JEE Advanced 2020 · Paper 2, Mathematics question 1

Simplify. Write $z=re^{i\theta}$ with $z\ne0$ ($z=0$ has real part 0, so it cannot be $z_1$ or $z_2$):
$$r^4(e^{4i\theta}-1)=4ir^2e^{2i\theta}\implies r^2e^{2i\theta}(e^{2i\theta}-e^{-2i\theta})=4ie^{2i\theta}\implies r^2\sin2\theta=2.$$
With $z=x+iy$ this says $2xy=2$, i.e. $xy=1$.

The set. $S\setminus\{0\}$ is the hyperbola $xy=1$. Points with $\operatorname{Re}z>0$ lie on the first-quadrant branch, and those with $\operatorname{Re}z<0$ on the third-quadrant branch.

Closest points. The two branches are nearest at $(1,1)$ and $(-1,-1)$:
$$|z_1-z_2|^2=2^2+2^2=8.$$
Answer: 8