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One such value is $\displaystyle\cos\left(2\pi\alpha\right)+i\sin\left(2\pi\alpha\right)$, by Euler's theorem. On the other hand, we can choose an arbitrary sequence $S=(a_n)_n$ of rational numbers converging to $\alpha$, pick up $a_n=p_n/q_n\in S$ and calculate possible values of $e^{2i\pi p_n/q_n}$. The $q_n$ part will open many branches, but eventually $p_n$ will make some of them become equal. It is not clear whether the number of solutions is growing.

Because multivaluedness can cause serious headaches sometimes, let's define some things. Cosine and sine are defined by the usual taylor series and exponentiation by a natural number remains untouched. Roots are calculated by this procedure: $\mathbf{1^{1/q_n}}$ is defined to be the set $$\mathbf{1^{1/q_n}}=\{z:z\in\mathbb{C}\wedge z^{q_n}=1\}.$$

We, then, define $\mathbf{1^{p_n/q_n}}$ to be the set $K_n=\{z^{p_n}:z\in\mathbb{C}\wedge z^{q_n}=1\}$. To avoid even more problems, we let $p_n$ and $q_n$ be coprime.

How is the cardinality of $K_n=\mathbf{1^{p_n/q_n}}$ growing over time? I have no intuition in this opening and closing branches game. I want to know the limit $\displaystyle\lim_{n\to\infty}|K_n|$ and a rigorous proof of it.

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    $\begingroup$ Very concretely, for any integer $n$: $1^\alpha = (e^{2\pi n i})^\alpha = e^{2\pi n \alpha i} = \cos(2\pi n \alpha) + i \sin(2\pi n \alpha)$. When $\alpha$ is irrational, all these values are distinct. So there are countably infinitely many "feasible" values of $1^\alpha$. $\endgroup$ – Slade Dec 29 '13 at 6:35
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    $\begingroup$ @User-33433 a bit late but I see, suppose integers $m$, $n$ and $\ell$ such that $2\pi m\alpha + 2\pi n=2\pi \ell\alpha$. This would imply $\alpha=\dfrac{n}{\ell -m},$ a contradiction to the fact that $\alpha$ is irrational. $\endgroup$ – Ian Mateus Jan 9 '14 at 23:44
  • $\begingroup$ It's probably also worth mentioning that this lets us think about complex powers as well. There's the classic "identity" $i^i = (e^{i\pi/2})^i=e^{-\pi/2}$. (Of course, there are infinitely many other values that would also make sense here.) $\endgroup$ – Slade Jan 10 '14 at 23:25
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Each set $\mathbf{1^{1/q_n}}$ is the $q_n$ $q_n$th roots of 1, which form a cyclic group under multiplication. Let a be a generator of this group.

$p_n$ and $q_n$ are coprime, so the order of $a^{p_n}$ is that of the group; $a^{p_n}$ is another generator. Hence define $\phi : b \mapsto b^{p_n}$ on $\mathbf{1^{1/q_n}}$. It is simple to check that this is an automorphism, hence is bijective.

This establishes that the cardinality of $\mathbf{1^{p_n/q_n}}$ equals that of $\mathbf{1^{1/q_n}}$, which is $q_n$.

Hence as $q_n$ diverges, so does the set of solutions (as corroborates with the infinite solutions to the irrational case).

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  • $\begingroup$ I think I understand it. Because $p$ and $q$ are coprime, each $2k\pi/q$ will be mapped to $2k\pi p/q$, but they are still a permutation of the original roots $\pmod{2\pi}$. The condition $(p, q)=1$ was, then, crucial for it to happen. Thank you. $\endgroup$ – Ian Mateus Dec 29 '13 at 5:39
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Since $p$ and $q$ are coprime, the set $\mathbf{1^{1/q}}$ is equal to $\mathbf{1^{p/q}}$. Why? In the way we defined $\mathbf{1^{1/q}}$, it is equal to $$\{\cos\left(2k\pi/q\right)+i\sin\left(2k\pi/q\right):k\in\{1,2,\ldots,q\}\}.$$ However, as $p$ and $q$ are coprime, the multiplication by $p$ just shuffles the roots, as $$\mathbb{Z}/q\mathbb{Z}=\{1, 2,\ldots,q\}=\{p,2p,\ldots,qp\}.$$ That way, we see $\mathbf{|1^{1/q}|}=\mathbf{|1^{p/q}|}=q$, so $\displaystyle\lim_{n\to\infty}|K_n|=\lim_{n\to\infty}q_n=\infty$ if $\alpha$ is irrational. Moreover, it can be proven that if $\gcd(p,q)=d$, then $\mathbf{|1^{p/q}|}=q/d$.

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