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

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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

\(\ds z^2 \paren {1 - z^2}\) \(=\) \(\ds 16\)
\(\ds \leadsto \ \ \) \(\ds z^4 - z^2 - 16\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds z^4 + 8 z^2 + 16 - 9 z^2\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {z^2 + 4}^2 - 9 z^2\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {z^2 + 4 + 3 z} \paren {z^2 + 4 - 3 z}\) \(=\) \(\ds 0\) Difference of Two Squares

Thus there are two quadratic equations:

\(\ds \paren {z^2 + 4 + 3 z}\) \(=\) \(\ds 0\)
\(\ds \paren {z^2 + 4 - 3 z}\) \(=\) \(\ds 0\)

which give rise, via the Quadratic Formula, to:

\(\ds z\) \(=\) \(\ds \dfrac 3 2 \pm \dfrac {\sqrt 7} 2 i\)
\(\ds z\) \(=\) \(\ds -\dfrac 3 2 \pm \dfrac {\sqrt 7} 2 i\)

$\blacksquare$


Sources