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:

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

$\blacksquare$



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