Let $S$ be the set of all complex numbers $z$ satisfying $\vert z-2+i\vert\ge\sqrt 5.$If the complex number $z_0$ is such that $\frac{1}{\vert z_0-1\vert}$ is the maximum of the set $\{\frac{1}{\vert z-1\vert}:z\in S\}$,then the principal argument of $\frac{4-z_0-\bar z_0}{z_0-\bar z_0+2i}$ is -
(A)$\frac{-\pi}{2}$
(B)$\frac{\pi}{4}$
(C)$\frac{\pi}{2}$
(D)$\frac{3\pi}{4}$
Solution
- $\vert z-2+i\vert\ge \sqrt 5\implies \vert z-(2-i)\vert\implies \vert (x+iy)-(2-i)\vert\ge \sqrt 5\implies (x-2)^2+(y+1)^2\ge \sqrt 5\implies (x-2)^2+(y+1)^2\ge (5)^{1/4}$
- $\frac{1}{\vert z_0-1\vert}$ is the maximum of the set $\{\frac{1}{\vert z-1\vert}:z\in S\}$ $\implies \frac{1}{\vert z-1\vert}\le \frac{1}{\vert z_0-1\vert}\implies \frac{1}{(x-1)^2+y^2}\le \frac{1}{(x_0-1)^2+y_0^{2}}$
From here i'm not getting how to proceed further. Please provide some hint. This question was asked in JEE(ADVANCED)$2019$ PAPER 1 is Mathematics section. Thanks!!
The center of the circle is at $2 - i$. The point on the circle closest to $1$ lies on the line through $2 - i$ and $1$, therefore $\operatorname{Im} z_0 = 1 - \operatorname{Re} z_0$ and $$\frac {4 - z_0 - \overline {z_0}} {z_0 - \overline {z_0} + 2 i} = -i.$$