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:

\(\displaystyle z^4 - 2 z^2 + 4\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle z^2\) \(=\) \(\displaystyle \dfrac {-\paren {-2} \pm \sqrt {\paren {-2}^2 - 4 \times 1 \times 4} } {2 \times 1}\) Quadratic Formula: $a = 1$, $b = -2$, $c = 4$
\(\displaystyle \) \(=\) \(\displaystyle 1 \pm \dfrac {\sqrt {4 - 16} } 2\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle 1 \pm \sqrt {-3}\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle 1 \pm i \sqrt 3\) Definition of Imaginary Unit

$\Box$


It remains to solve for $z$.

Let $z = x + i y$.

We have:

\(\displaystyle \paren {x + i y}^2\) \(=\) \(\displaystyle 1 \pm i \sqrt 3\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2\) \(=\) \(\displaystyle \dfrac {1 + \sqrt {1^2 + \paren {\sqrt 3}^2} } 2\) Square Root of Complex Number in Cartesian Form
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {1 + \sqrt 4} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 3 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \pm \sqrt {\dfrac 3 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \pm \dfrac {\sqrt 6} 2\)
\(\displaystyle y^2\) \(=\) \(\displaystyle \dfrac {-1 + \sqrt {1^2 + \paren {\sqrt 3}^2} } 2\) Square Root of Complex Number in Cartesian Form
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {-1 + \sqrt 4} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle \pm \dfrac {\sqrt 2} 2\)


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