Question

Find the greatest value assumed by the function $w=|z-\frac{4}{z}|=2$ where z is a complex variable.$.$

Answer 1

$|\left|z_1\right|-\left|z_2\right||\le |z_1-z_2|$

It follows that

$|\left|z\right|-\left|\frac{4}{2}\right||\le |z-\frac{4}{z}|$

$\Rightarrow |\left|z\right|-\left|\frac{4}{z}\right||<2$

$\Rightarrow |\left|z\right|-\frac{4}{\left|z\right|}|\le 2$

$\Rightarrow -z<\left|z\right|-\frac{4}{\left|z\right|}\le 2$

$\Rightarrow -2\left|z\right|\le {\left|z\right|}^2-4\le 2\left|z\right|$

$\Rightarrow {\left|z\right|}^2+2\left|z\right|-4\ge 0$ and ${\left|z\right|}^2-2\left|z\right|-4\le 0$

From $\left(i\right)$ we get $\left|z\right|\ge \frac{-2\pm \sqrt{4+16}}{2}=\frac{-2\pm 2\sqrt{5}}{2}=-1\pm \sqrt{5}$

$\Rightarrow \left|z\right|\ge -1+\sqrt{5}$

From $\left(ii\right)$ we get $\left|z\right|\le \frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm \sqrt{5}$

$\Rightarrow \left|z\right|\le 1+\sqrt{5}$

Hence maximum value is $\sqrt{5}+1$ and minimum value is $\sqrt{5}-1$