Complex Algebra/Examples/(z-1)^6 + (z+1)^6 = 0
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Example of Complex Algebra
The roots of the equation:
- $\paren {z - 1}^6 + \paren {z + 1}^6 = 0$
are:
- $\pm i \cot \dfrac \pi {12}, \pm i \cot \dfrac {5 \pi} {12}, \pm i$
Proof
\(\ds \paren {z - 1}^6 + \paren {z + 1}^6\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 z^6 + 30 z^4 + 30 z^2 + 2\) | \(=\) | \(\ds 0\) | expanding the terms, perhaps using Binomial Theorem | ||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds z^6 + 15 z^4 + 15 z^2 + 1\) | \(=\) | \(\ds 0\) | dividing by 2 on each side |
By substituting $z = i$ in $(1)$ above, it is seen that $i$ is a root of $(1)$:
\(\ds i^6 + 15 i^4 + 15 i^2 + 1\) | \(=\) | \(\ds -1 + 15 - 15 + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
From Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs it follows that $-i$ is also a root of $(1)$.
So it is now established $i$ and $-i$ are indeed roots of $(1)$.
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \paren {z^2 + 1} \paren {z^4 + 14 z^2 + 1}\) | \(=\) | \(\ds 0\) | factoring out the roots found above |
Substituting $x = z^2$ into the second factor of $(2)$ gives the quadratic equation:
- $x^2 + 14 x + 1 = 0$
which, from Solution to Quadratic Equation, has roots:
\(\ds x\) | \(=\) | \(\ds \dfrac {-14 \pm \sqrt {14^2 - 4} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -7 \pm 4 \sqrt 3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds z^2\) | \(=\) | \(\ds -7 \pm 4 \sqrt 3\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \pm \sqrt{ -7 \pm 4 \sqrt 3}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm i \sqrt{ 7 \pm 4 \sqrt 3}\) |
Since:
- $7 + 4 \sqrt 3 = \paren {2 + \sqrt 3}^2$
and:
- $7 - 4 \sqrt 3 = \paren {2 - \sqrt 3}^2$
it follows we have roots:
- $z = \pm i \paren {2 \pm \sqrt 3}$
From Particular Values of Cotangent Function:
- $\cot \dfrac {\pi} {12} = 2 + \sqrt 3$
and:
- $\cot \dfrac {5 \pi} {12} = 2 - \sqrt 3$
Therefore we can simplify the remaining roots as:
- $\pm i \cot \dfrac \pi {12}, \pm i \cot \dfrac {5 \pi} {12}$
as required.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity: Exercise $10$