Complex Algebra/Examples/z^8 + 1

Example of Complex Algebra

$z^8 + 1 = \paren {z^2 - 2 z \cos \dfrac \pi 8 + 1} \paren {z^2 - 2 z \cos \dfrac {3 \pi} 8 + 1} \paren {z^2 - 2 z \cos \dfrac {5 \pi} 8 + 1} \paren {z^2 - 2 z \cos \dfrac {7 \pi} 8 + 1}$

Proof

From Roots of $z^8 + 1 = 0$ and the corollary to the Polynomial Factor Theorem:

$z^8 + 1 = \ds \prod_{k \mathop = 0}^7 \paren {z - \paren {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8} }$

Hence:

\(\ds z^8 + 1\)

\(=\)

\(\ds \prod_{k \mathop = 0}^7 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 8}\)

Definition of Exponential Form of Complex Number

\(\ds \)

\(=\)

\(\ds \prod_{k \mathop = 0}^3 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 8} \paren {z - \exp \dfrac {-\paren {2 k + 1} i \pi} 8}\)

Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs

\(\ds \)

\(=\)

\(\ds \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} 8 + \exp \dfrac {-\paren {2 k + 1} i \pi} 8} + \exp \dfrac {\paren {2 k + 1} i \pi} 8 \exp \dfrac {-\paren {2 k + 1} i \pi} 8}\)

multiplying out

\(\ds \)

\(=\)

\(\ds \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} 8 + \exp \dfrac {-\paren {2 k + 1} i \pi} 8} + 1}\)

simplifying

\(\ds \)

\(=\)

\(\ds \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8 + \cos \dfrac {\paren {2 k + 1} \pi} 8 - i \sin \dfrac {\paren {2 k + 1} \pi} 8} + 1}\)

Definition of Exponential Form of Complex Number

\(\ds \)

\(=\)

\(\ds \prod_{k \mathop = 0}^3 \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} 8 + 1}\)

simplifying

Hence the result.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $8$