JEE Main 2013 (Online) 23rd AprilQuestion 1
Let $a=\operatorname{Im}\left(\frac{1+z^{2}}{2iz}\right)$, where $z$ is any non-zero complex number. The set $A=\{a:|z|=1\text{ and }z\neq\pm1\}$ is equal to :
Complex Numbers
Solution
$z=e^{i\theta}$: $\frac{1+z^2}{2iz}=\frac{z+\bar z}{2i}=-i\cos\theta$, so $a=-\cos\theta$ with $\theta\ne0,\pi$. Hence $A=(-1,1)$.