# 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

 $\displaystyle \paren {z - 1}^6 + \paren {z + 1}^6$ $=$ $\displaystyle 0$ $\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:

$z = \pm i \paren {2 \pm \sqrt 3}$
$\cot \dfrac {\pi} {12} = 2 + \sqrt 3$

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.

$\blacksquare$