Proof that sum of complex unit roots is zero

Solution 1:

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 $\omega^n = 1$.

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

Solution 2:

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.

Solution 3:

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 $0 + 0i$.

Solution 4:

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

Solution 5:

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.