Question

If $z=x+iy,\omega =\frac{2-iz}{2z-i}$ and $|\omega |=1$, find the locus of $z$ and illustrate it in the Argand Plane.

Answer 1

Given, $z=x+iy$
$\omega =\frac{2-iz}{2z-i}$ and $|\omega |=1$

$\Rightarrow \left|\begin{array}{c}\frac{2-i(x+iy)}{2(x+iy)-i}\end{array}\right|=1$

$\Rightarrow |2x+(2y-1)i|=|(y+2)-ix|$

$\sqrt{(2x{)}^2+(2y-1{)}^2}=\sqrt{(y+2{)}^2+(-x{)}^2}$

$\Rightarrow 4x^2+4y^2-4y+1=y^2+4y+4+x^2$

$\Rightarrow 3x^2+3y^2-8y-3=0$

$\Rightarrow x^2+y^2-\frac{8}{3}y-1=0$, which represents a circle at $\left(\begin{array}{c}0,\frac{4}{3}\end{array}\right)$ and radius

$r=\sqrt{(0{)}^2+{\left(\begin{array}{c}\frac{-4}{3}\end{array}\right)}^2-(-1)}=\frac{5}{3}$