Quadratic Equation/Examples/z^4 - 2z^2 + 4 = 0

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 $z^2$.

 $\ds z^4 - 2 z^2 + 4$ $=$ $\ds 0$ $\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:

$z = \begin{cases} \dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\ \dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \\ -\dfrac {\sqrt 6} 2 + i \dfrac {\sqrt 2} 2 \\ -\dfrac {\sqrt 6} 2 - i \dfrac {\sqrt 2} 2 \end{cases}$

$\blacksquare$