Question

if $\left|z-\frac{4}{z}\right|=2$, then the maximum of value of $\left|z\right|$ is equal to

Answer 1

$|z-\frac{4}{z}|=2$

Using inequality identity, we get

$\left|z\right|=|(z-\frac{4}{z})+\frac{4}{z}|\le |z-\frac{4}{z}|+\frac{4}{\left|z\right|}$

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

${(\left|z\right|-1)}^2\le 5$

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

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

as $1-\sqrt{5}<0$, therefore $0\le \left|z\right|\le 1+\sqrt{5}$

$0\le \left|z\right|\le \sqrt{5}+1$

Thus, maximum value of $\left|z\right|=1+\sqrt{5}$