Let $z \in \mathbb C$, show using de Moivre's that the product of the roots of $z^n=1$ is $1$ when $n$ is odd.

Question

I am taking a course which introduces complex numbers and some of the theorems relating to them. I wish to prove the following result:

Prove, using de Moivre's Theorem, that the product of the roots of $$z^n=1$$ is $1$ when $n$ is odd for $z \in \mathbb C$.

I do not see how de Moivre is very helpful here. Any ideas, since De Moivre only introduces a sine and cosine, I do not see how this helps met get farther. $$\cos(n\pi) + \sin(n\pi)i =1. $$

Answer 1

Using Vieta's formula on $$z^n-1=0,$$ We get that: $$P:=\prod_{r=0}^{n-1}z_r=(-1)^n\cdot\dfrac{-1}1= (-1)^{n+1}.$$ Where $z_r$ denotes the complex roots of unity.

If $n=2k+1$ for $k\in \mathbb Z$ we see using the rule $(x^a)^b= x^{ab}$: $$ P=(-1)^{2k+2}= \left((-1)^2 \right)^{k+1}= 1.$$

Answer 2

By looking at $z^n=1$ and taking the complex norm on both sides we see $|z^n|=|1| \implies |z|^n =1 \implies |z|=1$ we see that all solutions lie on the unit circle. This is why they are called the complex roots of unity. All that we need to determine are the arguments that satisfy this equation, so for $z= \cos(\theta) + i \sin (\theta)$ we want to know which $\theta$ solve the equation $z^n =1$?

Now it turns out that the roots are $\{ e^{\pi i k/n} | k \in \{1, \dots, n \} \}$ a result that follows from de Moivre's by considering the numbers of solutions of the real sine and cosine on the interval $[0,2\pi]$. Look for instance here. Essentially one wishes to look at: $$ z^n =1 \implies(\cos(n \theta)+ \sin(n\theta)i)^n=1.$$ Or using de Moivre we get: $$ \cos(n\theta) + \sin(n\theta)i=1+0i.$$ This gives us two very real equations, we pick one of them for instance $\cos(n\theta)=1$, this is equivalent to $\sin(n\theta)=0$. We see that we get equality when $n \theta= 2\pi k $ for $k\in \mathbb Z$, or equivalently $\theta= \frac{2 \pi k i}{n} $. The corresponding arguments on the interval of $[0,2\pi]$ are then given by: $$ \frac{2 \pi}{n}, \frac{4 \pi}{n}, \frac{6 \pi}{n}, \dots, \frac{2n \pi}{n}=1.$$ Each argument corresponds to the vertex of a regular $n$-gon that represents the solutions of the $n$-roots unity. We now have all information about the argument and the modulus/length of these numbers. See the picture below for $n=6$.

Since $n=6$ is even the product of these numbers in polar form becomes: $$ e^{\frac{2\pi i}{6}} e^{\frac{4\pi i}{6}}e^{\frac{6\pi i}{6}}e^{\frac{8\pi i}{6}}e^{\frac{10\pi i}{6}}e^{\frac{12\pi i}{6}}= e^{\frac{42\pi i}{6}}=e^{7 \pi i}=-1. $$ We will now look at the general product of complex roots of unity. We again use the exponential law $e^{a+b}=e^a \cdot e^b$, this conveniently changes a product into a sum of the arguments as observed above for the case $n=6$. $$ \prod_{k=1}^{n} e^{ 2\pi k i/ n}= e^{\sum_{k=1}^n 2\pi k i/ n }=e^{ 2 \pi i/n\sum_{k=1}^n k } .$$ Now observe that for this last term we have $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ which gives: $$ \prod_{k=1}^{n} e^{ 2\pi k i/ n}=e^{ \frac{2 \pi i}{n} \frac{n (n+1)}{2} }= e^{\pi(n+1)i }= \left( e^{\pi \ i}\right) ^{n+1}=(-1)^{n+1}.$$ If $n$ is odd, this gives $1$, if $n$ is even this gives $-1$.