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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 ago

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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")

Patrick Bourg - 2 years ago

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Thanks for the explanation

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l nhkhb

Lucy Oakley - 1 year, 10 months ago

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sorry that was my friend

Lucy Oakley - 1 year, 10 months ago

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