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If $z$ is a complex number satisfying $\displaystyle \left|z^3+\frac{1}{z^3}\right|\leq 2\;,$ Then Prove that $\max$ of $\displaystyle \left|z+\frac{1}{z}\right|\leq 2.$

$\bf{My\; Try::}$ Let $$\displaystyle k=\left(z+\frac{1}{z}\right)\;,$$ Then $$\displaystyle \left|z^3+\frac{1}{z^3}\right| = \left|\left(z+\frac{1}{z}\right)\cdot \left[\left(z+\frac{1}{z}\right)^2-3\right]\right|\leq 2.$$

So we get $$\displaystyle \left|k^3-3k\right|\leq 2\Rightarrow -2\leq k^3-3k\leq 2$$

So we get either $$\displaystyle k^3-3k+2\geq 0$$ or $$k^3-3k-2\leq 0$$

So we get $$(k-1)^2(k+2)\geq 0$$ or $$(k-2)(k+1)^2\leq 0$$

Now How can I proceed further, Help me

Thanks.

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    $\begingroup$ You cannot write $-2\leq k^3-3k\leq 2$ as $k$ is a complex number. Now an hint: suppose $|k|>2$. Then $|k^2-3|\geq |k|^2-3 >4-3=1$. $\endgroup$ – Kelenner Nov 1 '15 at 15:41
  • $\begingroup$ Take k=\mid(z+\frac{1}{z}\mid}.Clearly $k>0$. So the first inequality is true for all real $k$ but the second inequality holds true only for $k\in(0,2]$.Hence maximum value is 2. $\endgroup$ – Maverick Dec 12 '16 at 4:24
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We argue by contradiction.

Suppose $|w| > 2$ where $w = z + \frac1z$. Then we have $$ \left| z^3 + \frac{1}{z^3} \right| = \left| w^3 - 3w \right| = |w| |w^2-3| > |w| > 2 $$ because $|w^2-3| \geq |w^2| - |3| = |w|^2 - 3 > 4 - 3 = 1$ by the triangle inequality.

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