Roots of Unity

The \(n\)th roots of unity are given transcendentally by

\[ cos(2\pi k/n) + i sin(2\pi k/n) \]

Algebraic expressions take a little more effort to derive. Note that the general problem can be reduced to finding the \(p\)th roots of unity for every prime \(p\). The small cases \(p = 2,3,5,7\) are easy exercises (for \(p = 7\), after exploiting the fact that the polynomial is palindromic, a cubic must be solved). The prime \(p = 11\) is the smallest case that requires a different approach. The following method is due to Vandermonde, and it generalizes to all primes greater than eleven. (Compare with Gauss' method.)

We wish to solve the equation

\[x^{10} + x^9 + ... + 1 = 0 \]

Let \(\beta\) be a tenth root of unity. Pick a primitive root of 11, such as 2, and place the roots in the order

\[\alpha , \alpha^2, \alpha^4, \alpha^8, \alpha^5, \alpha^{10}, \alpha^9, \alpha^7, \alpha^3, \alpha^6 \]

Then the Lagrange resolvent is

\[ t = \alpha + \beta \alpha^2 + ... + \beta^9 \alpha^6 \]

We show that \(t^{10}\) is known. Let

\[ t^{10} = \rho_0(\beta) + \rho_1(\beta)\alpha +...+ \rho_{10}(\beta)\alpha^{10} \]

where the \(\rho_i\)'s are rational functions with known coefficients. Now if we replace \(\alpha\) by \(\alpha^2\), then \(t\) becomes \(\beta^{-1}t\), hence \(t^{10}\) is unchanged, which means

\[ \rho_0 + \rho_1\alpha^2 + ... + \rho_{10} \alpha^6 = \rho_0 + \rho_1\alpha +...+ \rho_{10}\alpha^{10} \]

where the \(\beta\)'s have been omitted for clarity. Thus

\[ 0 = (\rho_1 - \rho_{10})\alpha + (\rho_2 - \rho_1)\alpha^2 +...+ (\rho_{10} - \rho_9)\alpha^6 \]

But since \(\alpha ,..., \alpha^{10}\) are linearly independent (otherwise \(x^{10} +... + 1 = 0\) would be reducible), we must have \(\rho_1 - \rho_{10} = 0, ...\), that is, \(\rho_1 = \rho_2 = ... = \rho\) for some \(\rho\). Then

\[ t^{10} = \rho_0 + \rho(\alpha + ... + \alpha^{10}) = \rho_0 - \rho \]

which is independent of \(\alpha\) and thus known.

For \(i = 1,...,10\) let \(t_i\) be equal to \(t\) where every \(\beta\) has been replaced by \(\beta^i\). Then

\[ \alpha = \frac{t_1 + ... + t_{10}}{10} = \frac{\sqrt[10]{t_1^{10}} + ... + \sqrt[10]{t_{10}^{10}}}{10} \]

A similar argument to the one used above shows that \(t_i^{10}\) is known for all \(i\), which allows us to solve for \(\alpha\). However, this requires us to choose the correct 10th root 10 times. Instead, we can prove that \(t_i t_1^{10-i}\) is known using a similar argument, which means that once \(t_1\) has been chosen, the other \(t_i\)'s can be determined.


Ben Lynn blynn@cs.stanford.edu 💡