# Find $\sum_{i=0}^{k-1}w^{in}$ in terms of $n \in \Bbb N$ being $w \in \Bbb C$ a k-th primitive root of unity.

I need some help with the following problem:

Find $\sum_{i=0}^{k-1}w^{in}$ in terms of $n \in \Bbb N$ being $w \in \Bbb C$ a k-th primitive root of unity.

I thought of writing it as: $\frac{w^{nk}-1}{w^n-1}$ and then replacing $w$ with its exponential form $e^{\frac{2 q\pi i}{k}}$. Is this valid? $k$ would cancel out and then all powers would have a factor of $2 \pi$, equaling 1, right? Well, as you can see I have many doubts.

Any suggestions or conceptual remarks maybe?

Thanks!

It is a geometric progression. We have $$\sum_{j=0}^{k-1}w^{jn} = \frac{(w^n)^{k}-1}{w^n-1} = \frac{w^{nk}-1}{w^n-1} = 0,$$since $w^{nk} = (w^k)^n=1^n=1$ and $1-1=0$. The numerator vanishes.

• So I was right, thanks! – jrs Dec 11 '16 at 1:35
• @marty I don't follow. Can you elaborate please? :-) – Ivo Terek Dec 11 '16 at 1:41
• Naw. I wrote too quickly. – marty cohen Dec 11 '16 at 1:42