Prove that $\sum_{k=0}^{6} \zeta_7^k= 0$.

Question

I have tried this complex analysis problem for hours on my own but am still stumped.

Suppose $\zeta_7 = e^{2\pi i/7} = \cos(2\pi/7)+i\sin(2\pi/7)$. Prove that $$\sum_{k=0}^{6} \zeta_7^k = 0.$$

I have so far obtained that $$\sum_{k=0}^{6} (\zeta_7)^k = 1 + (\zeta_7) + (\zeta_7)^2 + (\zeta_7)^3 +(\zeta_7)^4 +(\zeta_7)^5 + (\zeta_7)^6,$$ but that's as far as I get.

Do you know how to sum a finite geometric series $1+r+r^{2}+\ldots+r^{n}$? — Brian Borchers, Jul 24 '19 at 01:06
MathJax would make your question look nicer. Did you want to put $k$ in the title? — J. W. Tanner, Jul 24 '19 at 01:22
I do, but I've tried it for this specific problem and am still confused. — Rey, Jul 24 '19 at 01:26

Mohammad Riazi-Kermani · Answer 1 · 2019-07-24T01:51:45.867

7

Note that $$\zeta ^7 = \cos 2\pi +i \sin 2\pi =1$$

Therefore, $$ \zeta ^7-1=0$$

That is $$ (\zeta -1 )(\zeta ^6 + \zeta ^5 +...+\zeta +1) =0$$

Since $\zeta \ne 1$ we have $\zeta -1 \ne 0$.

Thus $$\zeta ^6 + \zeta ^5 +...+\zeta +1 =0$$

edited Jul 24 '19 at 01:51

answered Jul 24 '19 at 01:32

Mohammad Riazi-Kermani

68,728

score 4 · Answer 2 · answered Jul 24 '19 at 01:25

4

Because $$\sum_{k=0}^6\zeta^k=\frac{\zeta^7-1}{\zeta-1}=0.$$

answered Jul 24 '19 at 01:25

Michael Rozenberg

194,933

score 1 · Answer 3 · answered Jul 24 '19 at 03:02

The equation $X^7-1=0$ has no repeated roots (can be checked by looking at the derivative of $X^7-1$) and $\zeta_7$ is one of the roots,

The 7 numbers, $\zeta_7^k,k=0,1,\ldots, 6$ are also solutions of the same equation.

The have to be all the solutions, as we are dealing with the a 7th degree equation.

Your question requires finding the sum of all the roots of the equation.

For any equation of degree $n$ sum of all its (real/complex) roots (counting multiplicity) is the negative of the coefficient of $X^{n-1}$, In the case we are discussing $X^6$ term is not present hence the sum is zero.

Prove that $\sum_{k=0}^{6} \zeta_7^k= 0$.

3 Answers3