Question

If \( \omega_1 \) and \( \omega_2 \) are two non-zero complex numbers and \( a, b \) are non-zero real numbers such that \[ |a\omega_1 + b\omega_2| = |a\omega_1 - b\omega_2|, \] then \( \dfrac{\omega_1}{\omega_2} \) is:

• A. a positive real number
• B. a negative real number
• C. zero
• D. purely imaginary number

Hint:

This type of question often tests your understanding of the geometric interpretation of complex numbers. If magnitudes of vector additions and subtractions are equal, the vectors are perpendicular.

Answer 1

The Correct Option is D

Given: \( |a\omega_1 + b\omega_2| = |a\omega_1 - b\omega_2| \)
Square both sides: \[ |a\omega_1 + b\omega_2|^2 = |a\omega_1 - b\omega_2|^2 \]
Use identity \( |z|^2 = z\overline{z} \): \[ (a\omega_1 + b\omega_2)(\overline{a\omega_1 + b\omega_2}) = (a\omega_1 - b\omega_2)(\overline{a\omega_1 - b\omega_2}) \]
Simplify: \[ a^2|\omega_1|^2 + ab(\omega_1\overline{\omega_2} + \overline{\omega_1}\omega_2) + b^2|\omega_2|^2 \] \[ = a^2|\omega_1|^2 - ab(\omega_1\overline{\omega_2} + \overline{\omega_1}\omega_2) + b^2|\omega_2|^2 \]
Subtract: \[ 2ab(\omega_1\overline{\omega_2} + \overline{\omega_1}\omega_2) = 0 \Rightarrow \omega_1\overline{\omega_2} + \overline{\omega_1}\omega_2 = 0 \]
This implies: \[ \Re(\omega_1\overline{\omega_2}) = 0 \Rightarrow \omega_1\overline{\omega_2} \text{ is purely imaginary} \Rightarrow \frac{\omega_1}{\omega_2} \text{ is purely imaginary} \] \[ \boxed{\text{Correct Option: (4)}} \] % Tip