Search any question & find its solution
Question:
Answered & Verified by Expert
If $\left|z-\frac{2}{z}\right|=2$, then the greatest value of $|z|$ is
Options:
Solution:
1422 Upvotes
Verified Answer
The correct answer is:
$\sqrt{3}+1$
$\begin{aligned}
& \text {Given }\left|z-\frac{2}{z}\right|=2 \\
& \Rightarrow|| z\left|-\frac{2}{|z|}\right| \leq\left|z-\frac{2}{z}\right|=2 \\
& \Rightarrow|z|-\frac{2}{|z|} \leq 2 \Rightarrow|z|^2-2|z|-2 \leq 0 \\
& \Rightarrow|z| \leq \frac{1 \pm \sqrt{4+8}}{2} \leq 1+\sqrt{3}
\end{aligned}$
So $\max$ value of $|z|=\sqrt{3}+1$
& \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 $|z|=\sqrt{3}+1$
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.