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: Understanding the Concept:
A complex number $z$ is purely imaginary if its Real part is zero.
We must rationalize the denominator to separate the real and imaginary parts.
Step 2: Key Formula or Approach:
For $z = \frac{a+bi}{c+di}$, $\text{Re}(z) = \frac{ac+bd}{c^2+d^2}$.
Here $a=3, b=2\sin\theta, c=1, d=-2\sin\theta$.
Step 3: Detailed Explanation:
$\text{Re}(z) = \frac{3(1) + (2\sin\theta)(-2\sin\theta)}{1^2 + (-2\sin\theta)^2} = \frac{3 - 4\sin^2\theta}{1 + 4\sin^2\theta}$.
For purely imaginary: $3 - 4\sin^2\theta = 0 \implies \sin^2\theta = 3/4$.
$\sin^2\theta = \sin^2(\pi/3) \implies \theta = n\pi \pm \pi/3$.
Step 4: Final Answer:
The solution is $\theta = n\pi \pm \frac{\pi}{3}$.