Q. $\left| \frac{1}{2}\left( z_{1} + z_{2} \right) + \sqrt{z_{1}z_{2}} \right| + \left| \frac{1}{2}\left( z_{1} + z_{2} \right) - \sqrt{z_{1}z_{2}} \right|$ is equal to

A) $|z_{1} + z_{2}|$
B) $|z_{1} - z_{2}|$
C) $\left| z_{1} \right| + |z_{2}|$
D) $\left| z_{1} \right| - |z_{2}|$
Explanation

(c)
$\left| \frac{1}{2}\left( z_{1} + z_{2} \right) + \sqrt{z_{1}z_{2}} \right| + \left| \frac{1}{2}\left( z_{1} + z_{2} \right) - \sqrt{z_{1}z_{2}} \right|$
$= \frac{1}{2}\left| \left( \sqrt{z_{1}} + \sqrt{z_{2}} \right)^{2} \right| + \frac{1}{2}|\left( \sqrt{z_{1}} - \sqrt{z_{2}} \right)^{2}|$
$= \frac{1}{2}\left| \sqrt{z_{1}} + \sqrt{z_{2}} \right|^{2} + \frac{1}{2}\left| \sqrt{z_{1}} - \sqrt{z_{2}} \right|^{2}\ \ \ \ \lbrack\because\ \ \left| z^{2} \right| = |z|^{2}\rbrack$
$= \frac{1}{2}.2\left\lbrack \left| \sqrt{z_{1}} \right|^{2} + \left| \sqrt{z_{2}} \right|^{2} \right\rbrack = \left| z_{1} \right| + |z_{2}|$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.