<p>Find the values of \ \(\theta\) if \ \(\dfrac{3 + 2i\sin\theta}{1 - 2i\sin\theta}\) is purely real or purely imaginary.</p>
Step-by-Step Solution
Key Concept: A complex number is purely real when its imaginary part equals zero, and purely imaginary when its real part equals zero. Rationalize by multiplying by the conjugate of the denominator, then separate real and imaginary parts.
<p><strong>Step 1:</strong> Multiply numerator and denominator by the conjugate of the denominator (1 + 2i sin θ):</p><p>$$\frac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \cdot \frac{1 + 2i\sin\theta}{1 + 2i\sin\theta}$$</p><p><strong>Step 2:</strong> Denominator becomes: $(1 - 2i\sin\theta)(1 + 2i\sin\theta) = 1 + 4\sin^2\theta$</p><p><strong>Step 3:</strong> Numerator becomes: $(3 + 2i\sin\theta)(1 + 2i\sin\theta) = 3 + 6i\sin\theta + 2i\sin\theta + 4i^2\sin^2\theta$</p><p>$$= 3 + 8i\sin\theta - 4\sin^2\theta = (3 - 4\sin^2\theta) + i(8\sin\theta)$$</p><p><strong>Step 4:</strong> The expression is: $$\frac{3 - 4\sin^2\theta + 8i\sin\theta}{1 + 4\sin^2\theta} = \frac{3 - 4\sin^2\theta}{1 + 4\sin^2\theta} + i\frac{8\sin\theta}{1 + 4\sin^2\theta}$$</p><p><strong>For purely real:</strong> Imaginary part = 0 → $8\sin\theta = 0$ → $\sin\theta = 0$ → $\theta = n\pi$</p><p><strong>For purely imaginary:</strong> Real part = 0 → $3 - 4\sin^2\theta = 0$ → $\sin^2\theta = \frac{3}{4}$ → $\sin\theta = \pm\frac{\sqrt{3}}{2}$ → $\theta = n\pi \pm \frac{\pi}{3}$</p><p>∴ <strong>Answer:</strong> Purely real: $\theta = n\pi$; Purely imaginary: $\theta = n\pi \pm \frac{\pi}{3}$</p>
Correct Answer: Purely real: θ = nπ; Purely imaginary: θ = nπ ± π/3