JEE Advanced 2019 · Paper 1, Mathematics question 1

The set. $S$ is the region on or outside the circle with centre $2-i$ and radius $\sqrt5$.

The point $z_0$. Maximising $\frac1{|z-1|}$ means finding the point of $S$ nearest to 1. The point 1 is inside the circle (its distance from the centre is $\sqrt2<\sqrt5$), so $z_0$ is on the circle along the ray from the centre through 1:
$$z_0=(2-i)+\sqrt5\cdot\frac{-1+i}{\sqrt2}.$$
With $z_0=x_0+iy_0$, $2-x_0=y_0+1=\sqrt{5/2}$.

The expression.
$$\frac{4-z_0-\bar z_0}{z_0-\bar z_0+2i}=\frac{4-2x_0}{2i(y_0+1)}=\frac{2-x_0}{i(y_0+1)}=\frac1i=-i.$$
Its principal argument is $-\frac\pi2$.

(A)