Search any question & find its solution
Question:
Answered & Verified by Expert
If $z$ is complex number such that $\left|z-\frac{4}{z}\right|=2$, then the greatest value of $|z|$ is
Options:
Solution:
2925 Upvotes
Verified Answer
The correct answer is:
$1+\sqrt{5}$
Given, $\left|z-\frac{4}{z}\right|=2$
$\begin{array}{cc}\therefore & \quad|z|=\left|z-\frac{4}{z}+\frac{4}{z}\right| \\ & \leq\left|z-\frac{4}{z}\right|+\left|\frac{4}{z}\right| \\ \Rightarrow & |z| \leq 2+\frac{4}{|z|} \\ \Rightarrow & |z|^2-2|z| \leq 4 \\ \Rightarrow & \left(|z|^2-1\right)^2 \leq 5 \\ \Rightarrow & (|z|-1) \leq \sqrt{5} \\ \Rightarrow & |z| \leq \sqrt{5}+1\end{array}$
$\begin{array}{cc}\therefore & \quad|z|=\left|z-\frac{4}{z}+\frac{4}{z}\right| \\ & \leq\left|z-\frac{4}{z}\right|+\left|\frac{4}{z}\right| \\ \Rightarrow & |z| \leq 2+\frac{4}{|z|} \\ \Rightarrow & |z|^2-2|z| \leq 4 \\ \Rightarrow & \left(|z|^2-1\right)^2 \leq 5 \\ \Rightarrow & (|z|-1) \leq \sqrt{5} \\ \Rightarrow & |z| \leq \sqrt{5}+1\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.