JEE PYQ: Complex Numbers Question 35
Question 35 - 2019 (09 Jan Shift 1)
Let $A = \left{\theta \in \left(-\frac{\pi}{2}, \pi\right) : \frac{3+2i\sin\theta}{1-2i\sin\theta} \text{ is purely imaginary}\right}$. Then the sum of the elements in $A$ is:
(1) $\frac{5\pi}{6}$
(2) $\pi$
(3) $\frac{3\pi}{4}$
(4) $\frac{2\pi}{3}$
Show Answer
Answer: (4)
Solution
$\frac{(3+2i\sin\theta)(1+2i\sin\theta)}{1+4\sin^2\theta}$. Real part $= \frac{3-4\sin^2\theta}{1+4\sin^2\theta} = 0 \Rightarrow \sin^2\theta = \frac{3}{4} \Rightarrow \sin\theta = \pm\frac{\sqrt{3}}{2}$. In $(-\frac{\pi}{2}, \pi)$: $\theta = \frac{\pi}{3}, \frac{2\pi}{3}, -\frac{\pi}{3}$. Sum $= \frac{\pi}{3} + \frac{2\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3}$.
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