Complex Algebra/Examples/(1+z)^5 = (1-z)^5

Example of Complex Algebra
The roots of the equation:
 * $\paren {1 + z}^5 = \paren {1 - z}^5$

are:
 * $z = \set {\dfrac {\omega^k - 1} {\omega^k + 1}: k = 0, 1, 2, 3, 4}$

That is:
 * $z = \set {0, \dfrac {\omega - 1} {\omega + 1}, \dfrac {\omega^2 - 1} {\omega^2 + 1}, \dfrac {\omega^3 - 1} {\omega^3 + 1}, \dfrac {\omega^4 - 1} {\omega^4 + 1} }$

where:
 * $\omega = \cis \dfrac {2 \pi i} 5$

Proof
We have:

Let $\cis \dfrac {2 \pi i} 5 = \omega$ as suggested.

Thus:

So:

When $n = 0$:
 * $\omega^n = 1$

and thus:
 * $z = \paren {1 - 1} / \paren {1 + 1}$

So:


 * $z = 0, \dfrac {\omega - 1} {\omega + 1}, \dfrac {\omega^2 - 1} {\omega^2 + 1}, \dfrac {\omega^3 - 1} {\omega^3 + 1}, \dfrac {\omega^4 - 1} {\omega^4 + 1}$