Q:

If $|z|=1$ , then prove that $\frac{z}{\bar z}$ lies on a circle

$$|z|^2=1$$
$$z.\bar z=1$$
$$\frac{z}{\bar z}=\frac{1}{ \bar z^2}$$
I couldn’t solve further. Can I get some insight into this question?

A:

Or you can let $z = e^{i \theta}$ and then go $$\frac{z}{\bar{z}} = e^{i \theta} / e^{-i \theta} = e^{2 i \theta}$$ which is of length $1$.

A:

The fact that $|z|=1$ is a red herring. You just need $z\ne0$, because
$$|\bar{z}|=|z|$$
so you have 
$$\left|\frac{z}{\bar{z}}\right|=\frac{|z|}{|\bar{z}|}=1$$

A:

If
$\vert z \vert = 1, \tag 1$
then
$\vert \bar z \vert = 1, \tag 2$
so
$\left \vert \dfrac{z}{\bar z} \right \vert = \dfrac{\vert z \vert}{\vert \bar z \vert } = 1; \tag 3$
thus
$\dfrac{z}{\bar z} \in S^1, \tag 4$
the unit circle.

