Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: $\sqrt[n] 1 = \sqrt[n]r\left(\cos\frac{2\pi }{n} + i\sin\frac{2\pi k}{n}\right)$

share|cite|improve this question
If $w^n = 1$ then $(w^k)^n = w^{kn} = (w^n)^k = 1^k = 1$ for any integer $k$. Thus $w^0,\dots,w^{n-1}$ are all $n$th roots of unity. – KCd May 19 '12 at 3:09
And $r = 1$, because the norm has to be $1$. – Arturo Magidin May 19 '12 at 3:12
up vote 3 down vote accepted

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.

share|cite|improve this answer
1)What is "x"? Is it just a random integer? – user26649 May 19 '12 at 3:40
@FarhadYusufali: $x$ in $x^n-1$ is the variable of a polynomial. The "indeterminate". $x^n-1$ is a polynomial, not an integer, not random. You are familiar with polynomials, are you not? – Arturo Magidin May 19 '12 at 3:40
Of course. I do not understand what you did after you came to the conclusion$w^r = w^s$ if $w^{r-s} =1$. Could you expand on that? – user26649 May 19 '12 at 3:45
Note: that a polynomial of degree $n$ has at most $n$ roots is not a consequence of the Factor Theorem alone. It also requires that the coefficient ring be a domain. Indeed, the Factor Theorem holds over any ring. But the root bound does not, e.g. $x^2-1$ has $4$ roots $\pm1,\pm3\:$ over $\mathbb Z/8.\ $ – Bill Dubuque May 19 '12 at 3:57
@Farhad: Note that I explicitly state that $\omega$ must be such that $\omega^n=1$, and $\omega^k\neq 1$ for all $k$, $1\leq k\lt n$. The only value of $n$ for which $\omega=1$ satisfies this condition is $n=1$. For $n=4$, the only $4$th roots of unity that satisfy this condition are $i$ and $-i$, and both have the corresponding property. Neither $1$ (which has $1^1=1$) nor $-1$ (which has $(-1)^2=1$) do. All conditions are important, not just some of them! – Arturo Magidin May 22 '12 at 2:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.