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

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## 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:

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

$\blacksquare$

## Sources

- 1981: Murray R. Spiegel:
*Theory and Problems of Complex Variables*(SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $50$