Quadratic Equation/Examples/z^6 + z^3 + 1 = 0/Proof 2

Example of Quadratic Equation

The sextic equation:

$z^6 + z^3 + 1 = 0$

has the solutions:

$z = \begin{cases} \cos \dfrac {2 \pi} 9 \pm i \sin \dfrac {2 \pi} 9 \\ \cos \dfrac {4 \pi} 9 \pm i \sin \dfrac {4 \pi} 9 \\ \cos \dfrac {8 \pi} 9 \pm i \sin \dfrac {8 \pi} 9 \end{cases}$

Proof

From Complex Algebra Examples $z^6 + z^3 + 1$:

$z^6 + z^3 + 1 = \paren {z^2 - 2 z \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {8 \pi} 9 + 1}$

Then from Complex Roots of Unity occur in Conjugate Pairs:

$\paren {z^2 - 2 z \cos \dfrac {2 k \pi} 9 + 1} = \paren {z - \paren {\cos \dfrac {2 k \pi} 9 + i \sin \dfrac {2 k \pi} 9} } \paren {z - \paren {\cos \dfrac {2 k \pi} 9 - i \sin \dfrac {2 k \pi} 9} }$

for each $k$.

The result follows.

$\blacksquare$