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

$\left|\frac{z}{z-\frac{i}{3}}\right|=1\Rightarrow \left|z\right|=\left|z-\frac{i}{3}\right|$

Clearly locus of $z$  is perpendicular bisector of line joining points having complex number $0+i0$  and $0+\frac{i}{3}$ . Hence, $z$ lies on a straight line.