Question

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

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

Answer 1

Answer: (B)

$1+\sqrt{5}$

$|z-\frac{4}{z}|\ge |z|-\frac{|4|}{|z|}\Rightarrow 2\ge |z|-\frac{4}{|z|}\Rightarrow |z{|}^2-2|z|-4\le 0\Rightarrow (|z|-(1+\sqrt{5}\left)\right)(|z|-(1-\sqrt{5}\left)\right)\le 0$

Therefore, $(1-\sqrt{5})\le |z|\le (1+\sqrt{5})$

Therefore, maximum value of $|z|=1+\sqrt{5}$