$i^{th}$ root(s) of unity If we define $S:=\{z\in\mathbb{C}:z^n=1\}$ (i.e. the $n^{th}$ roots of unity), then $|S|=n$ (i.e. we have $n$ of them). We can even go as far as to say: $$S=\{z_k:k\in\mathbb{N}\cap[1,n]\}=\{e^{2\pi i/k}:k\in\mathbb{N}\cap[1,n]\}$$
Plotting these points on an Argand diagram shows the above point, namely that the plot looks symmetric across the real axis, and has rotational symmetry order $n$. It follows that, if we let $n\to\infty$, the plot of points tends towards being a circle centred on the origin, radius $1$.
But $z^i$, and even $z^{1/i}$, are both (possibly) well-defined ideas, so can we talk about the $i^{th}$ roots of unity? How many of them are there? What geometric interpretation is there when we talk about them, and can we analogously look at the $2i^{th}$ roots, etc.?
I know that $z^{1/i}:=e^{(1/i)\log{z}}$, so $z^{1/i}=1$ means that we search for solutions to $(1/i)\log{z}=0 \Rightarrow \log{z}=0$.
I guess a similar question is: what geometric/insightful interpretation is there to the fact that $i^i=e^{(-\pi+4k)/2}$ (for $k\in\mathbb{Z})$?
Edit: turns out that I can't multiply things by $0$ correctly... mistake fixed, thanks.
 A: Define $\lg(z)$ to be the standard branch cut of $\ln$, discontinuous on the negative reals. Now since $\exp$ is periodic with period $2\pi i$, $\exp(y)=z\implies y=\lg(z)+(2\pi i)n$ for some $n\in\mathbb{Z}$.
So if you would like $z^i$ to mean $\exp(i\cdot\lg(z))$, then $$\begin{align}z^i=1&\implies\exp(i\cdot\lg(z))=1\\
&\implies i\cdot\lg(z)=\lg(1)+(2\pi i)n\\
&\implies \lg(z)= 2\pi n\\
&\implies z=\exp(2\pi n)\\
\end{align}$$
So the collection of $i$th roots of unity is countably infinite, consisting of the following positive real numbers: $$\{\ldots,e^{-4\pi},e^{-2\pi},1,e^{2\pi},e^{4\pi},\ldots\}$$
Repeating this for general complex roots, the $w$th roots of unity are the countable collection $$\{\exp(2\pi i n/w)\mid n\in\mathbb{Z}\}$$ (which may be merely finite if $w$ is such that there is overlap in this collection.)
A: One considers not so much the cyclotomic numbers $cis(2\pi m/n)$, but rather the table of chords $C(m/n) = 1-cis(2\pi m/n)$.  $cis() = cos() + i sin()$.
It's a pretty straight-forward proof that the intersection between the set of spans of the cyclotomics, and the set of fractional numbers (Z/Z), is the set of reals.  
Using the further item that the product of numbers $C(m/n)$ for 0

So if you determine that the chord of an angle is 4-sqrt(13), then it must be irrational.
