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

As given  $\mathrm{\omega }=\frac{\mathrm{z}}{\mathrm{z}-\frac{1}{3}\mathrm{i}}\Rightarrow \left|\mathrm{\omega }\right|=\frac{\left|\mathrm{z}\right|}{\left|\mathrm{z}-\frac{1}{3}\mathrm{i}\right|}=1$
$\Rightarrow$distance of z from origin and point $\left(0,\frac{1}{3}\right)$ is same ,hence z lies on bisector of the line joining points (0,0) and (0,1/3).

Hence Z lies on a straight line.