JEE Main 2026 (Online) 23rd January Evening ShiftQuestion 1
If $z=\frac{\sqrt3}{2}+\frac i2$, $i=\sqrt{-1}$, then $\left(z^{201}-i\right)^8$ is equal to :
Complex Numbers
Solution
$z=e^{i\pi/6}$, $z^{201}=e^{i\,33.5\pi}=e^{i\pi/2}\cdot e^{33\pi i}=-i$. So $(z^{201}-i)^8=(-2i)^8=256$.