JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
Let $A=\left\{\theta\in(0,2\pi):\frac{1+2i\sin\theta}{1-i\sin\theta}\text{ is purely imaginary}\right\}$. Then the sum of the elements in $A$ is
Complex Numbers
Solution
$\frac{1+2is}{1-is}=\frac{(1+2is)(1+is)}{1+s^2}=\frac{1-2s^2+3is}{1+s^2}$, $s=\sin\theta$. Purely imaginary: $1-2s^2=0$ (and $s\ne0$) so $\sin\theta=\pm\frac1{\sqrt2}$. $\theta=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4$, sum $=4\pi$. Option C.