Question

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

• A. $-\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $-\frac{3\pi }{4}$

Answer 1

Answer: (A)

$\left|z-2+i\right|\ge \sqrt{5}$
$\left|\frac{1}{z_0-1}\right|$ is maximum
when $|z_0-1|$ is minimum
Let $z_0=x+iy$
$x<1$ and $y>0$
$\frac{4-z_0-\overline{z_0}}{z_0-\overline{z_0+2i}}$
$=\frac{4-x-iy-x-iy}{x+iy-x+iy+2i}$
$=\frac{4-2x}{\left(y+1\right)2i}=\frac{-i\left(2-x\right)}{\left(y+1\right)}$
$\because \frac{2-x}{y+1}$ is a positive real number
$\Rightarrow arg\left(\frac{4-z_0-z_0}{z_0-z_0+2i}\right)=-\frac{\pi }{2}$