Q. For two complex numbers $z_{1},\ z_{2}$ the relation $\left| z_{1} + z_{2} \right| = \left| z_{1} \right| + |z_{2}|$ holds, if
A) $\arg{(z_{1})} = \arg{(z_{2})}$
B) $\arg{\left( z_{1} \right) + \arg{\left( z_{2} \right) = \frac{\pi}{2}}}$
C) $z_{1}z_{2} = 1$
D) $\left| z_{1} \right| = |z_{2}|$
Explanation
(a)
Since, $\left| z_{1} + z_{2} \right| = \left| z_{1} \right| + |z_{2}|$
$\Rightarrow \ \ \ \ \ \left| z_{1} \right|^{2} + \left| z_{2} \right|^{2} + 2\left| z_{1} \right|\left| z_{2} \right|\cos{(\theta_{1} - \theta_{2})}$
$= \left| z_{1} \right|^{2} + \left| z_{2} \right|^{2} + 2\left| z_{1} \right||z_{2}|$
$\Rightarrow \ \ \ \cos{(\theta_{1} - \theta_{2})} = 1 = \cos{0{^\circ}}$
$\Rightarrow \ \ \ \ \ \theta_{1} - \theta_{2} = 0\ \ \Rightarrow \ \ \ \ \theta_{1} = \theta_{2}$
$\Rightarrow \ \ \ \arg{\left( z_{1} \right) = \arg{(z_{2})}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.