$z\in C$ such that $\frac{z}{z-i}$ is real. Show that z is imaginary.

just at the first stage of intro to complex numbers and beginning to struggle

so far i have this but don't know how to proceed any further

$\frac{z}{z-i}=\frac{z-i+i}{z-i}$

$=\frac{z-i}{z-i}+\frac{i}{z-i}$

$=1+\frac{i}{z-i}$

$=1+\frac{i\times i}{i(z-i)}$

$=1-\frac{1}{iz+1}$

hence $\frac{1}{iz+1}$ is real

multiply by conjugate

$=\frac{1}{iz+1}\times\frac{1-i\overline{z}}{1-i\overline{z}}$

$=\frac{1-i\overline{z}}{(iz+1)(-i\overline{z}+1)}$ which is real

hence $i\overline{z}$ must also be real

not quite sure what to do to proceed now.

any help is enormously appreciated