If z is a complex number satisfying |z^3+z^{- | vedclass

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(A) Given $|z^3+z^{-3}| \leq 2$. Let $z = re^{i	heta}$. Then $|z^3+z^{-3}| = |r^3e^{i3	heta} + r^{-3}e^{-i3	heta}| \leq 2$. Using the triangle ineq

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(A) Given $|z^3+z^{-3}| \leq 2$. Let $z = re^{i	heta}$. Then $|z^3+z^{-3}| = |r^3e^{i3	heta} + r^{-3}e^{-i3	heta}| \leq 2$. Using the triangle inequality,$|z^3+z^{-3}| \geq | |z^3| - |z^{-3}| | = |r^3 - r^{-3}|$. However,we know that $|z^3+z^{-3}| \leq |z^3| + |z^{-3}| = r^3 + r^{-3}$. By the $AM$-$GM$ inequality,$r^3 + r^{-3} \geq 2$. Since $|z^3+z^{-3}| \leq 2$ and $r^3+r^{-3} \geq 2$,the only way for the condition to hold is if $|z|=1$. If $|z|=1$,then $z = e^{i	heta}$,so $|z+z^{-1}| = |e^{i	heta} + e^{-i	heta}| = |2\cos	heta|$. The maximum value of $|2\cos	heta|$ is $2$ when $\cos	heta = \pm 1$.

List-$I$	List-$II$
$(i)$ $a \bar{a}$	$(A)$ $-\frac{\pi}{3}$
$(ii)$ $\arg \left(\frac{1}{\bar{a}}\right)$	$(B)$ $-i \sqrt{3}$
$(iii)$ $a - \bar{a}$	$(C)$ $2i / \sqrt{3}$
$(iv)$ $\operatorname{Im}\left(\frac{4}{3a}\right)$	$(D)$ $1$
	$(E)$ $\pi / 3$
	$(F)$ $\frac{2}{\sqrt{3}}$

If $z$ is a complex number satisfying $|z^3+z^{-3}| \leq 2$,then the maximum possible value of $|z+z^{-1}|$ is

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