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

Let the given expression be denoted by \( z \):

\[ z = \frac{1 - i \sin \alpha}{1 + 2i \sin \alpha} \]

For \( z \) to be purely real, its imaginary part must be zero. We can find the real and imaginary parts by multiplying the numerator and denominator by the conjugate of the denominator:

\[ z = \frac{(1 - i \sin \alpha)(1 - 2i \sin \alpha)}{(1 + 2i \sin \alpha)(1 - 2i \sin \alpha)} = \frac{1 - 2i \sin \alpha - i \sin \alpha + 2i^2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} = \frac{1 - 3i \sin \alpha - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} \]

Separating the real and imaginary parts:

• Real part = \( \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} \)
• Imaginary part = \( \frac{-3 \sin \alpha}{1 + 4 \sin^2 \alpha} \)

For \( z \) to be purely real, the imaginary part must be zero:

\[ \frac{-3 \sin \alpha}{1 + 4 \sin^2 \alpha} = 0 \]

This implies that the numerator must be zero:

\[ -3 \sin \alpha = 0 \] \[ \sin \alpha = 0 \]

The general solution for \( \sin \alpha = 0 \) is given by:

\[ \alpha = n\pi \]

where \( n \) is an integer (\( n \in \mathbb{Z} \)).

Therefore, the real values of \( \alpha \) for which the given expression is purely real are \( \alpha = n\pi \), where \( n \) is an integer.

Final Answer: The final answer is \( {n\pi} \).