Let $\mathrm{A}=\left\{\theta \in\left(-\frac{\pi}{2}, \pi\right): \frac{3+2 i \sin \theta}{1-2 i \sin \theta}\right.$ is purely imaginary $\}$. Then the sum of the elements in A is:

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Complex Numbers

Let

A = {θ \in (- \frac{π}{2}, π) : \frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}

is purely imaginary

}

. Then the sum of the elements in A is:

Answer: (D)

Multiply top and bottom by the conjugate of the denominator:

z = \frac{( 3 + 2 i s i n θ ) ( 1 + 2 i s i n θ )}{1 + 4 s i n ^{2} θ} = \frac{3 - 4 s i n ^{2} θ + 8 i s i n θ}{1 + 4 s i n ^{2} θ}

. For

z

to be purely imaginary the real part must vanish,

3 - 4 sin^{2} θ = 0

, so

sin θ = \pm \frac{3}{2}

. Inside

(- \frac{π}{2}, π)

the solutions are

- \frac{π}{3}, \frac{π}{3}, \frac{2 π}{3}

, and their sum is

\frac{2 π}{3}

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option