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# The curious case of x^pi - 1 = 0

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.

Note by Janardhanan Sivaramakrishnan
2 years, 3 months ago

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l nhkhb

- 2 years, 1 month ago

sorry that was my friend

- 2 years, 1 month ago

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 \ U| is countable thus, upon integrating, the percentage is 100%. (Isolated points have no "weight")

- 2 years, 3 months ago

Thanks for the explanation

- 2 years, 3 months ago