<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] Here is one approach:
Let $z=r e^{i\theta}$, then we want to compute the $\min,\max$ of
$I=\{ r  \mid \, |r e^{i2\theta} + {2 \over r}| = 2 \}$. We see that $r \in I$ iff $|r \cos ( 2 \theta)+ {2 \over r} +i r \sin ( 2 \theta)|^2 = r^2 +{4 \over r^2} +  4 \cos(2 \theta)= 4$ for some $\theta$.
If we let $f(r) = 1-{1 \over 4} (r^2 +{4 \over r^2} )$, we see that $I = f^{-1} [-1,1]$.
The function $f$ is strictly concave, $\lim_{r \downarrow 0} f(r) = \lim_{r \to \infty} f(r) = -\infty$, $f$ has a maximum value of $0$ at $r=\sqrt{2}$. Hence $I = [r_0,r_1]$ where $r_k$ solve $f(r_k) = -1$.
Solving $f(r) = -1$ yields the equation $r^2 +{4 \over r^2} = 8$ which
has positive solutions $r = \sqrt{2} \sqrt{2 \pm  \sqrt{3}}$.
Hence the minimum value of $|z|$ is $\sqrt{2} \sqrt{2 -  \sqrt{3}}$ and the maximum value if $\sqrt{2} \sqrt{2 +  \sqrt{3}}$.
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