Let $\mathrm{A}=\left\{\theta \in(0,2 \pi): \frac{1+2 \mathrm{i} \sin \theta}{1-\mathrm{i} \sin \theta}\right.$ is purely imaginary $\}$. Then the sum of the elements in A is

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Complex Numbers

Let

A = {θ \in (0, 2 π) : \frac{1 + 2 i s i n θ}{1 - i s i n θ}

is purely imaginary

}

.

Then the sum of the elements in A is

Answer: (C)

Let

z = \frac{1 + 2 i s i n θ}{1 - i s i n θ}

. Multiply numerator and denominator by the conjugate:

z = \frac{( 1 + 2 i sin θ ) ( 1 + i sin θ )}{1 + sin ^{2} θ} = \frac{1 - 2 sin ^{2} θ + i ( 3 sin θ )}{1 + sin ^{2} θ}

For

z

to be purely imaginary, the real part must be zero:

1 - 2 sin^{2} θ = 0

sin^{2} θ = \frac{1}{2}

sin θ = \pm \frac{1}{2}

(0, 2 π)

θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}

Sum of elements:

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

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option