When reading a proof of why $x^3+y^3=z^3$ has no nontrivial integer solutions I came across following identity:

$$ y^3 = z^3-x^3 = (z-x)(z-\omega x)(z-\omega^2 x) \qquad \text{where } \omega = e^{2\pi i /3} \quad \text{i.e.}\quad \omega^3 = 1$$

Expanding the RHS results in:

$$ z^3-(1+\omega+\omega^2)z^2x+(\omega+\omega^2+\omega^3)zx^2-\omega^3x^3 = z^3-x^3,$$

since obviously $\omega+\omega^2+\omega^3 = 1+\omega +\omega^2 = 0$. Well then I thought about how obvious that is. I mean geometrically it is obvious that the sum of all $n$-th unity roots must equal $0$, but is there an analytical proof? I was not able to come up with one straight away.

  • 5
    $\begingroup$ Given any polynomial, the second coefficient is the sum of the roots of the polynomial. If we take $p(X) = X^n - 1$, then its roots are the $n^{th}$ roots of unity, and the second coefficient is the coefficient of $X^{n-1}$, which is $0$ as long as $n>1$. This is really Vieta's formula as mentioned below, but I find it easier to state it this way. $\endgroup$ Aug 9 '14 at 13:05

I think I just found one more time the answer myself just after submitting the question, it is so simple...

Let $\omega = e^{2 \pi i / n}$ which implies $\omega^n = 1$.

$$ 1 + \omega + \omega^2 + \ldots + \omega^{n-1} = \frac{\omega^n-1}{\omega-1} = 0 $$


Also consider $$\omega S=\omega(1+\omega+\omega^2)=\omega+\omega^2+\omega^3=\omega+\omega^2+1=S.$$ Unless $\omega=1$, $S=0$.

You needn't know the summation formula for geometric progressions.

  • $\begingroup$ You can easily extend this to a proof for the geometric progression. $\endgroup$
    – Christoph
    Jun 8 '20 at 9:18
  • 1
    $\begingroup$ This answer is the best one. Purely visual and conceptual. No algebra. $\endgroup$
    – Smithey
    Apr 1 at 19:24

Nongeometricrally, nth-roots of unity are the solutions to the equation $x^n - 1 = 0$. The $x^n$ coeff is $1$ and the $x^{n-1}$ coeff is $0$, so the sum of the roots is zero.

Geometrically, the n-th roots of unity are equally spaced vectors around a unit circle, so their sum is the center of the circle, which is $0 + 0i$.


Let $S$ denote the sum of the $n$ roots of unity. We have

$$\exp\bigg(\frac{2{\pi}i}{n}\bigg)S = \exp\bigg(\frac{2{\pi}i}{n}\bigg) \sum_{a=0}^n \exp\bigg(\frac{2{\pi}ia}{n}\bigg)$$

$$= \sum_{a=0}^n \exp\bigg(\frac{2{\pi}i(a+1)}{n}\bigg)$$

$$= \sum_{b=0}^n \exp\bigg(\frac{2{\pi}ib}{n}\bigg), \; b=a+1$$

$$= S$$

Because $a+1$ is just a cyclic shift of the roots, the sum still contains the same terms. So we have shown that

$$\exp\bigg(\frac{2{\pi}i}{n}\bigg)S = S$$

and therefore $S = 0$.

Terras, A. (1999). The Discrete Fourier Transform on the Finite Circle ℤ/nℤ. In Fourier Analysis on Finite Groups and Applications (London Mathematical Society Student Texts, pp. 30-45). Cambridge: Cambridge University Press. doi:10.1017/CBO9780511626265.004


You know that $$\omega = e^{\frac{2\pi i}{n}}$$

$$\Longrightarrow \omega^n = 1$$

Now if your sum (the one of all the roots up to n-1) is S, you can pose :

$$1+\omega S = \omega^n = 1$$

$$1+\omega S = 1$$

$$\omega S = 0$$

$$S = \frac{0}{\omega} = 0$$

Seems pretty ghetto but i'm fairly sure it's legit.


This is just a more general rewriting of Yves Daoust's answer:

$$ \begin{align*} S&=\sum_{k=0}^{n}\omega^{k}\\\\ \omega\cdot S&=\sum_{k=0}^{n}\omega^{k+1}\\\\ &=\omega+\sum_{k=1}^{n-1}\omega^{k+1}+\omega^{n+1}\\\\ &=1+\omega+\sum_{k=2}^{n}\omega^{k}=\sum_{k=0}^{n}\omega^{k}\\\\ &=S\\\\ \implies S&=0 \end{align*} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.