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

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

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

Proof
From Factorisation of $z^n + 1$:


 * $(1): \ds \quad z^{2 n} + 1 = \prod_{k \mathop = 0}^{2 n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} }$

From Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs, the roots of $(1)$ occur in conjugate pairs.

Hence we can express $(1)$ as:

Hence the result.

Also see

 * Factors of Sum of Two Even Powers