Suppose $z=e^{\frac{2\pi i}m}$ for $m\in\mathbb N$ and $m\neq1$. Is the following equality hold? $$\sum_{k=1}^mz^k=0\tag{1}$$
$(1)$ seems trivial geometrically ; it says that the sum of all vectors with equal magnitudes and uniform angle differences should be zero.
If $m$ is even ($m=2l$), then $$z^l=e^{i\pi}=-1$$ and $$z^k+z^{k+l}=z^k(1+z^l)=0$$ for every $k=1,2,\cdots,l$. So $(1)$ holds.
If $m$ is odd, then $(1)$ seems nontrivial. For $m=3$, the question becomes straightforward. If $m=5$, it is quite sophisticated but solvable by using elementary mathematics ; Let $\theta=\frac25\pi$. Then $(1)$ is equivalent to $(2)$ and $(3)$ where $$\sum_{k=1}^5\cos k\theta=0\tag{2}$$ and $$\sum_{k=1}^5\sin k\theta=0\tag{3}$$ $(3)$ is trivial from $\sin(2\pi-\phi)=-\sin\phi$. For $(2)$, we can use $\cos(2\pi-\phi)=\cos\phi$ and compute as follows \begin{align*} \sum_{k=1}^5\cos k\theta &=\cos\theta+(2\cos^2\theta-1)+(2\cos^2\theta-1)+\cos\theta+1\\ &=4\cos^2\theta+2\cos\theta-1 \tag{4} \end{align*} On the other hand, (denote $s=\sin\frac\pi{10}$ and $c=\cos\frac\pi{10}$) $$2sc=\sin\frac\pi5=\cos\left(\frac\pi2-\frac\pi5\right)=\cos\frac{3\pi}{10}=4c^3-3c$$ Dividing both sides by $c(\neq0)$ yields \begin{gather*} 2s=4c^2-3\\ 4s^2+2s-1=0\tag{5} \end{gather*} Since $s=\cos\theta$, the last expression of $(4)$ becomes zero, and this proves $(2)$.
So $(1)$ holds for every even $m$ and $m=3,5$. But what if $m$ is an odd number grater than $5$?. Is there a general way to explain $(1)$?
And if it is the case, can I conclude $(2)$ and $(3)$, $m$ in place of $5$? i.e. do $$\sum_{k=1}^m\cos\frac{2\pi k}m=0\tag{2*}$$ and $$\sum_{k=1}^m\sin\frac{2\pi k}m=0\tag{3*}$$ hold for every $m\in\mathbb N\setminus\{1\}$?
$\sum\limits_{k=0}^{m-1}z^{k}=\frac {z^{m}-1}{z-1}=0$ because $z^{m}=1$. Now just multiply both sides by $z$.