Q19 / 50 · Manipal (MET) 2022 · Mathematics

If $\left|z+\frac{2}{z}\right|=2$, then the minimum value of $|z|$ is

  • A
    $1+\sqrt{2}$
  • B
    $1+2 \sqrt{2}$
  • C
    $3 \sqrt{3}+1$
  • Correct answer
    D
    $1-\sqrt{3}$

Solution

We can write $\begin{aligned} & \quad|z|=\left|z+\frac{2}{|z|}-\frac{2}{|z|}\right| \leq\left|z+\frac{2}{z}\right|+\left|\frac{2}{|z|}\right| \\ & \quad=2+\frac{2}{|z|} \\ & \therefore \quad|z|^2-2|z|-2 \leq 0 \end{aligned}$
Here, $|z|$ lies between $\frac{2 \pm \sqrt{12}}{2}=1 \pm \sqrt{3}$ $\therefore$ Minimum value of $|z|=1-\sqrt{3}$

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

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