Q. The maximum value of $\left| \mathcal{z} \right|\ \ where\mathcal{\ z\ }$satisfies the condition $\left| \mathcal{z +}\frac{2}{\mathcal{z}} \right| = 2$, is
A) $\sqrt{3} - 1$
B) $\sqrt{3} + 1$
C) $\sqrt{3}$
D) $\sqrt{2} + \sqrt{3}$
Explanation
(b)
$\left| \mathcal{z +}\frac{2}{\mathcal{z}} \right| = 2\ \Rightarrow \left| \mathcal{z} \right| - \frac{2}{\left| \mathcal{z} \right|} \leq 2$
$\Rightarrow \left| \mathcal{z} \right|^{2} - 2\left| \mathcal{z} \right| - 2 \leq 0$
This is a quadratic equation in |$\mathcal{z|}$
$\mathcal{\therefore|z| \leq}\frac{2 \pm \sqrt{4 + 8}}{2} \leq 1 \pm \sqrt{3}$
Hence, maximum value of |$\mathcal{\ z|}$ is $1 + \sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.