Factorisation of z^(2n+1)+1 in Real Domain
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Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
- $\ds z^{2 n + 1} + 1 = \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} {2 n + 1} + 1}$
Proof
From Factorisation of $z^n + 1$:
- $(1): \quad \ds z^{2 n + 1} + 1 = \prod_{k \mathop = 0}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }$
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:
| \(\ds z^{2 n + 1} + 1\) | \(=\) | \(\ds \prod_{k \mathop = 0}^n \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \prod_{k \mathop = n + 1}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {z - \exp \dfrac {\paren {2 n + 1} i \pi} {2 n + 1} } \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \prod_{k \mathop = n + 1}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {z - e^{i \pi} } \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \prod_{k \mathop = n + 1}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \paren {z - \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} } + \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} } + 1}\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n + 1} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n + 1} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n + 1} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n + 1} } + 1}\) | Euler's Formula | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} {2 n + 1} + 1}\) | simplifying |
$\blacksquare$