# Roots of Unity

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

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

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

Then the Lagrange resolvent is

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

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

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

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

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For \(i = 1,...,10\) let \(t_i\) be equal to \(t\) where every \(\beta\) has been replaced by \(\beta^i\).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

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.

*blynn@cs.stanford.edu*💡