complex numbers - Why is the sum of the first $k$ powers of a $k$-th primitive root $varphi_k$ of $1$ always $0$?

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 $k=3$:

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