Complex Numbers
Real and Imaginary Parts
Grade 11
Question:
<p>The real value of <em>θ</em> for which \(\operatorname{Re}\left(\dfrac{2+3i\sin\theta}{1-2i\sin\theta}\right)=0\) is:</p>
<p>\(\theta = \sin^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\)</p>
<p>\(\theta = \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)\)</p>
<p>\(\theta = \sin^{-1}\left(\dfrac{1}{2}\right)\)</p>
<p>\(\theta = \sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: To find when the real part of a complex fraction is zero, multiply numerator and denominator by the conjugate of the denominator, then set the real part of the resulting numerator equal 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> Compute the denominator:</p><p>$$(1-2i\sin\theta)(1+2i\sin\theta) = 1 + 4\sin^2\theta$$</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 complex fraction becomes:</p><p>$$\frac{(2-6\sin^2\theta) + 7i\sin\theta}{1+4\sin^2\theta}$$</p><p><strong>Step 5:</strong> For Re(z) = 0, set the real part of numerator equal to 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: A</p>
Correct Answer: A