Answer to Question #91324 in Complex Analysis for Ra

Question

If ω is a cube root of unity show that (1-ω+ω²)(1+ω-ω²)=4

Answer 1

As $\omega$ is a cube root of unity, $\omega^3 = 1$. Hence

$0 = 1^3 - \omega^3 = (1-\omega) (1 +\omega +\omega^2).$

As $1\not =\omega$, $1-\omega\not =0$, $1 +\omega +\omega^2 = 0$. Therefore,

$1 +\omega = -\omega^2, \text{(1)}$

$1 +\omega^2 = -\omega. \text{(2)}$

$(1 -\omega +\omega^2)(1 +\omega -\omega^2) = (-\omega -\omega)(1 +\omega -\omega^2) \text{\# by (2)}$

$= (-\omega -\omega)(-\omega^2 -\omega^2) \text{\# by (1)}$

$= (-2)\omega (-2)\omega^2$

$= 4\omega^3 = 4.$