The $n$th roots of unity are the complex numbers: $1, w,w^2,…,w^{n-1}$, where $w=e^{\frac{2\pi i}{n}}$.
Why is this true? I understand why $w$ is 1 root of unity, but why are $w^0,…, w^{n-1}$ the other roots of unity? Why do the roots of unity consist of the exponents of $w$?
I am only aware that:
The $n$th roots of unity roots of unity are: $\sqrt[n] 1 = \sqrt[n]r\left(\cos\frac{2\pi }{n} + i\sin\frac{2\pi k}{n}\right)$
Best Answer
First, there are at most $n$ $n$th roots of unity, because $x^n-1$ can have at most $n$ roots (as a consequence of the Factor Theorem applied in $\mathbb{C}$).
Second, if $\omega$ is an $n$th root of unity, that means that $\omega^n = 1$. But then, for any integer $k$, we have $$(\omega^k)^n = \omega^{kn} = (\omega^n)^k = 1^k = 1,$$ so $\omega^k$ is also an $n$th root of unity.
So now the question is which ones are different? If $\omega$ is such that $\omega^n=1$ but $\omega^{\ell}\neq 1$ for any $0\lt \ell\lt n$, then $\omega^r=\omega^s$ if and only if $\omega^{r-s}=1$, if and only if $n|r-s$: indeed, using the division algorithm, we can write $r-s$ as $qn + t$, with $0\leq t\lt n$ (division with remainder). So then $$1=\omega^{r-s} = \omega^{qn+t} = \omega^{qn}\omega^t = (\omega^n)^q\omega^t = 1^q\omega^t = \omega^t.$$ But we are assuming that $\omega^t\neq 1$ if $0\lt t\lt n$; since $0\leq t\lt n$ and $\omega^t=1$, the only possibility left is that $t=0$; that is, that $n|r-s$.
Thus, if $\omega^{\ell}\neq 1$ for $0\lt \ell\lt n$ and $\omega^n=1$, then $\omega^r=\omega^s$ if and only if $n|r-s$, which is the same as saying $r\equiv s\pmod{n}$.
So it turns out that $\omega^0$, $\omega^1$, $\omega^2,\ldots,\omega^{n-1}$ are all different (take any two of the different exponents: the difference is not a multiple of $n$), and they are all roots of $x^n-1$, and so they are all the roots of $x^n-1$.
So it all comes down to finding an $\omega$ with the property that $\omega^k\neq 1$ for $0\lt k\lt n$, but $\omega^n=1$. And $$\large\omega = e^{2\pi i/n}$$ has that property.