3

Show that $$ \left\vert z^{3}+\frac{1}{z^{3}}\right\vert \leq1\Rightarrow\left\vert z+\frac{1}{z}\right\vert \leq1. $$ I have tried with the triangle inequality and the reverse triangle inequality, i.e. $$ |a+b|\le|a|+|b| \text{ and } ||a|-|b||\le|a-b|,\forall a,b\in \mathbb{C}. $$

Michael Hardy
  • 1
Mihai
  • 33
  • Are you sure the problem is correct? With $z = e^{i\pi/6}$, $z^3 = e^{i\pi/2} = i$, and $z^3+z^{-3} = i - i = 0 \le 1$, but $z+z^{-1} = \sqrt 3 \ge 1$. – hbp Dec 02 '15 at 04:21

1 Answers1

0

It is a little forced exercise...the original one is with $|...|\le 2\implies |...|\le 2$ (one can see Titu Andreescu, Complex numbers from A to Z...)...For this one, using the identity $$ \left(z+\dfrac{1}{z} \right)^3=z^3+\dfrac{1}{z^3}+3\left(z+\dfrac{1}{z} \right), $$ we have $$ \left|z+\dfrac{1}{z} \right|^3\le\left|z^3+\dfrac{1}{z^3}\right|+3\left|z+\dfrac{1}{z} \right| \le2+3\left|z+\dfrac{1}{z} \right|. $$ Letting $0<a=\left|z+\dfrac{1}{z} \right|$, we get $$ a^3-3a-1\le 0\Leftrightarrow (a-x_0)(a-x_1)(a-x_2)\le0, $$ where $x_0=-1,53...,x_1=-0,34...,x_2=1,87...$ (see Wolfram or using Rolle theorem we get rapidly $x_2$ and so on...). Since $(a-x_0)(a-x_1)>0$ when $a<x_0$ or $a>x_1$, it follows that $$ a-1,87\le 0 \Leftrightarrow a\le 1,87. $$ Finally we obteined a better estimation, a limitation and "the desired inequality". Something is rotten in the state of Denmark...

teo
  • 166