Concerning property of roots of unity

Question

Let $e_k = e^{2\pi i \frac{k}{17}}\;$ for $ 1 \le k \le 16$. How do I prove the following $$\sum_{k=1}^{k=16}e_k = -1.$$ The only thing I know is that $e_k$ are roots of the following $$\frac{x^{17}-1}{x-1}=x^{16}+x^{15}+.....+1.$$ But I don't know how to proceed further.

Answer 1

Do you know the formula relating the sum of the roots of a polynomial to its coefficients? It will make the problem quite easy, using the polynomial you've cited in your question.

Answer 2

Hint: Show that for a fixed $k$ and $i\in\{1,2,\ldots,16\}$ if, $$ik\equiv r_i\pmod {17}$$ then $r_i\ne r_j$ if $i\ne j$. Then use the fact that $e_k$ is a root of the equation $$\dfrac{x^{17}-1}{x-1}=0$$

Answer 3

We can generalize the above statement. Let $n\in \mathbf N$ be arbitrary and set $e_k:=\exp\left\{2\pi \textrm{i} \frac{k}{n} \right\}$ for $1\leq k\leq n$. Then we have $$1+\sum_{k=1}^n e_k=\sum_{k=0}^n e_k=\frac{e_k^n-1}{e_k-1}=0$$ and therefore $$\sum_{k=1}^n e_k=-1$$ where we used $$e_k^0=1, \qquad e_k^n=1 \qquad \text{and} \qquad e_k-1\neq 0 \ \text{for all } 1\leq k\leq n.$$