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For $i = 1,...,10$ let $t_i$ be equal to $t$ where every $\beta$ has been replaced by $\beta^i$.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