# Number of complex n'th roots of unity

I just started to read Freitag/Busam's book on complex analysis. They prove that there are exactly $n$ complex $n$th roots of unity (page 9). But I only understand the existence proof.

In it they argue that for $\varphi = k {2 \pi \over n}$, $(e^{i\varphi})^n = 1$. Furthermore they note that because as previously noted the map $\mathbb R_{\gt 0}\times (-\pi,\pi]\to\mathbb C \setminus \{0\}$ is bijective it is enough to consider $k \in \{0,\dots,n-1\}$.

But for $k \in \{0,\dots,n-1\}$ $\varphi$ is not in $(-\pi,\pi]$ it is in $[0,2\pi)$.

Is their argument correct even so?

• Surely they also covered that the period of $e^{i\varphi}$ is $2\pi$? "Modulo" this the proof is equivalent for either interval of $\varphi$. – hardmath Feb 13 '15 at 1:53
• It's a bit unclear what your question is. When you say the existence proof is understood, are you implying "uniqueness" of the $n$ roots is not understood? The machinery you describe certainly establishes (via "bijectivity") that uniqueness, but since $\mathbb{C}$ is a field, the purely algebraic proof that $z^n = 1$ has at most $n$ roots also applies. – hardmath Feb 13 '15 at 1:57

Yes, because you translate from one angle interval to the other without changing the scale, then the map $\Bbb R_{>0}\times [0;2\pi) \to \Bbb C\setminus\{0\}$ is just as sufficiently bijective as the map $\Bbb R_{>0}\times (-\pi;\pi] \to \Bbb C\setminus\{0\}$.

It's just the same angle with a different period.

Take $f: [0;2\pi) \to (-\pi; \pi]$ such that $f (\varphi) = \begin{cases} \varphi & : \phi \in [0; \pi]\\ \varphi-2\pi & : \phi \in (\pi; 2\pi)\end{cases}$

Yes it is correct, or rather, it can be saved by composing $\mathbb R_{\gt 0}\times (-\pi,\pi]\to\mathbb C \setminus \{0\}$ with map $M : (-\pi,\pi) \mapsto (0,2\pi] : M(x) = x + 2\pi * (1-\Theta(x))$ where $\Theta(x)$ is the Heavyside step function which is 1 for $x\geq 0$ and zero otherwise.

That composed map can be substituted in the proof without changing any other steps.

A simplistic proof:

suppose $z^n-1=0$ has a double root. (Let $n>1$ to avoid triviality) Then $z^n-1=(z-a_1)^2(z-a_2)\cdots(z-a_{n-1})$ .Now lets take derivative $$nz^{n-1}=2(z-a_1)(z-a_2)\cdots(z-a_{n-1})+(z-a_1)^2[(z-a_2)\cdots(z-a_{n-1})]'$$

Now lets evaluate at $z=a_1$ we find that $$na_1^{n-1}=0\,\,\!!!$$

Moral of the story is that every single root has to be simple and by fundamental theorem of algebra we should have exactly $n$ of them.so they are indeed distinct.