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Let $\varphi_k\in\mathbb{C}$ be a primitive root of $1$. It turns out, that $$\varphi_k^1+\ldots+\varphi_k^k=0\text{ .}$$ If I draw the roots for some fixed $k$, I can see that this seems evident. For every root there is a complex conjugated one, so that the imaginary parts cancel out (except for $1+0i$, but this doesn't matter). But why do the real parts always sum to $0$?

Example $k=3$: $$\varphi_3:=e^{i\frac{2}{3}\Pi}$$ $$\begin{split}\Rightarrow\varphi_3^1+\varphi_3^2+\varphi_3^3&=e^{i\frac{2}{3}\Pi}+e^{i\frac{4}{3}\Pi}+1\\&=\cos\left(\frac{2}{3}\Pi\right)+i\sin\left(\frac{2}{3}\Pi\right)+\cos\left(\frac{4}{3}\Pi\right)+i\sin\left(\frac{4}{3}\Pi\right)+\cos\left(2\Pi\right)+i\sin\left(2\Pi\right)\\&=-\frac{1}{2}+i\frac{\sqrt{3}}{2}-\frac{1}{2}-i\frac{\sqrt{3}}{2}+1+0\\&=0\end{split}$$

Rob
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1 Answers1

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Notice that $(x-a_1)(x-a_2)\cdot\ldots\cdot(x-a_n)=x^n-(a_1+a_2+\ldots+a_n)x^{n-1}+\ldots$. So if you are solving a polynomial equation of form $x^n-1=0$, then clearly the sum of all roots is the coefficient of $x^{n-1}$ with opposite sign, but in this case it is zero.

Stefan Gyürki
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