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, $-5 \leq|z|-\frac{6}{|z|} \leq 5$
$\Rightarrow -5|z| \leq|z|^{2}-6$
$\Rightarrow |z|^{2}-6+5|z| \geq 0$
$\Rightarrow |z|^{2}+6|z|-|z|-6 \geq 0$
$\Rightarrow |z|(|z|+6)-1(|z|+6) \geq 0$
$\Rightarrow (|z|-1)(|z|+6) \geq 0$
$\Rightarrow |z| \geq 1 \,\,\,(\because|z| \neq-6)$
and $|z|^{2}-6 \leq 5|z| $
$\Rightarrow |z|^{2}-5|z|-6 \leq 0$
$\Rightarrow (|z|-6)(|z|+1) \leq 0$
$\Rightarrow |z| \leq 6$
So, $1 \leq|z| \leq 6$
So, maximum value of $ (z)=6$