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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then:

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


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:

\(\ds \alpha\) \(=\) \(\ds e^{2 i \pi / \paren {2 n} }\)
\(\ds \) \(=\) \(\ds \cos \dfrac {2 \pi} {2 n} + i \sin \dfrac {2 \pi} {2 n}\)
\(\ds \) \(=\) \(\ds \cos \dfrac \pi n + i \sin \dfrac \pi n\)


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 $z^{2 n} - 1$:

\(\ds \paren {z - \alpha^k} \paren {z - \alpha^{2 n - k} }\) \(=\) \(\ds \paren {z - \alpha^k} \paren {z - \overline {\alpha^k} }\) Complex Roots of Unity occur in Conjugate Pairs
\(\ds \) \(=\) \(\ds z^2 - z \paren {\alpha^k + \overline {\alpha^k} } + \alpha^k \overline {\alpha^k}\)
\(\ds \) \(=\) \(\ds z^2 - z \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2\) Modulus in Terms of Conjugate
\(\ds \) \(=\) \(\ds z^2 - z \paren {\alpha^k + \overline {\alpha^k} } + 1\) Modulus of Complex Root of Unity equals 1
\(\ds \) \(=\) \(\ds z^2 - z \paren {\cos \dfrac {k \pi} n + i \sin \dfrac {k \pi} n + \cos \dfrac {k \pi} n - i \sin \dfrac {k \pi} n} + 1\) Definition of $\alpha$
\(\ds \) \(=\) \(\ds z^2 - 2 \cos \dfrac {k \pi} n z + 1\) simplification

Hence the result.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Factorisation_of_x%5E(2n)-1_in_Real_Domain&oldid=600992"