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

Question

$$|z|^2=1$$ $$z.\bar z=1$$ $$\frac{z}{\bar z}=\frac{1}{ \bar z^2}$$

I couldn’t solve further. Can I get some insight into this question?

Answer 1

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.

Answer 2

First, you can use \bar $z$ or \overline $z$ to write $\bar z$. Next, $$\lvert z \vert = 1 \iff \lvert \bar z \vert = 1 .$$

So, $$\lvert \frac{z}{\bar z} \vert = \frac{\lvert z \vert}{\lvert \bar z \rvert} = 1 \iff \frac{z}{\bar z} \quad \text{lies on a unit circle.}$$

Answer 3

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 $1$.

Answer 4

Since $|z| = 1$, one has $|\bar{z}| = 1$. Hence, you get :

$$|\frac{z}{\bar{z}} | = \frac{|z|}{|\bar{z}|} = 1$$

Hence, the point of affix $\frac{z}{\bar{z}}$ lies on the circle with center $O$ (the point of affix $0$) and radius $1$.

Answer 5

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