Question

If $\omega =\frac{z}{z-\frac{i}{3}}and\left|\omega \right|=1,$ then locus of z is:

Answer 1

Given,

$w=\frac{z}{z-\frac{i}{3}}$

$|w|=1$

$w=\frac{x+iy}{x+iy-\frac{i}{3}}$

$=\frac{x+iy}{x+i(y-\frac{1}{3})}$

$|w|=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+{(y-\frac{1}{3})}^2}}$

$1=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+{(y-\frac{1}{3})}^2}}$

$\sqrt{x^2+{(y-\frac{1}{3})}^2}=\sqrt{x^2+y^2}$

On squaring on both sides, we get,

$x^2+{(y-\frac{1}{3})}^2=x^2+y^2$

$y^2+\frac{1}{9}-\frac{2y}{3}=y^2$

$\frac{1}{9}=\frac{2y}{3}$

$\therefore y=\frac{1}{6}$ is the locus of $z$.

Hence, this is the answer.