Complex Numbers
Purely Imaginary Numbers
Grade None
Question:
<p>A value of \(\theta\) for which \(\dfrac{2 + 3i\sin\theta}{1 - 2i\sin\theta}\) is purely imaginary, is</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\sin^{-1}\left(\dfrac{\sqrt{3}}{4}\right)\)</p>
<p>\(\sin^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)</p>
<p>\(\dfrac{\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: For a complex number to be purely imaginary, its real part must equal zero. Multiply numerator and denominator by the conjugate of the denominator, then set the real part of the resulting expression to zero.
<p><strong>Step 1:</strong> Multiply numerator and denominator by the conjugate of the denominator (1 + 2i sin θ):</p><p>$$\frac{2 + 3i\sin\theta}{1 - 2i\sin\theta} \cdot \frac{1 + 2i\sin\theta}{1 + 2i\sin\theta}$$</p><p><strong>Step 2:</strong> Expand the denominator:</p><p>$$= \frac{(2 + 3i\sin\theta)(1 + 2i\sin\theta)}{1 + 4\sin^2\theta}$$</p><p><strong>Step 3:</strong> Expand the numerator:</p><p>$$(2 + 3i\sin\theta)(1 + 2i\sin\theta) = 2 + 4i\sin\theta + 3i\sin\theta + 6i^2\sin^2\theta$$</p><p>$$= 2 + 7i\sin\theta - 6\sin^2\theta = (2 - 6\sin^2\theta) + 7i\sin\theta$$</p><p><strong>Step 4:</strong> The result is purely imaginary when the real part equals zero:</p><p>$$2 - 6\sin^2\theta = 0$$</p><p>$$\sin^2\theta = \frac{1}{3}$$</p><p>$$\sin\theta = \pm\frac{1}{\sqrt{3}}$$</p><p>∴ Answer: C (θ = arcsin(±1/√3) or equivalent form)</p>
Correct Answer: C