If $z=e^{\frac{2\pi i}m}$, then $\sum_{k=1}^mz^k=0$ ($m\neq1$) 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 $z$.
