So I recently stumbled across the principle that the set of all solutions to the equation $z^n=1\mid n\in\mathbb{Q^\neq}$ has a cardinality equal to the denominator of $n$ as represented by an irreducible fraction of two integers.
When I tried to extend this principle to $n\in\mathbb{R^\neq}$ (a.k.a. including irrational numbers into the domain of $n$), I came to some... perplexing (and perhaps contradictory) conclusions. Namely, if there are infinite $\pi$th (or $e$th, or any irrational-number-th) roots of a complex number $a+bi$, that means each of these roots are not unique to $a+bi$, but rather every complex number with the same radius $\sqrt{a^2+b^2}$ will share the same infinite roots. This would, in turn, imply that $(a+bi)^\pi$ also has infinitely many solutions. None of it seems correct to me. My reasoning is as follows:
Definition
According to Wikipedia, the $n$th roots of unity are defined by a finite set of solutions to the equation $z^n=1$. These solutions are most easily defined using Euler's formula:
$$ z(n)= \sqrt[n]{1}= \left\{ e^{\frac{2i\pi k}{n}} \equiv \cos\left(\frac{2\pi k}{n}\right)+i\sin\left(\frac{2\pi k}{n}\right) \mid k\in\left\{0,1,...,n-2,n-1\right\} \right\} \mid n\in\mathbb{Z^>} $$
Shortcomings of this definition:
- $n$ must be nonnegative: $n\lt0\implies\left\{0,1,...,n-2,n-1\right\}=\emptyset$ when it should be a set with the same cardinality as the set implied by $-n$.
- $n$ must be an integer: $n\notin\mathbb{Z}$ implies that $\left\{0,1,...,n-2,n-1\right\}$ is ambiguous because $\left\{0,1,2,...\right\}$ are integers whereas $\left\{n-1,n-2,n-3,...\right\}$ are not.
My First Redefinition
$$ z(n)= \sqrt[n]{1}= \left\{ e^{\frac{i2\pi k}n} \mid k\in\mathbb{Z^\geq} \land k \lt \lvert n\rvert \right\} \mid n\in\mathbb{Q^\neq} $$
But this still doesn't cut it. While it does compensate for negative values of $n$, it still doesn't work for nonzero, non-integer values of $n$. For example, with $sqrt[\frac{3}{2}]{1}$, $n=\frac{3}{2}$. $\left\{k\mid k\in\mathbb{Z^\geq}\land k \lt\lvert n\rvert\right\}$ is a set of cardinality $2$, implying that $z^{\frac{2}{3}}=1$ has only two solutions, when in reality it has three.
My Second Redefinition
As far as I can tell, there are two ways to fix this issue:
You could define each nonzero rational number $n$ as an irreducible fraction of two nonzero integers: $\left\{n\mid n\in\mathbb{Q^\neq}\right\}=\left\{ \frac{p}{q} \mid p,q\in\mathbb{Z^\neq}\right\}$, and then define $k$ in terms of $p$: $\left\{k\mid k\in\mathbb{Z^\geq}\land k\lt\lvert p\rvert\right\}$
You could expand the domain of $k$ to the set of all natural numbers: $\left\{k\mid k\in\mathbb{Z^\geq}\right\}$. This should be acceptable because the period of $f(k)=\cos\left(\frac{2\pi k}{n}\right)+i\sin\left(\frac{2\pi k}{n}\right)$ is, by definition, $n$ (a.k.a. $\frac{p}{q}$). Therefore, the infinite sequence $z(n)$ will periodically repeat itself every $p$ elements, meaning that the cardinality of the resulting set $z(n)$ must have a cardinality of $p$.
These two methods are, as far as I can tell, functionally identical.
Second Redefinition: Method 2
$\pi$ is obviously irrational, meaning that there is no number $n \mid \pi\cdot n \in \mathbb{Z}$. Interestingly, means that the values of set $k$ will never match the period of the sine function, but a set with a cardinality of $\aleph_0$ would be able to encapsulate every distinct $\pi$th root.
Second Redefinition: Method 1
Just in case expanding the domain of $k$ to the set of all natural numbers is an invalid operation, I tried attacking it a different way. So once again, $\pi$ is not a rational number. However, John Wallis discovered that $\pi$ can be represented as the infinite product:
$$ \frac{\pi}{2}=\prod_{n=1}^{\infty}\frac{4n^2}{4n^2-1} $$
Noting that $\left(\forall n\right)\left[\left(4n^2\right),\left(4n^2-1\right)\in\mathbb{Q}\right]$, it is my understanding that $\pi$ can essentially be thought of as the ratio between two infinitely large integers. As such, I am led to believe that the numerator being an $\infty$ means that there are infinite $\pi$th roots of unity.
Finally, the Question
Is everything I've reasoned here correct?
It is not true that "every complex number with the same radius...will share the same infinite roots" but yes, an infinite number of them will. The term "Almost everywhere" comes up often in these circumstances. In this case the solutions on the circle are as dense as the rational numbers are on the real line (you are correct when you say "a set with a cardinality of $\aleph_0$ would be able to encapsulate every distinct $\pi$th root"). That does not mean that every real number here is a rational number. There are also an infinite number of points on the circle that aren't $\pi$-th roots. Any non trivial integer root for example.