How do you simplify (p+q*omega+r*omega^2)/(r+p*omega+q*omega^2) ?

1 Answer
Mar 20, 2016

See explanation...

Explanation:

$\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

So ${\omega}^{3} = 1$ and ${\omega}^{4} = \omega$

$\frac{p + q \cdot \omega + r \cdot {\omega}^{2}}{r + p \cdot \omega + q \cdot {\omega}^{2}}$

$= \frac{p \cdot {\omega}^{3} + q \cdot {\omega}^{4} + r \cdot {\omega}^{2}}{r + p \cdot \omega + q \cdot {\omega}^{2}}$

$= \frac{r \cdot {\omega}^{2} + p \cdot {\omega}^{3} + q \cdot {\omega}^{4}}{r + p \cdot \omega + q \cdot {\omega}^{2}}$

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(r + p \cdot \omega + q \cdot {\omega}^{2}\right)}}} {\omega}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(r + p \cdot \omega + q \cdot {\omega}^{2}\right)}}}}$

$= {\omega}^{2}$