Question

Find the maximum value of $|z|$ when $|z-\frac{3}{z}|=2$, where $z$ being a complex number.

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

Answer 1

Answer: (B)

$3$

We have,
$|z|=|z-\frac{3}{z}+\frac{3}{z}|$

$\Rightarrow |z|\le |z-\frac{3}{z}|+\left|\frac{3}{z}\right|$ [Using triangle inequality]

$\Rightarrow |z|\le 2+\left|\frac{3}{z}\right|$ $[\because |z-\frac{3}{z}|=2]$

$\Rightarrow |z|\le 2+\left|\frac{3}{z}\right|$

$\Rightarrow |z|\le 2+\frac{3}{|z|}$

$\Rightarrow |z{|}^2-2|z|-3\le 0$

$\Rightarrow (|z|-3)(|z|+1)\le 0$

$\Rightarrow -1\le |z|\le 3$
Thus, the maximum value of $|z|$ is $3$.