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:

Complex-8th-Roots-of-Minus-1.png


Sources