Question

The values of \(\theta\), for which \[ \frac{3+2i\sin\theta}{1-2i\sin\theta} \] is real are

• A. \(\theta=n\pi+\dfrac{\pi}{3}\) for \(n\in \mathbb{Z}\)
• B. \(\theta=n\pi+\dfrac{\pi}{4}\) for \(n\in \mathbb{Z}\)
• C. \(\theta=n\pi+\dfrac{\pi}{2}\) for \(n\in \mathbb{Z}\)
• D. \(\theta=n\pi\) for \(n\in \mathbb{Z}\)

Hint:

To check when a complex expression is real, rationalize the denominator and make the imaginary part equal to zero.

Answer 1

The Correct Option is D

Step 1: Rationalize the expression.
\[\frac{3+2i\sin\theta}{1-2i\sin\theta} \cdot \frac{1+2i\sin\theta}{1+2i\sin\theta} = \frac{(3+2i\sin\theta)(1+2i\sin\theta)}{1+4\sin^2\theta}.\] The numerator expands to \((3-4\sin^2\theta) + 8i\sin\theta\).

Step 2: Set imaginary part to zero.
For the expression to be real: \(8\sin\theta = 0 \Rightarrow \sin\theta = 0 \Rightarrow \theta = n\pi\).
\[\boxed{\theta = n\pi}\]