Question

If $|z-\frac{4}{z}|=2$, then the greatest value of $|z|$ is

• A. $1+\sqrt{2}$
• B. $2+\sqrt{2}$
• C. $\sqrt{3}+1$
• D. $\sqrt{5}+1$

Answer 1

The correct option is D

$\sqrt{5}+1$

We have
$|z|=|z-\frac{4}{z}+\frac{4}{z}|\le |z-\frac{4}{z}|+\frac{4}{|z|}=2+\frac{4}{|z|}\Rightarrow |z{|}^2\le 2|z|+4\Rightarrow (|z|-1{)}^2\le 5\Rightarrow |z|-1\le \sqrt{5}\Rightarrow |z|\le \sqrt{5}+1$
Also, for $z=\sqrt{5}+1,|z-\frac{4}{z}|=2$
Therefore, the greatest value of $|z|$ is $\sqrt{5}+1$ which is attained when $z=\sqrt{5}+1$