How many values does $1^{\alpha}$ have for $\alpha$ irrational?

Question

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.

Answer 1

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).

Answer 2

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 $\mathbf{|1^{p/q}|}=q/d$.