Proving a result related to $100^{th}$ roots of unity. If we take the $100^{th}$ roots of unity ie. all complex roots of the equation $z^{100}-1=0$ and denote them as $\alpha_{1},\alpha_{2},...,\alpha_{100}$ then we are required to prove that $$\alpha_{1}^r+\alpha_{2}^r+...+\alpha_{100}^r=0$$ for $r\neq100k$ where $k$ is an integer.
I tried using the Euler form by denoting $$\alpha_{t}=e^{\frac{i2t\pi}{100}}$$ and trying to evaluate it as a GP but it was pretty lengthy and I couldn't simplify it out to zero so now I am confused.
Any help would be appreciated.
 A: A non-classical approach:
We know, $z^{100}-1=0$ will have it's sum of roots as $0$ (using Vieta's theorem; $\frac{-b}{a}$)
So, we know that $$\alpha_1 + \alpha_2 ...+\alpha_{100}=0$$
Another point to note is, $\alpha^r$ will have the same modulus as $\alpha$ which is $1$, but the angle in the complex plane will be $r$ times as it was before.
So, applying this here, we can be confident to say that,
$${\alpha_1}^r + {\alpha_2}^r ...+{\alpha_{100}}^r=0$$
where $r$ is not an integral multiple of $100$.
For a simplified case of how this works, a graph is attached
Number of roots=5

https://www.desmos.com/calculator/hgdiophi2c
Since $$\alpha_{t}=e^{\frac{i2t\pi}{100}}$$
$${\alpha_{t}}^{100k}=e^{{i2t\pi}k}=1$$
And each of ${\alpha_i}^r$ will jump to ${\alpha_{ir}}$ as can be seen from the graph or proved as per convenience. $\alpha_{ir}$ can be reduced recursively by reducing $ir$ by 100, i.e. $ir-100$, sometimes the value will be coincident but would always add up to 0.
For 100 roots, you can observe that the vertices do not change except when r=50 , 40 etc.

https://www.desmos.com/calculator/rbi5ybugm1
(Don't miss this graph, it's worth it)

Why are we certain that it will work?
1.Whenever $G.C.D(r,100)>1$ we will loose some vertices, Why?

This is because ${\alpha_{t}}^r=e^{\frac{i2t\pi r}{100}}$ would reduce to roots of unity of some lower power.
2.What happens when $G.C.D(r,100)=1$?

Converse to the previous statement, the roots no longer roots of unity of some lower power. Hence the vertices are retained. If unsatisfied (consider $a$ as $\alpha$) using modulus
$\left[a_{1},a_{2},a_{3},...,a_{[\frac{100}r]+1,...}\right]$ when raised to the power $r$
would shift to $\left[{a_{r},a_{2r},...,a_{r-1}...}\right]$ and the same manipulation can be
done for $a_{[\frac{100}{r^2}]+1,...}$ if applicable. (in this case, it would change to ${a_{r-2}}$ )

here square brackets denote the greatest integer function
One can rigorously prove that the Argand plane method provides a visual solution by iterating the mentioned points 1. and 2. for different lower powered roots of unity
A: So, the roots of $$z^n-1=0$$ are $$a_k=e^{2i\pi k/n},0\le k<n$$
$$\sum_{t=0}^{n-1}a_k^r=a_0\dfrac{a_1^{nr}-1}{a_1^r-1}=0$$
if $n\nmid r$
A: In general we have that
$$z^{n}-1=0 \iff (z-1)(1+z+z^2+\ldots+z^{n-1})=0$$
the for any $z\neq 1$
$$1+z+z^2+\ldots+z^{n-1}=0$$
and using $z=\alpha_1=e^{i\frac{2\pi}{n}}$ we obtain
$$\alpha_1+\alpha_2+\ldots+\alpha_{n-1}+\alpha_{n}=0$$
and since $\alpha_1^{100k+r}=\alpha_1^r$ we obtain the result.
