Complex Numbers
Purely Imaginary Complex Numbers
Grade 11
Question:
<p>A value of <span>\(\theta\)</span> for which <span>\(\dfrac{2+3i\sin\theta}{1-2i\sin\theta}\)</span> is purely imaginary is</p>
<p>\(\sin^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\sin^{-1}\left(\dfrac{\sqrt{3}}{4}\right)\)</p>
Step-by-Step Solution
Key Concept: A complex number is purely imaginary when its real part equals zero and imaginary part is non-zero. Multiply numerator and denominator by the conjugate of the denominator, then equate the real part to zero.
<p><strong>Step 1:</strong> Multiply numerator and denominator by the conjugate of 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 denominator: $(1-2i\sin\theta)(1+2i\sin\theta) = 1 + 4\sin^2\theta$</p><p><strong>Step 3:</strong> Expand numerator: $(2+3i\sin\theta)(1+2i\sin\theta) = 2 + 4i\sin\theta + 3i\sin\theta + 6i^2\sin^2\theta$</p><p>$= 2 + 7i\sin\theta - 6\sin^2\theta = (2-6\sin^2\theta) + i(7\sin\theta)$</p><p><strong>Step 4:</strong> The complex number becomes: $$\frac{(2-6\sin^2\theta) + i(7\sin\theta)}{1+4\sin^2\theta}$$</p><p><strong>Step 5:</strong> For purely imaginary, real part = 0:</p><p>$$\frac{2-6\sin^2\theta}{1+4\sin^2\theta} = 0$$</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: A (θ = sin⁻¹(±1/√3) or equivalent)</p>
Correct Answer: A