Question

If \( z \) and \( \omega \) are two non-zero complex numbers such that \( |z\omega| = 1 \) and

\[ \arg(z) - \arg(\omega) = \frac{\pi}{2}, \]

Then the value of \( \overline{z\omega} \) is:

• A. \( i \)
• B. \( -1 \)
• C. \( 1 \)
• D. \( -i \)

Hint:

For complex number problems involving arguments and moduli, convert to exponential form using Euler's formula: \( z = re^{i\theta} \).

Answer 1

The Correct Option is D

Let \( z = r_1 e^{i\theta_1} \), \( \omega = r_2 e^{i\theta_2} \)
Then \( z\omega = r_1 r_2 e^{i(\theta_1 + \theta_2)} \)
Given: \[ |z\omega| = 1 \Rightarrow r_1 r_2 = 1 \] \[ \arg(z) - \arg(\omega) = \frac{\pi}{2} \Rightarrow \theta_1 - \theta_2 = \frac{\pi}{2} \] So: \[ \theta_1 + \theta_2 = \theta_2 + \frac{\pi}{2} + \theta_2 = 2\theta_2 + \frac{\pi}{2} \] But for the product: \[ \arg(z\omega) = \theta_1 + \theta_2 = (\theta_2 + \frac{\pi}{2}) + \theta_2 = 2\theta_2 + \frac{\pi}{2} \] We can simplify this more directly: Let: \[ z = \frac{1}{\omega} e^{i\pi/2} \Rightarrow z\omega = e^{i\pi/2} \Rightarrow \overline{z\omega} = \overline{e^{i\pi/2}} = e^{-i\pi/2} = -i \] \[ \boxed{\overline{z\omega} = -i} \] % Tip