Question

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

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

Answer 1

The correct option is B $\sqrt{5}+1$

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

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

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

$(\because |z-\frac{4}{z}|=2)$

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

$\Rightarrow (\left|z\right|-(\sqrt{5}+1))(\left|z\right|-(1-\sqrt{5}))\le 0$

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