JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Complex Numbers

Writing $z=x+iy$ and clearing the denominator, the imaginary part of $\frac{2z-3i}{4z+2i}$ is proportional to $-16x$, so it vanishes exactly when $x=0$, provided $4y+2\ne 0$, i.e. $y\ne -\frac{1}{2}$. Thus $S$ is the set of points with $x=0$ and $y\ne -\frac{1}{2}$; this makes $x=0$, the stated range of $y$, and $y+x^2+y^2\ne -\frac{1}{4}$ all hold. The point $\left(0,-\frac{1}{2}\right)$ carries the excluded value $y=-\frac{1}{2}$ and so does not lie in $S$, which is the statement that is not correct.