Quadratic Equation/Examples/z^6 + z^3 + 1 = 0
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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 1
Although this is a sextic in $z$, it can be solved as a quadratic in $z^3$.
From the Quadratic Formula:
| \(\ds z^6 + z^3 + 1\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z^3\) | \(=\) | \(\ds \dfrac {-1 \pm \sqrt {1^2 - 4 \times 1 \times 1} } {2 \times 1}\) | Quadratic Formula: $a = 1$, $b = 1$, $c = 1$ | ||||||||||
| \(\ds \) | \(=\) | \(\ds -1 \pm \dfrac {\sqrt {-3} } 2\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\dfrac 1 2 \pm i \dfrac {\sqrt 3} 2\) | Definition of Imaginary Unit |
$\Box$
$-\dfrac 1 2 \pm i \dfrac {\sqrt 3} 2$ are recognised as the complex cube roots of unity, and so:
| \(\ds z^3\) | \(=\) | \(\ds e^{\pm 2 i \pi / 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cos \dfrac {2 \pi} 3 \pm i \sin \dfrac {2 \pi} 3\) |
It remains to solve for $z$.
- $z = \set {w \alpha^k: k \in \set {1, 2, \ldots, n - 1} }$
where:
- $w$ is a cube root of $e^{\pm 2 i \pi / 3}$
- $\alpha$ is one of the complex cube roots of unity.
Taking $z^3 = e^{2 i \pi / 3}$:
| \(\ds z\) | \(=\) | \(\ds e^{2 i \pi / 9} e^{2 k i \pi / 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds e^{2 i \pi / 9 + 2 k i \pi / 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cos \paren {\dfrac {2 \pi} 9 + \dfrac {2 k \pi} 3} + i \sin \paren {\dfrac {2 \pi} 9 + \dfrac {2 k \pi} 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{cases} \cos \dfrac {2 \pi} 9 + i \sin \dfrac {2 \pi} 9 \\ \cos \dfrac {8 \pi} 9 + i \sin \dfrac {8 \pi} 9 \\ \cos \dfrac {14 \pi} 9 + i \sin \dfrac {14 \pi} 9 \end{cases}\) |
The roots obtained by taking $z^3 = e^{-2 i \pi / 3}$ can be calculated as above:
| \(\ds z\) | \(=\) | \(\ds e^{-2 i \pi / 9} e^{2 k i \pi / 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds e^{-2 i \pi / 9 + 2 k i \pi / 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cos \paren {\dfrac {-2 \pi} 9 + \dfrac {2 k \pi} 3} + i \sin \paren {\dfrac {-2 \pi} 9 + \dfrac {2 k \pi} 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \begin{cases} \cos \dfrac {2 \pi} 9 - i \sin \dfrac {2 \pi} 9 \\ \cos \dfrac {4 \pi} 9 + i \sin \dfrac {4 \pi} 9 \\ \cos \dfrac {10 \pi} 9 + i \sin \dfrac {10 \pi} 9 \end{cases}\) |
We note that:
| \(\ds \cos \dfrac {14 \pi} 9 + i \sin \dfrac {14 \pi} 9\) | \(=\) | \(\ds \cos \paren {2 \pi - \dfrac {4 \pi} 9} + i \sin \paren {2 \pi - \dfrac {4 \pi} 9}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cos \dfrac {4 \pi} 9 - i \sin \dfrac {4 \pi} 9\) |
and:
| \(\ds \cos \dfrac {10 \pi} 9 + i \sin \dfrac {10 \pi} 9\) | \(=\) | \(\ds \cos \paren {2 \pi - \dfrac {8 \pi} 9} + i \sin \paren {2 \pi - \dfrac {8 \pi} 9}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cos \dfrac {8 \pi} 9 - i \sin \dfrac {8 \pi} 9\) |
$\blacksquare$
Proof 2
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.
$\blacksquare$