Q. If $\frac{2z_{1}}{3z_{2}}$ is purely imaginary, then $\left| \frac{z_{1} - z_{2}}{z_{1} + z_{2}} \right|$ is
A) $\frac{2}{3}$
B) $\frac{3}{2}$
C) $\frac{4}{9}$
D) 1
Explanation
(d)
Let $\frac{2z_{1}}{3z_{2}} = ik\ \ \ \ \Rightarrow \ \ \ \ \frac{z_{1}}{z_{2}} = \frac{3ik}{2}$
$\therefore\ \ \ \ \ \left| \frac{z_{1} - z_{2}}{z_{1} + z_{2}} \right| = \left| \frac{(z_{1}/z_{2}) - 1}{(z_{1}/z_{2}) + 1} \right| = \left| \frac{(3ik/2) - 1}{(3ik/2) + 1} \right| = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.