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 May 29th, 2017, 12:30 PM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 irrational power of complex number What is $\displaystyle z^{a}$ when a is irrational? Formally, $\displaystyle z^{a}=r^{a}e^{i(a\theta \pm an2\pi) }$, n=0,1,2,.... but an is never an integer, so it looks like an infinite number of roots: points on the circle of radius $\displaystyle r^{a}$.
 May 29th, 2017, 12:52 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,393 Thanks: 749 $e^{i \theta}$ is a point on the unit circle. If $\theta$ is a rational multiple of $2 \pi$, the points $e^{n i \theta}$ are a finite set. That is, they repeat after a while. You can see this because $(e^{\frac{n}{m} 2 \pi i})^m = 1$. But if $\theta$ is an irrational multiple of $2 \pi$, the $n$-th powers never repeat. Amazingly, they are dense on the circle. They get arbitrarily close to every point on the circle. https://en.wikipedia.org/wiki/Irrational_rotation https://math.stackexchange.com/quest...irrational-aro Thanks from zylo Last edited by Maschke; May 29th, 2017 at 01:00 PM.
 May 29th, 2017, 05:21 PM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 If $\displaystyle \alpha$ is angle of arbitary point on cricle, can I find n and m so that I can come arbitrarily close to $\displaystyle (\alpha + m2\pi)$, ie, can i find m and n st $\displaystyle |(a\theta+na2\pi)-(\alpha + m2\pi)|<\epsilon$

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