JEE Main 2023 (Online) 13th April Evening ShiftQuestion 1
Let $S=\left\{z\in\mathbb{C}:\bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}$. Then $\sum_{z\in S}|z|^{2}$ is equal to
Complex Numbers
Solution
With $z=x+iy$: $x-iy=i(x^2-y^2+x)-2xy$. Real: $x=-2xy$; imaginary: $-y=x^2-y^2+x$. $x=0$: $y^2=y\Rightarrow z=0,\ i$. $y=-\frac12$: $x^2+x-\frac34=0\Rightarrow x=\frac12,-\frac32$. $\sum|z|^2=0+1+\frac12+\frac52=4$. Option D.