Quadratic Equation/Examples/z^4 - 2z^2 + 4 = 0
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Example of Quadratic Equation
The quartic equation:
- $z^4 - 2 z^2 + 4 = 0$
has the solutions:
- $z = \begin{cases} \dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\ \dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \\ -\dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\ -\dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \end{cases}$
Proof
Although this is a quartic in $z$, it can be solved as a quadratic in $z^2$.
From the Quadratic Formula:
\(\ds z^4 - 2 z^2 + 4\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds z^2\) | \(=\) | \(\ds \dfrac {-\paren {-2} \pm \sqrt {\paren {-2}^2 - 4 \times 1 \times 4} } {2 \times 1}\) | Quadratic Formula: $a = 1$, $b = -2$, $c = 4$ | ||||||||||
\(\ds \) | \(=\) | \(\ds 1 \pm \dfrac {\sqrt {4 - 16} } 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \pm \sqrt {-3}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \pm i \sqrt 3\) | Definition of Imaginary Unit |
$\Box$
It remains to solve for $z$.
Let $z = x + i y$.
We have:
\(\ds \paren {x + i y}^2\) | \(=\) | \(\ds 1 \pm i \sqrt 3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2\) | \(=\) | \(\ds \dfrac {1 + \sqrt {1^2 + \paren {\sqrt 3}^2} } 2\) | Square Root of Complex Number in Cartesian Form | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 + \sqrt 4} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \pm \sqrt {\dfrac 3 2}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \dfrac {\sqrt 6} 2\) | ||||||||||||
\(\ds y^2\) | \(=\) | \(\ds \dfrac {-1 + \sqrt {1^2 + \paren {\sqrt 3}^2} } 2\) | Square Root of Complex Number in Cartesian Form | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {-1 + \sqrt 4} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \pm \dfrac {\sqrt 2} 2\) |
Thus we have:
- $z = \begin{cases} \dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\ \dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \\ -\dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\ -\dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \end{cases}$
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1$. Algebraic Theory of Complex Numbers: Exercise $10$