Question

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

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

Answer 1

Answer: (B)

$\sqrt{5}+1$

$|z-\frac{4}{z}|=2$
Now, $|z|=|z-\frac{4}{z}+\frac{4}{z}|\le$ $2+\frac{4}{|z|}$
$\Rightarrow |z{|}^2-2|z|-4\le 0$

Solving quadratic inequality

$1-\sqrt{5}\le |z|\le 1+\sqrt{5}$
$\Rightarrow 0<|z|\le 1+\sqrt{5}$.(|z| can't be negative)