Question

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

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

Hint:

A complex number \(Z = a+ib\) is real if its imaginary part \(b=0\).
To find the real and imaginary parts of a fraction \(\frac{N_1+iN_2}{D_1+iD_2}\), multiply the numerator and denominator by the conjugate of the denominator (\(D_1-iD_2\)).
General solution for \(\sin\theta = 0\) is \(\theta = n\pi, n \in Z\).

Answer 1

The Correct Option is D

Let \(Z = \frac{3+2i\sin\theta}{1-2i\sin\theta}\). For Z to be real, its imaginary part must be zero. Multiply the numerator and denominator by the conjugate of the denominator: Conjugate of \(1-2i\sin\theta\) is \(1+2i\sin\theta\). \(Z = \frac{(3+2i\sin\theta)(1+2i\sin\theta)}{(1-2i\sin\theta)(1+2i\sin\theta)}\) Denominator = \(1^2 - (2i\sin\theta)^2 = 1 - (4i^2\sin^2\theta) = 1 - (-4\sin^2\theta) = 1+4\sin^2\theta\). Numerator = \(3(1+2i\sin\theta) + 2i\sin\theta(1+2i\sin\theta)\) \( = 3 + 6i\sin\theta + 2i\sin\theta + 4i^2\sin^2\theta \) \( = 3 + 8i\sin\theta - 4\sin^2\theta \) \( = (3-4\sin^2\theta) + i(8\sin\theta) \). So, \(Z = \frac{(3-4\sin^2\theta) + i(8\sin\theta)}{1+4\sin^2\theta} = \frac{3-4\sin^2\theta}{1+4\sin^2\theta} + i \frac{8\sin\theta}{1+4\sin^2\theta}\). For Z to be real, the imaginary part must be zero: \( \text{Im}(Z) = \frac{8\sin\theta}{1+4\sin^2\theta} = 0 \). Since \(1+4\sin^2\theta\) is always positive (as \(\sin^2\theta \ge 0\)), we must have the numerator equal to zero: \(8\sin\theta = 0 \Rightarrow \sin\theta = 0\). The general solution for \(\sin\theta = 0\) is \(\theta = n\pi\), where \(n\) is an integer (\(n \in Z\)). This matches option (d). \[ \boxed{\theta = n\pi \text{ for } n \in Z} \]