Let $S$ be the set of all complex numbers $z$ satisfying $|z - 2 + i| \geq \sqrt{5}$. If the complex number $z_0$ is such that $\frac{1}{|z_0 - 1|}$ is the maximum of the set $\left\{ \frac{1}{|z - 1|} : z \in S \right\}$, then the principal argument of $\frac{4 - z_0 - \overline{z_0}}{z_0 - \overline{z_0} + 2i}$ is

$S = \{ z : |z - 2 + i| \geq \sqrt{5} \}$ $\dfrac{1}{|z_0 - 1|} = \max\left\{ \dfrac{1}{|z - 1|} : z \in S \right\}$ $\text{Principal argument of } \dfrac{4 - z_0 - \overline{z_0}}{z_0 - \overline{z_0} + 2i}$ $|z - (2 - i)| \geq \sqrt{5}$ $\text{Center } C = 2 - i, \quad \text{radius } \sqrt{5}$ $\text{maximise } \dfrac{1}{|z - 1|} \iff \text{minimise } |z - 1|$ $1.\ \text{identify the region } S$ $2.\ \text{find } z_0 \text{ on the boundary closest to } 1$ $3.\ \text{compute the expression and its argument}$ $|A - C| = |1 - (2 - i)| = |-1 + i| = \sqrt{2}$ $\sqrt{2} < \sqrt{5} \ \Rightarrow \ 1 \text{ lies inside the circle}$ $z_0 = C + \sqrt{5}\,\dfrac{A - C}{|A - C|}$ $z_0 = (2 - i) + \sqrt{5}\,\dfrac{-1 + i}{\sqrt{2}}$ $z_0 = \left(2 - \sqrt{\tfrac{5}{2}}\right) + i\!\left(\sqrt{\tfrac{5}{2}} - 1\right)$ $z_0 = x_0 + i y_0, \quad x_0 = 2 - \sqrt{\tfrac{5}{2}}, \quad y_0 = \sqrt{\tfrac{5}{2}} - 1$ $z_0 + \overline{z_0} = 2x_0, \quad z_0 - \overline{z_0} = 2i y_0$ $4 - z_0 - \overline{z_0} = 4 - 2x_0$ $= 4 - 2\!\left(2 - \sqrt{\tfrac{5}{2}}\right) = 2\sqrt{\tfrac{5}{2}} = \sqrt{10}$ $z_0 - \overline{z_0} + 2i = 2i y_0 + 2i = 2i(y_0 + 1)$ $= 2i\sqrt{\tfrac{5}{2}} = i\sqrt{10}$ $\dfrac{4 - z_0 - \overline{z_0}}{z_0 - \overline{z_0} + 2i} = \dfrac{\sqrt{10}}{i\sqrt{10}}$ $= \dfrac{1}{i} = -i$ $-i = 0 + (-1)i \ \Rightarrow \ \arg(-i) = -\dfrac{\pi}{2}$ $\boxed{\arg\!\left(\dfrac{4 - z_0 - \overline{z_0}}{z_0 - \overline{z_0} + 2i}\right) = -\dfrac{\pi}{2}}$ $\text{Answer: option (4)}$