Question

If $z$ is a complex number such that $\left|z-\frac{6}{z}\right|=5$, then the maximum value of $|z|$ is

• A. 3
• B. 2
• C. 6
• D. $\frac{3}{2}$

Answer 1

Answer: (C)

Given,
$\left|z-\frac{6}{z}\right|=5$
Now, $-5\le |z|-\frac{6}{|z|}\le 5$
$\Rightarrow -5|z|\le |z{|}^2-6$
$\Rightarrow |z{|}^2-6+5|z|\ge 0$
$\Rightarrow |z{|}^2+6|z|-|z|-6\ge 0$
$\Rightarrow |z|(|z|+6)-1(|z|+6)\ge 0$
$\Rightarrow (|z|-1)(|z|+6)\ge 0$
$\Rightarrow |z|\ge 1(\because |z|\ne -6)$
and $|z{|}^2-6\le 5|z|$
$\Rightarrow |z{|}^2-5|z|-6\le 0$
$\Rightarrow (|z|-6)(|z|+1)\le 0$
$\Rightarrow |z|\le 6$
So, $1\le |z|\le 6$
So, maximum value of $(z)=6$