Sum of the powers of roots of unity

Question

I was reading this proof of the roots of unity filter and I point can't understand is why:

$$ S = \sum_{j=0}^{n-1} w^{kj} =0$$

If $ \omega $ is the nth root of unity and $k$ is some positive integer not equal to zero and n doesn't divide k

I understand that the sum of nth roots of unity are zero as in:

$$ S = \sum_{j=0}^{n-1} w^j = 0$$

But I can't understand the powers of them should be as well. The reason I find it strange is that the nth roots of unity solves the following polynomial equation

$$ P(x) =(x-1)( \sum_{i=0}^{n-1} x^i)$$

So why should it be that any of their powers solve the above equation as well?

They don't solve the same equation, but they solve a similar equation, following which you can use Vieta's formula. Basically, find a polynomial whose roots are exactly $\omega^{kj}, j = 0 ,...,n-1$, (should not be too difficult) and then use Vieta. — Sarvesh Ravichandran Iyer, Jan 03 '21 at 10:34
Hmm how would I find a polynomial who's roots are that? I could possibly write down the factors and multiply it but that seems too brute force @TeresaLisbon — tryst with freedom, Jan 03 '21 at 10:42

Sofía Marlasca Aparicio · Accepted Answer · 2021-01-03T10:42:02.103

5

You can use the typical trick for geometric series. You have that if

$$S = \sum_{j=0}^{n-1} w^{kj}$$

then

$$w^k S = \sum_{j=1}^n w^{kj} \implies S = \frac{1-w^{nk}}{1-w^k} = 0 $$

edited Jan 03 '21 at 10:42

answered Jan 03 '21 at 10:38

Sofía Marlasca Aparicio

3,238

How did you get that last equality? – tryst with freedom Jan 03 '21 at 10:39
$w^{n} = 1 \implies w^{nk} = (w^n)^k = 1^k = 1$ – Sofía Marlasca Aparicio Jan 03 '21 at 10:40
Ahh good point, thank you @marlasca23 – tryst with freedom Jan 03 '21 at 10:41
I don't quite understand what your "trick" is doing here. From the first sum, with $z=w^k$, you immediately get that $S=\sum_{j=0}^{n-1}z^j=\frac{z^n-1}{z-1}$, if $z\ne1$ of course. – Jean-Claude Arbaut Jan 03 '21 at 10:41
1

@Jean-ClaudeArbaut by "trick" I mean what would usually be the proof of the geometric series formula: multiply by the ratio and take the difference. – Sofía Marlasca Aparicio Jan 03 '21 at 10:42
@marlasca23 Ah, then I understand. I was taking this for granted as it's rather common, but if you need the proof, fine. – Jean-Claude Arbaut Jan 03 '21 at 10:43
@Jean-ClaudeArbaut I can never remember it so I always just do that, which is why I decided to include it – Sofía Marlasca Aparicio Jan 03 '21 at 10:45

score 1 · Answer 2 · answered Jan 03 '21 at 10:48

1

Here's the polynomial which solves it:

$$ 1 + x^k + x^{2k}..+x^{(n-1)k} = \frac{x^{nk} -1}{x-1}$$

Or,

$$ (x-1)\left[ 1 + x^k + x^{2k}..+x^{(n-1)k} \right]= x^{nk}-1 = P(x)$$

Clear to see power of nth roots of unity solves this equation

answered Jan 03 '21 at 10:48

tryst with freedom

11,538

score 0 · Answer 3 · answered Jan 03 '21 at 10:55

$w=e^{2i j\pi/n}, j=0,1,2,....n-1,$ $$S_n=\sum_{j=0}^{n-1} w^{k}= \sum_{j=0}^{n-1} e^{2\pi k j/n}~~~~(1) $$ Use sun of a finite GP, to get $$\implies S_n=\frac{e^{2i\pi k }-1}{e^{2i\pi k/n}-1}$$ If $k\ne pn$, $S_n=0$,

If $k=p n$ then $S_n=0/0$. So go (1) and put $k=pn$, where $p$ is an integer. We get $S_n=n$ when $k$ is an integer multiple of $n$.

Sum of the powers of roots of unity

3 Answers3