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

1. Rationalize the denominator:

To determine when the expression is purely real, we need to eliminate the imaginary part from the denominator. We do this by multiplying both the numerator and the denominator by the complex conjugate of the denominator:

$\frac{1 - i \sin \alpha}{1 + 2i \sin \alpha} \cdot \frac{1 - 2i \sin \alpha}{1 - 2i \sin \alpha}$

2. Expand and Simplify:

Expanding the numerator and the denominator:

Numerator: $(1 - i \sin \alpha)(1 - 2i \sin \alpha) = 1 - 2i \sin \alpha - i \sin \alpha + 2i^2 \sin^2 \alpha = 1 - 3i \sin \alpha - 2 \sin^2 \alpha$

Denominator: $(1 + 2i \sin \alpha)(1 - 2i \sin \alpha) = 1 - (2i \sin \alpha)^2 = 1 + 4 \sin^2 \alpha$

3. Combine and Separate Real and Imaginary Parts:

Therefore, the expression becomes:

$\frac{1 - 2 \sin^2 \alpha - 3i \sin \alpha}{1 + 4 \sin^2 \alpha} = \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} - i \frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha}$

4. Condition for Purely Real:

For the expression to be purely real, the imaginary part must be zero. This means:

$\frac{3 \sin \alpha}{1 + 4 \sin^2 \alpha} = 0$

5. Solve for α:

This implies that $\sin \alpha = 0$. The solutions to this equation are:

$\alpha = n\pi$, where n is an integer.

6. Consider the domain (n ∈ N):

Since n ∈ N (natural numbers), the solution remains $\alpha = n\pi$.

Conclusion: The real value of 'α' for which the given expression is purely real is $\alpha = n\pi$, where n is a natural number.

Correct Answer: (C) $n\pi$, n ∈ N

Answer 2

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