Question

$z = \frac{3+2i \sin \theta}{1-2i \sin \theta}, (i = \sqrt{-1})$ will be purely imaginary if $\theta =$

• A. $2n\pi \pm \frac{\pi}{8}, \text{where } n \in \mathbb{Z}$
• B. $n\pi + \frac{\pi}{8}, \text{where } n \in \mathbb{Z}$
• C. $n\pi \pm \frac{\pi}{3}, \text{where } n \in \mathbb{Z}$
• D. $n\pi, \text{where } n \in \mathbb{Z}$

Hint:

Purely imaginary $\implies \text{Re}(z) = 0$; Purely real $\implies \text{Im}(z) = 0$.

Answer 1

The Correct Option is C

Step 1: Rationalize
$z = \frac{(3+2i \sin \theta)(1+2i \sin \theta)}{1+4 \sin^2 \theta} = \frac{3 + 6i \sin \theta + 2i \sin \theta - 4 \sin^2 \theta}{1+4 \sin^2 \theta}$.
Step 2: Purely Imaginary Condition
$\text{Real part} = 0 \implies 3 - 4 \sin^2 \theta = 0$.
Step 3: Solve for $\theta$
$\sin^2 \theta = \frac{3}{4} \implies \sin \theta = \pm \frac{\sqrt{3}}{2}$.
$\theta = n\pi \pm \frac{\pi}{3}$.
Final Answer: (C)