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