Number distinct powers of $cos(\alpha \pi) + i\sin(\alpha \pi)$ How many distinct powers of $\cos(\alpha \pi) + i\sin(\alpha \pi)$ are there if $\alpha$ is rational?  Irrational?
Here's my thoughts.  The answer is probably some finite number based on $\alpha$ if $\alpha$ is rational and infinitely many if $\alpha$ is irrational.  So I'd like to confirm (or disprove) that guess.
First we start with the formula $$(\cos(\alpha \pi) + i\sin(\alpha \pi))^n = \cos(n\alpha \pi) + i\sin(n\alpha \pi)$$
If $\alpha$ is rational, then I can see that some of the powers would be the same.  Let $\alpha = \frac pq$, then we'd get $\cos(n\alpha \pi) + i\sin(n\alpha \pi) = \cos(m\alpha \pi) + i\sin(m\alpha \pi) \iff n\alpha \pi = m\alpha \pi + 2\pi k$.  This implies that $$(n-m)p=2kq$$
So whenever that condition is satisfied, then the $n$th power and $m$th power will be the same.  But I don't see how to count those possibilities.
Then if $\alpha$ is irrational, we'd still need $n\alpha \pi = m\alpha \pi + 2\pi k = (m\alpha + 2k)\pi$, but this time $\alpha \ne \frac pq$.  So we'd just get $$\alpha = \frac{2k}{n-m}$$ The right hand side is always rational, so there is no pair $(n,m)$ for which this holds.  So there are infinitely many solutions if $\alpha$ is irrational.
So it seems I've gotten the irrational case, but I can't see how to count up the solutions in the rational case.
 A: Remeber that you may write $\Bbb e ^{\Bbb i \alpha} = \cos \alpha + \Bbb i \sin \alpha$.


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*If $\alpha = \frac p q \pi$ with $p,q \in \Bbb Z, \ q \ne 0$ and $p$ even, then $(\Bbb e ^{\Bbb i \alpha}) ^0, \dots, (\Bbb e ^{\Bbb i \alpha}) ^{q-1}$ are all distinct, the next power being $(\Bbb e ^{\Bbb i \alpha}) ^{q} = \Bbb e ^{\Bbb i p \pi} = 1$ (so starting from here the powers begin to repeat in a cyclical way), so you have $q$ distinct powers.

*If $\alpha = \frac p q \pi$ with $p,q \in \Bbb Z, \ q \ne 0$ and $p$ odd, then $(\Bbb e ^{\Bbb i \alpha}) ^0, \dots, (\Bbb e ^{\Bbb i \alpha}) ^{2q-1}$ are all distinct, the next power being $(\Bbb e ^{\Bbb i \alpha}) ^{2q} = \Bbb e ^{2 \Bbb i p \pi} = 1$ (so starting from here the powers begin to repeat in a cyclical way), so you have $2q$ distinct powers.

*If $\alpha \notin \Bbb Q \pi$, then assume there are $m \ne n \in \Bbb Z$ with $(\Bbb e ^{\Bbb i \alpha})^m = (\Bbb e ^{\Bbb i \alpha})^n$ or, equivalently, $(\Bbb e ^{\Bbb i \alpha})^N = 1$ (where $N = m-n \ne 0$). Then $\alpha N = 2 k \pi$ for some $k \in \Bbb Z$, so $\alpha = \frac {2k} N \pi \in \Bbb Q \pi$, which is a contradiction, therefore all the powers are distinct.
