The curious case of $x^{\pi} - 1 = 0$

Question

I was thinking of the solution(s) for the fractional order algebraic equation $ x^{\pi} - 1 = 0$

Obviously, the solution set is of the form $x = e^{2nj}, n \in \mathbb{Z}$. A countably infinite set of complex numbers, all of unit magnitude.

Let $U$ be the set of unit magnitude complex numbers, defined as $U = \{ x | x \in \mathbb{C}, |x|=1, x^{\pi} \ne 1\}$

The question is, What percentage of the unit circle does $U$ constitute?

It seems that the answer is $100\%$. But, that is something that I have a hard time getting my head around.

Answer 1

It is easy to see that : $ |x|=1 \Leftrightarrow x = e^{i \theta} $. Thus, equating this with the solution you provide, we see that, while sticking to the domain $ [0,2 \pi] $, we see that $\theta = 2n $ to find the points on the unit circle which satisfy $ x^\pi =1 $. That is, $ \theta $ is an even number. Thus the set of solution are finite (in $ [0.2 \pi]$ ), in particular countable (countable in $ \mathbb R $ !!!!), and thus, if you know a bit about measure theory, the percentage is 100% indeed.

To make it simply but condensated, there are uncountably many points on the unit circle,$C,$ and $|C \setminus U|$ is countable thus, upon integrating, the percentage is 100%. (Isolated points have no "weight")

Answer 2

Define the multivalued function $x\mapsto x^{\pi}$ on $S^1$ as:

$$e^{iy}\mapsto e^{\pi i(y+2\pi k)},\,k\in\mathbb Z$$

As the other answer notes, the set of $x\in S^1$ such that $1$ is one of the values of $x^{\pi}$ is countable, and hence measure zero.

Namely: $x^\pi,$ has, as one of its values, the value $1$ if and only if $y+2\pi k$ is an even integer for some integer $k.$

However, for any $x=e^{iy}$ and any $\epsilon>0,$ we can find a $k$ so that $|x^{\pi}-1|<\epsilon.$ We also have that the set of possible values for $x^{\pi}$ is dense in $S^1.$

However, dense subsets can have measure $0.$ For example, the rational numbers in $[0,1]$ have measure zero, and countable sets in $S^1$ have measure zero.

Here, the set $\{x\in S^1\mid 1\text{ is a value of } x^{\pi}\}$ has measure $0.$ But given any open neighborhood, $N$, of $1$ in $S^1$, the set $\{x\in S^1\mid \text{a value of } x^{\pi}\text{ is in }N\}$ is $S^1.$