Question

The real value of $\alpha$ for which $\frac{1 - i \sin \alpha}{1 + 2i \sin \alpha}$ is purely real is:

• A. $(n + 1)\frac{\pi}{2}, n \in \mathbb{N}$
• B. $(2n + 1)\frac{\pi}{2}, n \in \mathbb{N}$
• C. $n\pi, n \in \mathbb{N}$
• D. $(2n - 1)\frac{\pi}{2}, n \in \mathbb{N}$

Answer 1

The Correct Option is C

For the fraction to be purely real, the imaginary part of the complex number must be zero.

Let $\sin \alpha = k$: $\frac{1 - ik}{1 + 2ik}$ must be purely real. 

Rationalizing, we get: $\text{Imaginary part} = 0 \implies k = 0$. 

Thus, $\sin \alpha = 0 \implies \alpha = n\pi, n \in \mathbb{N}$.