If $\omega =\frac{z}{z-(i/3)}$ and $|\omega |=1$, then $z$ lies on:

Given $|\omega |=1$:$\left|\frac{z}{z-i/3}\right|=1⟹|z|=|z-i/3|$ Let $z=x+iy$:$\sqrt{x^2+y^2}=\sqrt{x^2+(y-1/3{)}^2}$ Squaring both sides:$x^2+y^2=x^2+y^2-\frac{2}{3}y+\frac{1}{9}$$0=-\frac{2}{3}y+\frac{1}{9}⟹\frac{2}{3}y=\frac{1}{9}⟹y=\frac{1}{6}$ This is the equation of a straight line parallel to the x-axis.