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.

#Algebra #ComplexNumbers #UnitCircle #Infinity

Easy Math Editor

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Comments

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 - 5 years, 8 months ago

Thanks for the explanation

Janardhanan Sivaramakrishnan - 5 years, 8 months ago

l nhkhb

Lucy Oakley - 5 years, 6 months ago

sorry that was my friend

Lucy Oakley - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$