Sum of increasing powers

Question

In the answer to a different question someone wrote:

Let $\omega = e^{2 \pi i / n}$ which implies $\omega^n = 1$. $$ 1 + \omega + \omega^2 + \ldots + \omega^{n-1} = \frac{\omega^n-1}{\omega-1} = 0 $$

I tried to understand this for at least two hour now, and thought, that it maybe had something to do with geometric series but I can't quite figure it out.

I think what's confusing me most, is that in other places (Wikipedia and yet another math.stackexchange question) the formula for geometric series looks like this:

for $N$ finite $$ \sum_{n=0}^Nar^n=a\frac{1-r^N}{1-r},\:\:\:\: r\ne1$$

So I don't really see, why the intermediate result is

$$\frac{\omega^n-1}{\omega-1}$$

instead of

$$\frac{1-\omega^n}{1-\omega}$$

Answer 1

Observe that $$(\omega-1)(1+\omega+\omega^2+\ldots+\omega^{n-1})=$$ $$=(\omega+\omega^2+\omega^3+\ldots+\omega^n)-(1+\omega+\omega^2+\ldots+\omega^{n-1})$$ All the terms cancel out except $\omega^n-1$, so $$(\omega-1)(1+\omega+\omega^2+\ldots+\omega^{n-1})=\omega^n-1 \Leftrightarrow 1+\omega+\omega^2+\ldots+\omega^{n-1}=\frac{\omega^n-1}{\omega-1}$$

Answer 2

This is exactly a geometric series of n terms and of second term $\omega$. Replacing $\omega^n = 1 \rightarrow S(n) = 0$.

Answer 3

The simplest way to see this is noticing that $\omega$ is a solution of $x^n-1 =0$, which can be factorized like $(x-1)(1+x+\ldots+x^{n-1})$, and so as $\omega$ is not $1$ then $1+\omega + \ldots + \omega^{n-1}=0$