JEE Main 2026 (Online) 6th April Evening ShiftQuestion 1
Let $S=\{z\in\mathbb{C}:z^{2}+\sqrt{6}\,iz-3=0\}$. Then $\sum_{z\in S}z^{8}$ is equal to :
Complex Numbers
Solution
$z=\frac{-\sqrt6 i\pm\sqrt{-6+12}}{2}=\frac{\sqrt6}{2}(\pm1-i)=\sqrt3e^{-i\pi/4},\ \sqrt3e^{-3i\pi/4}$. $z^8=81e^{-2\pi i}=81$ and $81e^{-6\pi i}=81$. Sum $=162$. Option (A).