JEE Main 2016 (Offline)Question 1
A value of $\theta$ for which $\frac{2+3i\sin\theta}{1-2i\sin\theta}$ is purely imaginary, is
Complex Numbers
Solution
$\frac{(2+3is)(1+2is)}{1+4s^2}$ has real part $\frac{2-6s^2}{1+4s^2}=0$, so $\sin^2\theta=\frac13$, $\theta=\sin^{-1}\frac1{\sqrt3}$.