Factorisation of x^(2n+1)-1 in Real Domain

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:
 * $z^{2 n + 1} - 1 = \paren {z - 1} \displaystyle \prod_{k \mathop = 1}^n \paren {z^2 - 2 \cos \dfrac {2 \pi k} {2 n + 1} + 1}$

Proof
From Power of Complex Number minus 1:


 * $z^{2 n + 1} - 1 = \displaystyle \prod_{k \mathop = 0}^{2 n} \paren {z - \alpha^k}$

where:

From Complex Roots of Unity occur in Conjugate Pairs:
 * $U_{2 n + 1} = \set {1, \tuple {\alpha, \alpha^{2 n} }, \tuple {\alpha^2, \alpha^{2 n - 1} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k + 1} }, \ldots, \tuple {\alpha^n, \alpha^{n + 1} } }$

where $U_{2 n + 1}$ denotes the complex $2 n + 1$th roots of unity:
 * $U_{2 n + 1} = \set {z \in \C: z^{2 n + 1} = 1}$

The case $n = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $z - 1$.

Taking the product of each of the remaining factors of $z^{2 n + 1} - 1$ in pairs:

Hence the result.