Let S be the set of all complex numbers z satisfying |z - 2 + i| ≥...

The region represented by $\left|z - 2 + i\right| \ge \sqrt{5}$ will be region outside and on circle with centre \left(2, - 1\right)$and radius$\sqrt{5} . \left|z - 2 + i\right| \ge \sqrt{5}$Let$z = x + iy ⇒ \left|x + iy - 2 + i\right| \ge \sqrt{5} ⇒ \left|\left(x - 2\right) + i\left(y + 1\right)\right| \ge \sqrt{5} ⇒ \sqrt{{\left(x - 2\right)}^{2} + {\left(y + 1\right)}^{2}} \ge \sqrt{5} ⇒ {\left(x - 2\right)}^{2} + {\left(y + 1\right)}^{2} \ge 5$Now, for$\dfrac{1}{\left|{z}_{0} - 1\right|}$to be maximum, \left|{z}_{0} - 1\right|$ must be minimum.

We need to find the $B\left({z}_{0}\right)$ which is in given region and nearest to point $A\left(1, 0\right),$ hence nearest point from $A\left(1, 0\right)$ will be on the line joining $A$ and $C .$ Method 1 Let ${z}_{0} = x + iy$, then $x < 2$ and $y > 0$ (from diagram) Consider w = \dfrac{4 - {z}_{0} - \overset{¯}{z}0}{{z}_{0} - \overset{¯}{z}0 + 2i} = \dfrac{4 - \left(x +\text{ iy }\right) - \left(x -\text{ iy }\right)}{\left(x + iy\right) - \left(x - iy\right) + 2i} = \dfrac{4 - 2x}{i\left(y + 2\right)} = \dfrac{1}{i}\left(\dfrac{4 - x}{y + 2}\right)$w = - i\left(\dfrac{4 - x}{y + 2}\right)$w = i\left(K\right),$where$K = \dfrac{4 - x}{y + 2}$As$x < 2$and$y > 0 ⇒ K > 0 w = - iK,$will lie on negative imaginary axis Arg$\left(w\right) = - \dfrac{\pi}{2}$Alternate: As$B\left({z}_{0}\right)$lies on line$AC,$Equation of$AC: \left(y + 1\right) = \dfrac{0 - \left(- 1\right)}{1 - 2}\left(x - 2\right)$⇒ y + 1 = \dfrac{1}{- 1}\left(x - 2\right)$⇒ x + y = 2 . . . . \left(i\right)$Let${z}_{0} = x + iy$Consider$w = \dfrac{4 - {z}_{0} - \overset{¯}{z}0}{{z}_{0} - \overset{¯}{z}0 + 2i} w = \dfrac{4 - x}{i\left(y + 2\right)}$From equation$\left(i\right)$w = \dfrac{4 - x}{i\left(2 - x + 2\right)}$ $w = \dfrac{1}{i}$ $w = - i$ Arg\left(w\right) = - \dfrac{\pi}{2}