Roots of $z^r=1,r\notin\mathbb{Q}$ 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?
 A: 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).
A: 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.
