Complex Numbers
Algebra of Complex Numbers
Grade Class 11
Question:
<p>The real value of \(\alpha\) for which \(\dfrac{1-i\sin\alpha}{1+2i\sin\alpha}\) is purely real is (\(n\in\mathbb{Z}\)):</p>
(n+1)\pi/2
n\pi
(2n+1)\pi/2
(2n+1)\pi
Step-by-Step Solution
Key Concept: Multiply by conjugate of denominator; set Im = 0. Im part contains sin \alpha (2 - 1) = sin \alpha. So Im = 0 \Rightarrow sin \alpha = 0 \Rightarrow \alpha = n\pi.
<p>$\dfrac{1-i\sin\alpha}{1+2i\sin\alpha}\times\dfrac{1-2i\sin\alpha}{1-2i\sin\alpha}=\dfrac{(1-i\sin\alpha)(1-2i\sin\alpha)}{1+4\sin^2\alpha}$. Numerator Im part: $-\sin\alpha - 2\sin\alpha = -3\sin\alpha$. Set to 0: $\sin\alpha=0\Rightarrow\alpha=n\pi$. ✓</p>
Correct Answer: B