Can $x^n = 1$ describe the equation of a unit circle?

Question

The complex roots of the equation $x^n = 1$ lie on the unit circle.

Suppose $n$ goes large. Is it correct to say that all the roots of $x^n=1$ form a circle of radius one?

Answer 1

Well, strictly speaking, the set of points $x$ in the complex plane satisfying $x^n-1=0$ for some $n$ is exactly the set of points $e^{i\pi r}$ for rational $r$ - that is, any point on the unit circle whose argument is a rational number times $\pi$. This set is dense in the unit circle, but it's not the whole unit circle by a long shot.

However, we can characterize the unit circle as the points $x$ such that $$\liminf_{n\rightarrow\infty}|x^n-1| = 0$$ since the argument of $x^n$, as $n$ increases, will be dense mod $2\pi$ if it is not a rational multiple of $2\pi$, and it's trivial to show that if $|x|<1$, then the above limit is $1$, and if $|x|>1$, the above limit is $\infty$, sufficing to show that the above describes exactly the unit circle.

Answer 2

No luck there. Note that for every $n$, $x^n-1$ has $n$ roots, so it only generates $n$ points on the unit circle.

Even if we take the union over all integer $n$ of the roots of $x^n-1$, we only have a countable number of points while there are an uncountable number of points on the unit circle.

Answer 3

Assuming $n\in\mathbb{N}$, the answer is no.

The length of every root of $x^n=1$ is $1$.

The larger $n$ gets, the more roots you have.

So your question makes quite good sense in that aspect.

But the roots of $x^n=1$ are $(k\cos\frac{2\pi}{n},k\sin\frac{2\pi}{n})$ for $k\in[0,n-1]$.

In other words, only complex numbers whose angle with the axises is an integer multiple of $\frac{2\pi}{n}$.

Since there are infinitely many points on the unit circle whose angle with the axises is not an integer multiple of $\frac{2\pi}{n}$, these roots will never "cover it completely" (no matter how large $n$ gets).