<span class="math">\(z^{n}=1\)</span>

Say we want the roots of \(z^{n}=1\).

Let \(z=re^{i\theta}\). Then \(z^{n}=r^{n}e^{in\theta}=1\). We immediately see that \(r=1\). Thus this boils down to solving \(e^{in\theta}=1\).

We have:

\begin{equation*} \cos n\theta+i\sin n\theta=1 \end{equation*}

This gives us \(n\theta=2\pi k\) where \(k=0,\dots,n-1\). Note that the \(k=0\) is just the root 1.

Thus, the roots are:

\begin{equation*} e^{2\pi ki/n}=\cos\left(\frac{2\pi k}{n}\right)+i\sin\left(\frac{2\pi k}{n}\right),k=0,\dots,n-1 \end{equation*}

Note that these are the points on the unit circle, equally spaced apart, starting from 1.

The sum of all the roots is 0

Claim: The sum of all the roots is 0.

To see this, note that \(z^{n}-1=\left(z-1\right)\left(z^{n-1}+z^{n-2}+\cdots+z+1\right)=0\). The first expression gives us \(z=1\). Thus all the other roots satisfy \(z^{n-1}+z^{n-2}+\cdots+z+1=0\). Let’s examine these terms. Let \(\omega\) be a complex (non-real) root. Let’s start with 1. Multiplying it by \(\omega\) gives us the first root. Multiplying again with \(\omega\) gives us \(\omega^{2}\) - but this is simply \(\omega\) rotated by the same angle - in other words \(\omega^{2}\) is also one of the roots (as are all powers of \(\omega\).

Thus, given a complex root \(\omega\), we have \(\omega^{n-1}+\omega^{n-2}+\cdots+\omega+1=0\).

As a corollary, we have

\begin{equation*} \sum_{k=1}^{n-1}\cos\left(\frac{2\pi k}{n}\right)=-1 \end{equation*}

Note that if \(z\) is a root, then so is \(\overline{z}\), and thus all the imaginary parts cancel out. They have to, even if you don’t note this as the sum is real.

Below is an example using Sage:

n = 11
s = sum(cos(2*pi*k/n) for k in range(1, n))
print(s)
s.n()

Example Of Root Finding With Sage

x = var('x')
solve([x^3==1], x)