Question

Consider the following complex numbers
\[ z_1 = r_1(\cos\theta_1 + i\sin\theta_1) \]\[ z_2 = r_2(\cos\theta_2 + i\sin\theta_2) \]
where \(r_1, r_2\) are real numbers, \(0 \le \theta_1 \le \dfrac{\pi}{2}\), \(0 \le \theta_2 \le \dfrac{\pi}{2}\), and \(i=\sqrt{-1}\)
If \(|z_1+z_2| = |z_1|+|z_2|\), which one of the following conditions is necessarily CORRECT?

• A. \(\theta_1 = 0,\ \theta_2 = \dfrac{\pi}{2}\)
• B. \(\theta_1 = \dfrac{\pi}{2},\ \theta_2 = 0\)
• C. \(r_1 = r_2\)
• D. \(\theta_1 = \theta_2\)

Hint:

Equality in the triangle inequality |z1+z2|=|z1|+|z2| happens only when z1 and z2 point in the same direction; work out what that means for the arguments given that r1, r2 can be negative.

Answer 1

The Correct Option is D

Square both sides: $\text{Re}(z_1\bar z_2) = |z_1||z_2|$. Computing $z_1\bar z_2 = r_1r_2\cos(\theta_1-\theta_2)+i\,r_1r_2\sin(\theta_1-\theta_2)$ gives condition $r_1r_2\cos(\theta_1-\theta_2)=|r_1||r_2|$. For same-sign $r_1,r_2$, this reduces to $\cos(\theta_1-\theta_2)=1$, forcing $\theta_1=\theta_2$ within the allowed range.\[ \boxed{\theta_1=\theta_2} \]