JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i sin θ}{1 - i sin θ} is purely imaginary}$ , then the sum of the elements in $A$ is

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Explanation

Here,
$z = \frac{1 + 2 i sin θ}{1 - i sin θ} \times \frac{1 + i sin θ}{1 + i sin θ}$

$= \frac{1 + i sin θ + 2 i sin θ - 2 sin ^{2} θ}{<} b r > 1 - i^{2} sin^{2} θ$

$= \frac{( 1 - 2 sin ^{2} θ ) + i ( 3 sin θ )}{<} b r > 1 + sin^{2} θ$

$∵ z$ is purely imaginary, so $ℜ (z) = 0$

$\Rightarrow \frac{1 - 2 sin ^{2} θ}{1 + sin ^{2} θ} = 0$

$\Rightarrow 1 - 2 sin^{2} θ = 0$

$\Rightarrow 2 sin^{2} θ = 1 \Rightarrow sin^{2} θ = \frac{1}{2}$

$\Rightarrow sin θ = \pm \frac{1}{2}$

$∴ A = {\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}}, θ \in (0, 2 π)$

$\Rightarrow Sum = \frac{π + 3 π + 5 π + 7 π}{4} = \frac{16 π}{4} = 4 π$

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