Quadratic Equation/Examples/z^4 + z^2 + 1 = 0
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Example of Quadratic Equation
The quartic equation:
- $z^4 + z^2 + 1 = 0$
has the solutions:
- $z = \dfrac {\pm 1 \pm i \sqrt 3} 2$
Proof
Although this is a quartic in $z$, it can be solved as a quadratic in $z^2$.
| \(\ds z^4 + z^2 + 1\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z^2\) | \(=\) | \(\ds \dfrac {-1 \pm \sqrt {1^2 - 4 \times 1 \times 1} } {2 \times 1}\) | 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:
| \(\ds z^2\) | \(=\) | \(\ds e^{\pm 2 i \pi / 3}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cos \dfrac {2 \pi} 3 \pm i \sin \dfrac {2 \pi} 3\) |
Taking each in turn:
| \(\ds z^2\) | \(=\) | \(\ds \cos \dfrac {2 \pi} 3 + i \sin \dfrac {2 \pi} 3\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \cis {\dfrac {2 \pi} 3 + 2 k \pi\) | for $k = 0, 1$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \map \cis {\dfrac {\pi} 3 + k \pi}\) | for $k = 0, 1$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \map \cis {\dfrac {\pi} 3} \text { or } \map \cis {\dfrac {4 \pi} 3}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac {\sqrt 3} 2 \text { or } -\dfrac 1 2 - \dfrac {\sqrt 3} 2\) |
and:
| \(\ds z^2\) | \(=\) | \(\ds \cos \dfrac {2 \pi} 3 - i \sin \dfrac {2 \pi} 3\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \cos \dfrac {4 \pi} 3 + i \sin \dfrac {4 \pi} 3\) | for $k = 0, 1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \cis {\dfrac {4 \pi} 3 + 2 k \pi}\) | for $k = 0, 1$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \map \cis {\dfrac {2 \pi} 3 + k \pi}\) | for $k = 0, 1$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \map \cis {\dfrac {2 \pi} 3} \text { or } \map \cis {\dfrac {5 \pi} 3}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds -\dfrac 1 2 + \dfrac {\sqrt 3} 2 \text { or } \dfrac 1 2 - \dfrac {\sqrt 3} 2\) |
$\blacksquare$
Illustration
The roots of the quartic equation:
- $z^4 + z^2 + 1 = 0$
can be depicted as follows:
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $102$