Question

If $|z-\frac{3}{z}|=2$, then the greatest value of $\left|z\right|$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$
• E. $5$

Answer 1

Answer: (C)

$3$

Given, $|z-\frac{3}{z}|=2$
$\because \left|z\right|=|(z-\frac{3}{z})+\frac{3}{z}|$
$\Rightarrow \left|z\right|\le |z-\frac{3}{z}|+\frac{3}{\left|z\right|}$
$[\because |a+b|\le \left|a\right|+\left|b\right|]$
$\Rightarrow \left|z\right|\le 2+\frac{3}{\left|z\right|}$
$\Rightarrow {\left|z\right|}^2\le 2\left|z\right|+3$
$\Rightarrow {\left|z\right|}^2-2\left|z\right|-3\le 0$
$\Rightarrow -1\le \left|z\right|\le 3$
Hence the greatest value of $\left|z\right|$ is $3$