Roots of Complex Number/Examples/z^8 + 1 = 0
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Theorem
The roots of the polynomial:
- $z^8 + 1 = 0$
are:
- $\set {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8: k \in \set {0, 1, \ldots, 7} }$
Proof
\(\ds z^8 + 1\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \paren {-1}^{1/8}\) |
From Euler's Identity:
- $-1 = e^{i \pi}$
Let $b$ be defined as:
\(\ds b\) | \(=\) | \(\ds \sqrt [8] 1 \paren {\cos \dfrac \pi 8 + i \sin \dfrac \pi 8}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos \dfrac \pi 8 + i \sin \dfrac \pi 8\) |
From Roots of Complex Number: Corollary:
- $\paren {-1}^{1/8} = \cos \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8} + i \sin \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8}$
for $k = 0$ to $7$.
\(\ds \paren {-1}^{1/8}\) | \(=\) | \(\ds \set {\cos \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8} + i \sin \paren {\dfrac \pi 8 + \dfrac {2 k \pi} 8}: k \in \set {0, 1, \ldots, 7} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8: k \in \set {0, 1, \ldots, 7} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\exp \dfrac {\paren {2 k + 1} i \pi} 8: k \in \set {0, 1, \ldots, 7} }\) |
$\blacksquare$
Illustration
The roots of the polynomial:
- $z^8 + 1 = 0$
are illustrated below:
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $8$