Complex Algebra/Examples/(z-1)^6 + (z+1)^6 = 0

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

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 $(1)$.

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:

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.