# 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!

• – lhf
Feb 28, 2021 at 16:53

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 $$r_k^n=e^{2i\pi k}=1$$.
$$P=\prod _{k=1}^n r_k = \prod _{k=1}^n e^{2i\pi k / n}= e^{2i\pi / n \sum_{k=1}^n k}= e^{(2i\pi / n) \frac12 n(n+1)}$$ $$= e^{i\pi (n+1)}= (-1)^{n+1} = (-1)^{n-1} \;\;\blacksquare$$
$$S=\sum_{k=1}^n r_k = \sum_{k=1}^n e^{2i\pi k / n}$$ $$= e^{2i\pi / n} \frac{1-(e^{2i\pi / n})^n} {1-e^{2i\pi / n}} = e^{2i\pi / n} \frac{1-1} {1-e^{2i\pi / n}} = 0 \;\;\blacksquare$$
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
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.