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

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

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

Proof
From Power of Complex Number minus 1:


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

where:

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

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

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

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

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

Hence the result.

Also see

 * Factors of Difference of Two Even Powers