# 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

\(\displaystyle \paren {z - 1}^6 + \paren {z + 1}^6\) | \(=\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 2 z^6 + 30 z^4 + 30 z^2 + 2\) | \(=\) | \(\displaystyle 0\) | expanding the terms, perhaps using Binomial Theorem | |||||||||

\(\text {(1)}: \quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle z^6 + 15 z^4 + 15 z^2 + 1\) | \(=\) | \(\displaystyle 0\) | dividing by 2 on each side |

By substituting $z = i$ in $(1)$ above, it is seen that $i$ is a root of $(1)$:

\(\displaystyle i^6 + 15 i^4 + 15 i^2 + 1\) | \(=\) | \(\displaystyle -1 + 15 - 15 + 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) |

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 $(1)$.

\(\text {(2)}: \quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle \paren {z^2 + 1} \paren {z^4 + 14 z^2 + 1}\) | \(=\) | \(\displaystyle 0\) | 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:

\(\displaystyle x\) | \(=\) | \(\displaystyle \dfrac {-14 \pm \sqrt {14^2 - 4} } 2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -7 \pm 4 \sqrt 3\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle z^2\) | \(=\) | \(\displaystyle -7 \pm 4 \sqrt 3\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle z\) | \(=\) | \(\displaystyle \pm \sqrt{ -7 \pm 4 \sqrt 3}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \pm i \sqrt{ 7 \pm 4 \sqrt 3}\) |

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 $10$