Roots of Complex Number/Examples/z^8 + 1 = 0

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
From Euler's Identity:
 * $-1 = e^{i \pi}$

Let $b$ be defined as:

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 $7$.