Question

If $\left|\begin{array}{c}Z-\frac{4}{Z}\end{array}\right|=2$ 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

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

$|Z|=\left|\begin{array}{c}(Z-\frac{4}{Z})+\frac{4}{Z}\end{array}\right|\le \left|\begin{array}{c}Z-\frac{4}{Z}\end{array}\right|+\frac{4}{|Z|}\le 2+\frac{4}{|Z|}\Rightarrow |Z{|}^2-2|Z|-4\le 0\left(quadraticin|z|\right)\Rightarrow (|Z|-(\sqrt{5}+1\left)\right)(|Z|-(1-\sqrt{5}\left)\right)\le 0\Rightarrow 0\le |Z|\le \sqrt{5}+1$