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.
$\blacksquare$