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