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

Let $z=\color{red}{\sqrt{2}}re^{i\theta}$, then $\left|z+\frac{2}{z}\right|=2$ reduces to $$r^2+\frac{1}{r^2}=2-2\cos{2\theta} \tag{1}$$
Observe that $(1)$(considering it as a quadratic in $r^2$) will have two real roots of the form $\alpha$ and $\frac{1}{\alpha}$ only if $\cos{2\theta} \le 0$. Thus if the maximum value is $\alpha$, the minimum would be $\frac{1}{\alpha}$.
Without loss of generality assume maximum(of $r^2$) is $\ge 1$, notice that $r^2+\frac{1}{r^2}$ is an increasing function when $r\ge 1$, hence maximum is attained when R.H.S is maximum, i.e. when $\cos{2\theta}=-1$. Solving for $r$, get $r=\sqrt{2+\sqrt{3}}\quad,\quad\frac{1}{\sqrt{2+\sqrt{3}}}$
Maximum value of $|z|=\sqrt{2}\cdot \sqrt{2+\sqrt{3}}=1+\sqrt{3}$ and minimum value of $|z|=\frac{2}{1+\sqrt{3}}=\sqrt{3}-1$.
