Show that $\prod_{i=1}^{n} r_i = (−1)^{n−1}$ and $\sum_{i=1}^{n} r_i = 0.$ Suppose that $n\geq 2$ is a natural number and $r_1 . . . , r_n$ are all the distinct $n$th roots of 1.
Show that $\prod_{i=1}^{n} r_i = (−1)^{n−1}$ and $\sum_{i=1}^{n} r_i = 0.$
I am unsure of how to attempt this question any advice or help will be great thanks!
 A: We can write the n-th roots of unity as $  r_k=e^{2i\pi k / n}$ for $1 \leqslant k \leqslant n $. Proof: they each have different arguments so are distinct and raising them to the power of $n$ gives $r_k^n=e^{2i\pi k}=1$.
$$P=\prod _{k=1}^n r_k = \prod _{k=1}^n e^{2i\pi k / n}= e^{2i\pi / n  \sum_{k=1}^n k}= e^{(2i\pi / n)  \frac12 n(n+1)}$$
$$= e^{i\pi (n+1)}= (-1)^{n+1} = (-1)^{n-1} \;\;\blacksquare$$
$$S=\sum_{k=1}^n r_k = \sum_{k=1}^n e^{2i\pi k / n}$$
$$=  e^{2i\pi / n} \frac{1-(e^{2i\pi / n})^n} {1-e^{2i\pi / n}} = e^{2i\pi / n} \frac{1-1} {1-e^{2i\pi / n}} = 0  \;\;\blacksquare$$
A: Suppose $a,b,c$ are $3$rd roots of $1$. In other words they are solutions of
$$x^3-1=0\tag{1}$$
Then the following holds
$$(x - a) (x - b) (x - c)=0$$
expand and collect $x$
$$x^3-x^2 (a+b+c)+x (a b+a c+b c)-a b c=0$$
Compare with $(1)$. We have
$$a+b+c=0;\;abc=1$$
If the exponent is even, let's say $4$, then in a similar way we can prove that
$$a+b+c+d=0;\;abcd=-1$$
Hope this helps
A: Note that all roots of unity can be written as potences of one root of unity: W. l. o. g., $$r_i = r_1^i.$$
You can insert this into the product and apply Gauss' summation formula.
This proves at least the first equality.
