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If $\mathrm{z}$ is a complex number satisfying $\left|z^{3}+z^{-3}\right| \leq 2$, then the maximum possible value of $\left|z+z^{-1}\right|$ is
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Verified Answer
The correct answer is:
2
$$
\left|z^{3}+\frac{1}{z^{3}}\right| \leq 2
$$
But we know that $\left|z^{3}+\frac{1}{z^{3}}\right| \leq|z|^{3}+\frac{1}{|z|^{3}}$
So when $|\mathrm{z}|=1$ by $\mathrm{AM} \geq \mathrm{GM}$
$$
|\mathrm{z}|^{3}+\frac{1}{|\mathrm{z}|^{3}}=2
$$
Now $\left|\mathrm{z}+\frac{1}{\mathrm{z}}\right| \leq|\mathrm{z}|+\frac{1}{|\mathrm{z}|}=2$
Maximum value $=2$
\left|z^{3}+\frac{1}{z^{3}}\right| \leq 2
$$
But we know that $\left|z^{3}+\frac{1}{z^{3}}\right| \leq|z|^{3}+\frac{1}{|z|^{3}}$
So when $|\mathrm{z}|=1$ by $\mathrm{AM} \geq \mathrm{GM}$
$$
|\mathrm{z}|^{3}+\frac{1}{|\mathrm{z}|^{3}}=2
$$
Now $\left|\mathrm{z}+\frac{1}{\mathrm{z}}\right| \leq|\mathrm{z}|+\frac{1}{|\mathrm{z}|}=2$
Maximum value $=2$
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