Difference of Even Powers of z + a and z - a

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $m \in \Z$ be an integer such that $m > 1$.

Then for all complex number $z$:

$\paren {z + a}^{2 m} - \paren {z - a}^{2 m} = 4 m a z \displaystyle \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \dfrac {k \pi} {2 m} }$


Proof

From Factors of Difference of Two Even Powers:

$x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \displaystyle \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$

Substituting $z + a$ for $x$, $z - a$ for $y$, and $m$ for $n$ we get:

\(\displaystyle \) \(\) \(\displaystyle \paren {z + a}^{2 m} - \paren {z - a}^{2 m}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {z + a} - \paren {z - a} } \paren {\paren {z + a} + \paren {z - a} } \displaystyle \prod_{k \mathop = 1}^{m - 1} \paren {\paren {z + a}^2 - 2 \paren {z + a} \paren {z - a} \cos \dfrac {k \pi} n + \paren {z - a}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {2 a} \paren {2 z} \displaystyle \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + 2 a z + a^2 - 2 \paren {z^2 - a^2} \cos \dfrac {k \pi} m + \paren {z^2 - 2 a z + a^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle 4 a z \displaystyle \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 + 2 a^2 - 2 \paren {z^2 - a^2} \cos \dfrac {k \pi} m}\)
\(\displaystyle \) \(=\) \(\displaystyle 4 a z \displaystyle \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 \paren {1 - \cos \dfrac {k \pi} m} + 2 a^2 \paren {1 + \cos \dfrac {k \pi} m} }\)
\(\displaystyle \) \(=\) \(\displaystyle 4 a z \displaystyle \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 \paren {2 \sin^2 \dfrac {k \pi} {2 m} } + 2 a^2 \paren {2 \cos^2 \dfrac {k \pi} {2 m} } }\) Double Angle Formula for Cosine: Corollary 1 and Corollary 2



Sources