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I got stuck on this seemingly simple question:

If $z$ is a complex number satisfying $|z^3+z^{-3}| \le 2$, then the maximum possible value of $|z+z^{-1}|$ is:
(A) $2$
(B) $2^{1/3}$
(C) $2\sqrt 2$
(D) $1$

Using the AM-GM inequality, I showed that $|z|^3 + |z|^{-3} \ge 2$ and so I got: $$ |z^3+z^{-3}| \le 2 \le |z|^3 + |z|^{-3}$$

I don't inderstand how I can remove the cubic power, and bring any of this in terms of $|z|$ and $|z|^{-1}$. What am I missing?

Thank you.

FreezingFire
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2 Answers2

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We can write $$z^3+\frac{1}{z^3} = \left(z+\frac{1}{z}\right)^3-3\left(z+\frac{1}{z}\right).$$

Given $$2 \geq \left|z^3+\frac{1}{z^3}\right| = \left|\left(z+\frac{1}{z}\right)^3-3\left(z+\frac{1}{z}\right)\right|\geq \left|z+\frac{1}{z}\right|^3-3\left|z+\frac{1}{z}\right|.$$

Now put $$\left|z+\frac{1}{z}\right|=y.$$

So we get $y^3-3y\leq 2\Rightarrow y^3-3y-2\leq 0\Rightarrow y^3-4y+y-2\leq0$.

So $$\displaystyle y(y-2)(y+2)+1(y-2)\leq 0\Rightarrow (y-2)(y+1)^2\leq0.$$

So we get $$y\leq 2\Rightarrow \left|z+\frac{1}{z}\right|\leq 2$$

because $$(y+1)^2\geq 0.$$

juantheron
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  • Thank you for the answer! I also liked how you factorized the cubic, how did you think of it? – FreezingFire Nov 05 '16 at 10:55
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    Using Integral root Theorem, i have just factorize constant term $2$ as $\pm 1;, \pm 2$ and the put $y=1$ or $y=-1$ or $y=2$ or $y=-2$ ect – juantheron Nov 05 '16 at 10:58
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$(z+z^{-1})^3=z^3 + 3z + 3z^{-1}+z^{-3}$ therefore $|z+z^{-1}|^3 \le |z^3 + z^{-3}| + 3 |z+z^{-1}| \le 2 + 3|z+z^{-1}|$ and using notation $|z+z^{-1}| = x$ we have:

$x^3 \le 2 + 3x$ equivalent $(x-2)(x+1)^2 \le 0$ therefore $|z+z^{-1}| \le 2 $ is the answer