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

If $\left|z-\frac{2}{z}\right|=2$, then the greatest value of $|z|$ is
  1. $\sqrt{3}-1$
  2. $\sqrt{3}$
  3. $\sqrt{3}+1$
  4. $\sqrt{3}+2$

Solution

$\begin{aligned} & \text {Given }\left|z-\frac{2}{z}\right|=2 \\ & \Rightarrow|| z\left|-\frac{2}{|z|}\right| \leq\left|z-\frac{2}{z}\right|=2 \\ & \Rightarrow|z|-\frac{2}{|z|} \leq 2 \Rightarrow|z|^2-2|z|-2 \leq 0 \\ & \Rightarrow|z| \leq \frac{1 \pm \sqrt{4+8}}{2} \leq 1+\sqrt{3} \end{aligned}$ So $\max$ value of $|z|=\sqrt{3}+1$

Asked in: AP EAMCET 2023 (15 May Shift 2)

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