# Complex Algebra/Examples/z^2 (1 - z^2) = 16/Proof 2

## Example of Complex Algebra

The roots of the equation:

$z^2 \paren {1 - z^2} = 16$

are:

$\pm \dfrac 3 2 \pm \dfrac {\sqrt 7} 2 i$

## Proof

Let $w := z^2$.

Then:

 $\displaystyle w \paren {1 - w}$ $=$ $\displaystyle 16$ $\displaystyle \leadsto \ \$ $\displaystyle w^2 - w + 16$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle w$ $=$ $\displaystyle \dfrac {1 \pm \sqrt {1^2 - 4 \times 1 \times 16} } 2$ Quadratic Formula $\displaystyle$ $=$ $\displaystyle \dfrac {1 \pm 3 \sqrt {-7} } 2$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 2 \pm \dfrac 3 2 \sqrt 7 i$

Let $\paren {p + i q}^2 = \dfrac 1 2 \pm \dfrac 3 2 \sqrt 7 i$

Then:

 $\displaystyle \paren {p + i q}^2$ $=$ $\displaystyle p^2 - q^2 + 2 p q i$ $\dfrac 1 2 \pm \dfrac 3 2 \sqrt 7 i$ $\displaystyle \leadsto \ \$ $\displaystyle p^2 - q^2$ $=$ $\displaystyle \dfrac 1 2$ $\displaystyle 2 p q$ $=$ $\displaystyle \pm \dfrac 3 2 \sqrt 7$ $\displaystyle \leadsto \ \$ $\displaystyle q$ $=$ $\displaystyle \dfrac {1 \pm 3 \sqrt {-7} } {4 p}$ $\displaystyle \leadsto \ \$ $\displaystyle p^2 - \paren {\dfrac {1 \pm 3 \sqrt {-7} } {4 p} }^2$ $=$ $\displaystyle 16$

which leads after unpleasant algebra to:

$p + i q = \pm \dfrac 3 2 \pm \dfrac {\sqrt 7} 2 i$

$\blacksquare$