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

Step 1: Recall the equality condition of the triangle inequality.
Equality \(|a+b|=|a|+|b|\) occurs if and only if \(a\) and \(b\) point in exactly the same direction.
Step 2: Work out the argument of z1 and z2, allowing r1, r2 to be negative.
If r is positive, argument is theta; if r is negative, argument is theta+pi.
Step 3: Apply the equality condition.
Same-sign r1,r2 forces theta1=theta2 in both cases (positive or negative); opposite signs are impossible since the argument ranges are disjoint.
Step 4: Rule out the other options with a counter-example.
Take r1=1, r2=2, theta1=theta2=pi/4: equality holds yet r1≠r2, showing option (C) is not necessary.
Step 5: Conclusion.
\[ \boxed{\theta_1 = \theta_2} \]