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May 29th, 2017, 12:30 PM   #1
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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 $\displaystyle r^{a}$.
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May 29th, 2017, 12:52 PM   #2
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$e^{i \theta}$ 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
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Last edited by Maschke; May 29th, 2017 at 01:00 PM.
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May 29th, 2017, 05:21 PM   #3
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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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