Does $1^0$, as a limiting form of the $n$ $n$-th roots of unity $1^{1/n}$, represent every complex value on the unit circle? In the context of algebraic numbers, the $n$-th root has $n$ solutions in the complex plane. The equations look like: $1=x^n$, which gives $1^{1/n}=x$ (which can be rotated around the complex plane to generate all the solutions).  And then, if we let $n$ become large, we get many solutions. But then, when we let $n$ become arbitrarily large, we get $1^0 = x$, which only has one solution, namely $1$, because anything to the zeroth power gives $1$. This seems discontinuous to me, and it confuses me.
Does anyone know what I'm missing?

I have a feeling that perhaps $1^0$ actually is every value around the complex unit circle, which would be the natural extension of the limiting process. But, if that's true, there are tons of implications that don't seem right to me.

Maybe the following two facts are useful? I'm just spitballling on these two.
This reminds me of the fact that $a*0=0$ has infinitely many solutions.
And perhaps the following fact is useful: $2^0 = 2 ^ {100 * 0} =  (2 ^ {100})  ^ 0$
 A: The $n$th roots of $1$ are
$$
\exp\left(\frac{2\pi i k}{n}\right), \quad k \in \mathbb Z
$$
If $e^{i\theta}$ is any point on the unit circle, then as $n \to \infty$ we can choose $k_n \in \mathbb Z$ so that $\frac{2\pi k_n}{n} \to \theta$.  That is: if, for each $n$, we choose an appropriate $n$th root of $1$, then those roots converge to $e^{i\theta}$.
A: We can argue for this result geometrically to build intuition. For $n>1$ there are always two primitive $n$-th roots of unity that are closest to $1$ in the complex plane. Under multiplication they act on the plane by rotating each complex number $1/n$-th of a turn around the origin in either direction. As $n$ gets large, these rotations get very small. When $n$ gets infinitely large, the rotations get infinitely small, and so don't rotate the plane at all. This is the same action that $1$ has on the plane, in accordance with our intuition.
A: I think I might have a solution to the problem, and I think the solution is that yes, the solution set to $z^0 = 1$ is $\{(x,y) | x^2 + y^2 = 1\}$
First of all, we can start with the following: $z = 1^n$, which gives $z^{1/n} = 1$.  We know that the solution set to this problem starts with one point at $(1,0)$, and then it has $n-1$ points evenly distributed throughout the rest of the unit circle.  We will call this set $S_n$
Now we take the limit as n approaches infinity.  In other words, we want $\lim_{n\to \infty} S_n$.  To get this, we are going to use the Hausdorff metric (has a wikipedia page) on the subsets of the unit circle, then we will show that we can make this metric arbitrarily small as $n\to \infty$.  The Hausdorff metric is actually a pseudometric, but it can be treated as a metric if we treat the space as the quotient space of everything which is equal under the pseudometric.  What this means is that because the unit circle of real numbers and the unit circle of algebraic numbers are indistinguishable under this pseudometric, they are the same element of the quotient space. And you can see that as $n\to \infty$, the hausdorff metric goes to zero, and thus the solution to our problem is indeed this element of the quotient space.  So this answers the question that the limit is equal to the unit circle, but we cannot tell whether it is the real unit circle or the algebraic number unit circle.
