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

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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:

\(\ds z^{2 n} + 1\) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} } \prod_{k \mathop = n}^{2 n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} }\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} } \paren {z - \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} }\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} } + \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} }\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} } + 1}\) simplifying
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} } + 1}\) Euler's Formula
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} {2 n} + 1}\)


Hence the result.

$\blacksquare$


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