If $z$ is complex number such that $\left|z-\frac{4}{z}\right|=2$, then the greatest value of $|z|$ is

If $z$ is complex number such that $\left|z-\frac{4}{z}\right|=2$, then the greatest value of $|z|$ is
  1. $1+\sqrt{2}$
  2. $\sqrt{2}$
  3. $\sqrt{3}+1$
  4. $1+\sqrt{5}$

Solution

Given, $\left|z-\frac{4}{z}\right|=2$ $\begin{array}{cc}\therefore & \quad|z|=\left|z-\frac{4}{z}+\frac{4}{z}\right| \\ & \leq\left|z-\frac{4}{z}\right|+\left|\frac{4}{z}\right| \\ \Rightarrow & |z| \leq 2+\frac{4}{|z|} \\ \Rightarrow & |z|^2-2|z| \leq 4 \\ \Rightarrow & \left(|z|^2-1\right)^2 \leq 5 \\ \Rightarrow & (|z|-1) \leq \sqrt{5} \\ \Rightarrow & |z| \leq \sqrt{5}+1\end{array}$

Asked in: AP EAMCET 2012

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