Let S be the set of all complex numbers z satisfying z - 2 + i ≥ 5 . If the complex number z 0 is such…

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 $z=x+iy$ $\Rightarrow \left|x+iy-2+i\right|\ge \sqrt{5}$ $\Rightarrow \left|\left(x-2\right)+i\left(y+1\right)\right|\ge \sqrt{5}$ $\Rightarrow \sqrt{{\left(x-2\right)}^2+{\left(y+1\right)}^2}\ge \sqrt{5}$ $\Rightarrow {\left(x-2\right)}^2+{\left(y+1\right)}^2\ge 5$ Now, for $\frac{1}{\left|z_0-1\right|}$ 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 $w=\frac{4-z_0-{\overline{z}}_0}{z_0-{\overline{z}}_0+2i}=\frac{4-\left(x+iy\right)-\left(x-iy\right)}{\left(x+iy\right)-\left(x-iy\right)+2i}=\frac{4-2x}{i\left(y+2\right)}=\frac{1}{i}\left(\frac{4-x}{y+2}\right)$ $w=-i\left(\frac{4-x}{y+2}\right)$ $w=i\left(K\right),$ where $K=\frac{4-x}{y+2}$ As $x<2$ and $y>0\Rightarrow K>0$ $w=-iK,$ will lie on negative imaginary axis $Arg\left(w\right)=-\frac{\pi }{2}$ Alternate: As $B\left(z_0\right)$ lies on line $AC,$ Equation of $AC$: $\left(y+1\right)=\frac{0-\left(-1\right)}{1-2}\left(x-2\right)$ $\Rightarrow y+1=\frac{1}{-1}\left(x-2\right)$ $\Rightarrow x+y=2 ....\left(\mathrm{i}\right)$ Let $z_0=x+iy$ Consider $w=\frac{4-z_0-{\overline{z}}_0}{z_0-{\overline{z}}_0+2i}$ $w=\frac{4-x}{i\left(y+2\right)}$ From equation $\left(\mathrm{i}\right)$ $w=\frac{4-x}{i\left(2-x+2\right)}$ $w=\frac{1}{i}$ $w=-i$ $Arg\left(w\right)=-\frac{\pi }{2}$