't work

Thanks for any help.

Let us notice that this you get the same number whether you use

or $\overline{z}$ in this question, since $|\cdot |$ commutes with complex conjugation.

This means $n|1+z|+|1+z^2|+\ldots + |1+z^{2n+1}|=n|1+\overline{z}|+|1+\overline{z}^2|+\ldots + |1+\overline{z}^{2n+1}|$

So it is sufficient to show that:

$n|1+z|+|1+z^2|+\ldots + |1+z^{2n+1}|=n|1+\overline{z}|+|1+\overline{z}^2|+\ldots + |1+\overline{z}^{2n+1}|\ge 4n$

by the triangle inequality, we have:

$n|1+z|+|1+z^2|+\ldots + |1+z^{2n+1}|=n|1+\overline{z}|+|1+\overline{z}^2|+\ldots + |1+\overline{z}^{2n+1}|\ge$

$|n(2+z+\overline{z})+2+(z^2+\overline{z}^2)+\ldots +2+(z^{2n+1}+\overline{z}^{2n+1})|$

and I think from here it's not too hard to get this to be $\ge 4n$ , but I'm in a rush, so I cannot finish right now.

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hard complex inequality

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