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:
\(\ds w \paren {1 - w}\) | \(=\) | \(\ds 16\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds w^2 - w + 16\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds w\) | \(=\) | \(\ds \dfrac {1 \pm \sqrt {1^2 - 4 \times 1 \times 16} } 2\) | Quadratic Formula | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 \pm 3 \sqrt {-7} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 2 \pm \dfrac 3 2 \sqrt 7 i\) |
Let $\paren {p + i q}^2 = \dfrac 1 2 \pm \dfrac 3 2 \sqrt 7 i$
Then:
\(\ds \paren {p + i q}^2\) | \(=\) | \(\ds p^2 - q^2 + 2 p q i\) | $\dfrac 1 2 \pm \dfrac 3 2 \sqrt 7 i$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds p^2 - q^2\) | \(=\) | \(\ds \dfrac 1 2\) | |||||||||||
\(\ds 2 p q\) | \(=\) | \(\ds \pm \dfrac 3 2 \sqrt 7\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds q\) | \(=\) | \(\ds \dfrac {1 \pm 3 \sqrt {-7} } {4 p}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds p^2 - \paren {\dfrac {1 \pm 3 \sqrt {-7} } {4 p} }^2\) | \(=\) | \(\ds 16\) |
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: $50$