Question

If $|z+\frac{2}{z}|=2,$ then the maximum value of $|z|$ is $\sqrt{m}+1$.Find $m$

Answer 1

$\to |z+\frac{2}{z}|=2$

$\to \left|z\right|=|(z+\frac{2}{z})+\left(\frac{-2}{z}\right)|$

Using triangle inequality, we get

$\to \left|z\right|\le |z+\frac{2}{z}|+\left|\frac{-2}{z}\right|$

$\to \left|z\right|\le 2+\frac{2}{\left|z\right|}$

${\left|z\right|}^2\le 2\left|z\right|+2$

${\left|z\right|}^2-2\left|z\right|-2\le 0$

$\to \left|z\right|ϵ(1-\sqrt{3},1+\sqrt{3})$

As $1-\sqrt{3}<0$

Therefore, $\left|z\right|ϵ(0,1+\sqrt{3})$

Maximum of $\left|z\right|=1+\sqrt{3}=1+\sqrt{3}$

$m=3$