Roots of $z^r=1,r\notin\mathbb{Q}$

Question

If $a,b\in\mathbb{Z}$, and $\frac a b$ is in lowest terms, then $$z^{\frac a b}=1\\\implies z=\exp\left(\frac{2\pi in b}{a}\right)\forall n\in\mathbb{Z}$$ This means that $z$ has exactly $a$ distinct values on the unit circle, as $n\pm a$ gives the same $z$ value as $n$, as the exponent is rational. However, what if the exponent is irrational? If we have $$z^{r}=1,r\notin\mathbb{Q}\\\implies z=\exp\left(\frac{2\pi in }{r}\right)\forall n\in\mathbb{Z}$$ then as this does not loop as with a rational number, does $z$ cover the entire unit circle in the complex plane? Does it cover a dense set of the unit circle? Or is there something wrong with taking irrational roots of unity?

Answer 1

Let me bring you an example

$$z^\pi=1$$

I guess an obvious solution which is

$$z_0=e^{\frac{2\pi i}\pi}=e^{2i}=\cos(2)+i \sin(2)$$

In addition, any other number which is natural power of $e^{2i}$ must match the equation. So, $e^{2in}$ where $n\in N$ is a solution. How do you imagine locus of $e^{2i n}$? It covers infinite points of unit circle but not all points of them.

As an example, is $1\angle \frac{\pi}6$ a solution? then you need to find out if there is a $n$ where $$n \times 2 =\frac{\pi}6+2 k\pi$$

So, it has infinite solution on unit circle while they do not cover the whole points of unit circle(Although they are infinite, still they are a very small minority since they are countable).

Answer 2

The complex solutions of $z^r=1$ are $$ \exp((2\pi i n)/r),\qquad n \in \mathbb Z $$ as you said. If $r$ is not rational (for example, if $r$ is not real) then these are all distinct. But there are only countably many of them. If $r$ is real but irrational, these values are all on the unit circle. Countably many of them, dense in the circle, but not the whole circle.