JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Complex Numbers

Writing $z=x+iy$ gives $z(1+i)+\bar{z}(1-i)=2(x-y)$, so $S_2$ is the half-plane $x-y\ge 2$, while $S_1$ is the closed disc of radius $1$ centred at $(2,0)$. The line $x-y=2$ passes through that centre, so $S_1\cap S_2$ is a half-disc. The square of a distance is a convex function, so over this convex region it is largest at a point of the circular boundary.
A point of the circle is $(2+\cos \theta ,\sin \theta )$, and it satisfies $x-y\ge 2$ when $\cos \theta -\sin \theta \ge 0$, that is $-\frac{3\pi }{4}\le \theta \le \frac{\pi }{4}$. For such a point,

This is largest where $\cos \theta$ is least on that range, namely at $\theta =-\frac{3\pi }{4}$ with $\cos \theta =-\frac{1}{\sqrt{2}}$. The maximum value is