If $\omega =\frac{z}{z-\frac{1}{3}i}$ and $\left|\omega \right|=1,$ where $i=\sqrt{-1},$ then $z$ lies on 

$\left|\frac{z}{z-\frac{i}{3}}\right|=1\Rightarrow \left|z\right|=|z-\frac{i}{3}|$ 
Clearly, locus of $z$ is perpendicular bisector of line joining points having complex number $0+i$ 0  and $0+\frac{i}{3}.$ Hence, $z$ lies on a straight line.