<s>[INST] Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow  \left|z+\frac{2}{z}\right|^2 = 2^2=4$.
So $\displaystyle \left(z+\frac{2}{z}\right)\cdot \left(\bar{z}+\frac{2}{\bar{z}}\right) = 4\Rightarrow \left|z\right|^2+2\left(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right)+\frac{1}{|z|^2} = 4$.
Now how can I find the max. and min. values of $|z|$?
Help me please.
Thanks
 [/INST] $$|z|-\frac{2}{|z|}\le \left|z+\frac{2}{z}\right|=2 \implies\\ |z|^2-2|z|-2\le0\implies\\ 1-\sqrt{3}<0\le|z|\le1+\sqrt{3}$$
Maximum value is $1+\sqrt{3}$ provided we show that equality holds. 
Recall that equality holds when $z$, $-\frac{2}{z}$ have same argument. Hence $\arg{z}=-\frac{\pi}{2}$.
Similarly using $\frac{2}{|z|}-|z|\le \left|z+\frac{2}{z}\right|=2$ we can show that the minimum is $\frac{2}{1+\sqrt{3}}=\sqrt{3}-1$
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