Question

If \(z_1\) and \(z_2\) are two complex numbers such that \[ |z_1-a|=|z_2-a| \] for \(a\in\mathbb R\), and \[ Arg(z_1-a)+Arg(z_2-a)=\frac{\pi}{2}, \] then \[ \frac{z_1-a}{z_2-a}= \]

• A. \(a\)
• B. \(i\)
• C. \(-i\)
• D. \(ia\)

Hint:

For complex numbers written as \(re^{i\theta}\), division preserves modulus ratio and subtracts arguments.

Answer 1

The Correct Option is B

Step 1: Represent the numbers in polar form.
Let \[ z_1-a=re^{i\theta_1} \] and \[ z_2-a=re^{i\theta_2} \] since their moduli are equal. Also \[ \theta_1+\theta_2=\frac{\pi}{2} \]

Step 2: Find the ratio.
\[ \frac{z_1-a}{z_2-a} = e^{i(\theta_1-\theta_2)} \] Using symmetry with equal modulus and angle sum \(\frac{\pi}{2}\), \[ \theta_1-\theta_2=\frac{\pi}{2} \] Therefore \[ \frac{z_1-a}{z_2-a} = e^{i\pi/2} = i \] \[ \boxed{i} \]