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Prove the following inequality $\forall n>0$ $\forall z \in \mathbb{C}$ such that $|z|=1$:

$$\vert z+\frac{1}{z} \vert <\vert z^{n} + i \vert + \vert \overline{z}^{n} + i \vert \leq 2\sqrt{2} $$

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hint: for $|z|=1$ you have $\frac{1}{z}=\bar{z}$. – yohBS Feb 26 '13 at 18:46
I tried that, but didn't go far. Would appreciate a precise solution (since I found one myself, but not very satisfying) ;) – Cornelis Feb 26 '13 at 20:26
up vote 1 down vote accepted

Let $z=e^{i\phi}$, then your inequalities are equivalent to $$2|\cos\phi|<\sqrt{2}(\sqrt{1+\sin n\phi}+\sqrt{1-\sin n\phi})\le2\sqrt{2}.$$ Since all quantities are positive, it is ok to take the square which gives $$\cos^2\phi<1+|\cos n\phi|\le2,$$ where $1+|\cos n\phi|\le2$ is trivial and $\cos^2\phi<1+|\cos n\phi|$ is satisfied since $\cos^2\phi\le1$ and $\cos^2\phi=1$ for $\phi=\pi k$ only, but then $1+|\cos n\phi|=2$.

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$$z=e^{i \phi} \implies |z^n+i|=\sqrt{2} \sqrt{1+\sin{n \phi}}$$

Then $$|z^n+i|+|\bar{z}^n+i| = \sqrt{2} (\sqrt{1+\sin{n \phi}}+\sqrt{1-\sin{n \phi}})$$

Easy to show (i.e., take derivative wrt $\phi$, set equal to zero) that this is maximized when $\sin{n \phi}=0$. Therefore

$$|z^n+i|+|\bar{z}^n+i| \le 2 \sqrt{2}$$

I'm not so sure the first part of the inequality is true, because

$$\left | z + \frac{1}{z} \right |^2 = 4 \cos^2{\phi}$$ while

$$(|z^n+i|+|\bar{z}^n+i|)^2 = 8 \cos^2{\frac{n \phi}{2}}$$

If we take $\phi = \pi/n$ for some sufficiently large $n$, then I can see that $\cos^2{\frac{n \phi}{2}} = 0$, while $\cos^2{\phi} > 0$.

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You made something wrong in the last part: it comes out to me that $(\vert z^{n} + i \vert + \vert \overline{z}^{n} + i \vert)^{2}= 4+2\cos(n\phi)$, which proves the inequality. Good job on the other parts! – Cornelis Feb 26 '13 at 20:24
@Cornelis: I don't see it. The expression I have above, if correct, leads to this quantity being $4 + 4 \cos{n \phi}$. But if I am wrong, please point out where I goofed. – Ron Gordon Feb 26 '13 at 20:38
@Cornelis: I guess it must be late because you forgot the factor of 2 in front of the square roots. – Ron Gordon Feb 26 '13 at 21:09
Anyhow I don't get how you come to $8cos^{2}(\frac{n\phi}{2}$ (I am really sorry to upset your evening, but I am quite sure the inequality is verified) – Cornelis Feb 26 '13 at 21:23
@Cornelis: hey, it takes a lot more than that to upset my evening. (Actually, I am in CA right now and it is lunchtime.) But use $1+\cos{y} = 2 \cos^2{(y/2)}$. – Ron Gordon Feb 26 '13 at 21:25

$$\vert z^{n} + i \vert + \vert \overline{z}^{n} + i \vert \leq 2\sqrt{2} $$ $$w_1=z^{n} + i \rightarrow w_1=r^n(cos(n\theta)+i\space sin(n\theta))+i$$ $$ \vert w_1\vert ^2=1+r^{2n}+2r^n \space sin(n \theta) \rightarrow \vert z \vert =1 \rightarrow \vert w_1\vert^2=2+2sin(n\theta)$$ $$w_2=z^{-n} + i \rightarrow w_1=r^{-n}(cos(n\theta)-i\space sin(n\theta))+i$$ $$ \vert w_2\vert ^2=1+r^{-2n}-2r^{-n} \space sin(n \theta) \rightarrow \vert z \vert =1 \rightarrow \vert w_2\vert^2=2-2sin(n\theta) $$ $$\vert w_1 \vert + \vert w_2 \vert \leq 2\sqrt{2} \rightarrow ( \vert w_1 \vert + \vert w_2 \vert )^2 \leq (2\sqrt{2})^2 \rightarrow \vert w_1 \vert^2+\vert w_2 \vert^2 +2\vert w_1 \vert\vert w_2 \vert \leq8$$ $$\vert w_1 \vert^2+\vert w_2 \vert^2 +2\vert w_1 \vert\vert w_2 \vert = 4+2 \sqrt{1-sin^2(n \theta)}$$ $$ 4 \leq 4+2 \sqrt{1-sin^2(n \theta)} \leq 6 \rightarrow 4 \leq\vert w_1 \vert^2+\vert w_2 \vert^2 +2\vert w_1 \vert\vert w_2 \vert \leq 6 $$ $$So : 2 \leq \space \vert w_1 \vert + \vert w_2 \vert \leq \sqrt{6} \rightarrow \vert w_1 \vert + \vert w_2 \vert\leq2\sqrt{2}$$ $$\vert z + \frac{1}{z}\vert=\vert re^{i \theta} +re^{-i \theta}\vert \rightarrow \vert z\vert=1 \rightarrow r=1 $$ $$\vert e^{i \theta} +e^{-i \theta}\vert = \vert cos(\theta)+i \space sin(\theta) +cos(\theta)-i \space sin(\theta)\vert =\vert2cos(\theta) \vert \leq 2 $$ $$ $$

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