Miscellaneous Mathematics

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 $p=\mathrm{2,3,5,7}$ 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.

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For $i=\mathrm{1,...,10}$ let $t_i$ be equal to $t$ where every $\beta$ has been replaced by ${\beta }^i$.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