JEE Main 2025 Apr 7 Shift 2, Mathematics Q11: Introduction to Complex Numbers And Their Conjugates

Let $w=\frac{z-1}{2z+i}$. Since $\overline{w}=\frac{\bar{z}-1}{2\bar{z}-i}$ and $\text{Re}(\overline{w})=\text{Re}(w)$ for any complex number, the given condition simplifies to $2\text{Re}(w)=2$, i.e. $\text{Re}(w)=1$.

Let $z=x+iy$. Then $w=\frac{(x-1)+iy}{2x+i(2y+1)}$; multiplying numerator and denominator by the conjugate of the denominator gives $\text{Re}(w)=\frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1{)}^2}$.

Setting this equal to 1 and simplifying: $2x(x-1)+y(2y+1)=4x^2+(2y+1{)}^2\Rightarrow 2x^2+2y^2+2x+3y+1=0\Rightarrow x^2+y^2+x+\frac{3}{2}y+\frac{1}{2}=0$.

Completing the square: ${\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{3}{4}\right)}^2=\frac{5}{16}$, so the centre is $(a,b)=\left(-\frac{1}{2},-\frac{3}{4}\right)$ and $r^2=\frac{5}{16}$.

$\frac{15ab}{r^2}=\frac{15\left(-\frac{1}{2}\right)\left(-\frac{3}{4}\right)}{5/16}=\frac{45/8}{5/16}=18$.