If = zz-13i and ||=1, then z lies on

Question

If $\omega=\dfrac z{z-\frac13i}$ and $|\omega|=1$, then $z$ lies on

Accepted Answer

$|\omega|=1 \implies |z|=\left|z-\dfrac13i\right|$, meaning $z$ is equidistant from the origin and $\left(0,\dfrac13\right)$ --- the perpendicular bisector of that segment, a straight line.

Solutions · 1