JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Complex Numbers

The condition $|z+i|=|z-1|$ is the perpendicular bisector of $-i$ and $1$, which simplifies to imaginary part equal to minus the real part. The condition $|z-1+i|\ge |z|$ simplifies to $y\ge x-1$, and $|z|<2$ is the open disc of radius $2$. For $w=2x+iy$ to belong to $S$, its real part $2x$ and imaginary part $y$ must meet all three: the bisector gives $y=-2x$; the inequality $y\ge 2x-1$ then forces $-2x\ge 2x-1$, i.e. $x\le \frac{1}{4}$; and $(2x{)}^2+y^2<4$ with $y=-2x$ gives $8x^2<4$, i.e. $-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$. Intersecting these, the admissible real parts form $\left(-\frac{1}{\sqrt{2}},\frac{1}{4}\right]$.