Q180 / 270 · NDA 2017 (Phase 2) · Mathematics

If $\left|z-\frac{4}{z}\right|=2$, then the maximum value of $|z|$ is equal to

  • A
    $1+\sqrt{3}$
  • Correct answer
    B
    $1+\sqrt{5}$
  • C
    $1-\sqrt{5}$
  • D
    $\sqrt{5}-1$

Solution

$\left|z-\frac{4}{z}\right|=2 .$ We know $|a-b| \geq|a|-|b|$ $\therefore\left|z-\frac{4}{z}\right| \geq|z|-\left|\frac{4}{z}\right|$ $\Rightarrow 2 \geq|z|-\frac{4}{|z|} \Rightarrow|z|^2-2|z|-4 \leq 0$ $\Rightarrow|z|=\frac{2 \pm \sqrt{4-4(1)(-4)}}{2(1)}=\frac{2 \pm \sqrt{20}}{2}=\frac{2 \pm 2 \sqrt{5}}{2}=1 \pm \sqrt{5}$ So, $z=1+\sqrt{5}$.

Hence, the correct answer is option: 'B'.

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