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 $\frac{ 1}{\left|z_{0}-1\right|}$ is the maximum of the set $\left\{\frac{ 1}{\left|z_{0}-1\right|}: z\,\in\,S\right\},$ 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 $y>0$
$\frac{4-z_{0}-\bar{z_{0}}}{z_{0}-\bar{z_{0}+2i}}$
$=\frac{4-x-iy-x-iy}{x+iy-x+iy+2i}$
$=\frac{4-2x}{\left(y+1\right)2i}=\frac{-i\left(2-x\right)}{\left(y+1\right)}$
$\because \frac{2-x}{y+1}$ is a positive real number
$\Rightarrow arg \left(\frac{4-z_{0}-z_{0}}{z_{0}-z_{0}+2i}\right)=-\frac{\pi}{2}$