JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Complex Numbers

Write $Z=x+iy$, so $\bar{Z}=x-iy$, $Z^2=x^2-y^2+2ixy$ and $\mathrm{Re}(\bar{Z})=x$. The condition $\bar{Z}=i(Z^2+\mathrm{Re}(\bar{Z}))$ becomes $x-iy=-2xy+i(x^2-y^2+x)$. Comparing real parts, $x=-2xy$, so $x=0$ or $y=-\frac{1}{2}$; comparing imaginary parts, $-y=x^2-y^2+x$. If $x=0$, then $-y=-y^2$ gives $y=0$ or $y=1$, yielding $Z=0$ and $Z=i$. If $y=-\frac{1}{2}$, then $x^2+x-\frac{3}{4}=0$ gives $x=\frac{1}{2}$ or $x=-\frac{3}{2}$, yielding $Z=\frac{1}{2}-\frac{i}{2}$ and $Z=-\frac{3}{2}-\frac{i}{2}$. The values of $|Z{|}^2$ are $0,1,\frac{1}{2},\frac{10}{4}$, whose sum is ${\sum }_{z\in S}|z{|}^2=0+1+\frac{1}{2}+\frac{5}{2}=4$.