Sum of nth Roots of 1: Solving for k = 1 to n-1

[Benny]

Hi, can someone please help me with the following question?

Q. Let $\omega _0 ,...,\omega _{n - 1}$ be the nth roots of 1. Show that

$$\sum\limits_{j = 0}^{n - 1} {\omega _j ^k } = \left\{ {\begin{array}{*{20}c} {0,1 \le k \le n - 1} \\ {n,k = n} \\ \end{array}} \right.$$

The case of k = n is fairly easy but I don't know where to start when I attempt the case of 1 <= k <= n-1. For k = 1 I can use a previous result but for 'most' cases I can't. Any help would be good thanks.

Edit: For k =1, the result follows from a previous part of the question and k = n is an easy case. For 2 <= k <= n-1 I just said $\omega _j ^k \ne \omega _j$ but (w_j)^k must be equal to one of the other roots. For clarity, I considered (w_0)^k != w_0 for 2 <= k <= n-1 and then said (w_0)^k must be equal to one of the other roots. Applying the same reasoning to the other w_j and using that the sum of the roots is equal to zero leads to the given result.

That's about all I've been able to come up with. That's not correct because one of the nth roots of 1 is 1 and 1^2 = 1. I just thought that I'd put something up so that someone will help me out.

 

[AKG]

There are 6 sixth roots of 1. Let $w_j = \exp (2\pi ji/6)$. Now what happens if you, say, sum the squares of the roots. The 1st and 4th roots go to the 2nd, the 2nd and 5th roots go to the fourth, and the 0th and 3rd go to the 0th. So you're summing:

2 x (2nd root) + 2 x (4th root) + 2 x (0th root)
= 2 (2nd + 4th + 0th)

But the 2nd, 4th, and 0th roots are just the third roots of 1 (i.e. the solutions to x³ = 1). In particular, you get:

2 x (the sum of the first powers of the third roots of 1)

You've already proved that the sum of first powers of roots of any order is 0, so you get 2 x 0 = 0.

Prove that, in general, the sum of kth powers of the nth roots of unity equals m x (the sum of the first powers of the pth roots of unity) where r > 1, and you should get something like p | k, or p | n, or m | n, or m | k, or some combination thereof. I'm sure that if you think about it clearly for 10 seconds, you'll figure out what combination - I have to run to breakfast so I can't do it for you right now.

 

[Benny]

Thanks for the help, I'll see what I can come up with.

 

[Benny]

I'm still having trouble with this question. I considered some more specific cases; summing the cubes of the 6th roots of unity and I found the sum in this case to be 3(sum of square roots of 1) = 3(0) = 0. I also considered the sum of the squares of the 5th roots of unity but I couldn't find a simple relation between that sum and the sum of the 5th roots of unity.

I then considered the general case:

$$\omega _j ^k = \exp \left( {ik\left[ {\frac{{2j\pi }}{n}} \right]} \right),1 \le j,k \le n - 1$$

But I can't think of what to do with it. I've thought about adding integer multiples of 2pi and other things like that but I haven't been able to put anything substantial together.

 

[AKG]

The squares of the fifth roots of unity are just the fifth roots of unity:

1st -> 2nd
2nd -> 4th
3rd -> 6th (mod 5) = 1st
4th -> 8th (mod 5) = 3rd
0th -> 0th

If a given root makes an angle of X, then taking it's kth power will make an angle of kX (mod 2pi).

In general, if you are looking at the nth roots of unity, and of those n such roots, you are looking at the jth (labeling them in order counterclockwise, w_j = e(ij2$\pi$/n). Then the kth power of w_j will be the w_{kj (mod n)}.

let m = gcd(k,n). Then looking at the kth powers of the nth roots will give you m*(the sum of the (n/m)-th roots of unity).

So your question is to prove an equation. On the left side you have a sum, on the right side, you have a brace bracket with 0 in some cases and 1 in others. Consider the expression:

$$m\sum_{\omega^m = 1}\omega$$

where m = gcd(k,n). Prove that this expression equals the left side of your given equation, and also prove separately that it equals the right side.

 

[Benny]

Thanks for the help AKG, hopefully I can get it now.