How do you simplify $\frac{p + q \cdot ω + r \cdot ω^{2}}{r + p \cdot ω + q \cdot ω^{2}}$ $(p+qomega+romega^2)/(r+pomega+qomega^2)$ ?

Question

How do you simplify $\frac{p + q \cdot ω + r \cdot ω^{2}}{r + p \cdot ω + q \cdot ω^{2}}$ $(p+qomega+romega^2)/(r+pomega+qomega^2)$ ?

1 Answer

Impact of this question

4483 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-20T22:47:56

See explanation...

Explanation:

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

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

$\frac{p + q \cdot ω + r \cdot ω^{2}}{r + p \cdot ω + q \cdot ω^{2}}$ $(p+q*omega+r*omega^2)/(r+p*omega+q*omega^2)$

$= \frac{p \cdot ω^{3} + q \cdot ω^{4} + r \cdot ω^{2}}{r + p \cdot ω + q \cdot ω^{2}}$ $=(p*omega^3+q*omega^4+r*omega^2)/(r+p*omega+q*omega^2)$

$= \frac{r \cdot ω^{2} + p \cdot ω^{3} + q \cdot ω^{4}}{r + p \cdot ω + q \cdot ω^{2}}$ $=(r*omega^2+p*omega^3+q*omega^4)/(r+p*omega+q*omega^2)$

$=(color(red)(cancel(color(black)((r+p*omega+q*omega^2))))omega^2)/color(red)(cancel(color(black)((r+p*omega+q*omega^2))))$

$=omega^2$

Science

Math

Humanities

... and beyond

How do you simplify $\frac{p + q \cdot ω + r \cdot ω^{2}}{r + p \cdot ω + q \cdot ω^{2}}$ $(p+qomega+romega^2)/(r+pomega+qomega^2)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you simplify (p+q*omega+r*omega^2)/(r+p*omega+q*omega^2)p+q⋅ω+r⋅ω2r+p⋅ω+q⋅ω2(p+q*omega+r*omega^2)/(r+p*omega+q*omega^2) ?

1 Answer

Explanation:

Related questions

Impact of this question

How do you simplify $\frac{p + q \cdot ω + r \cdot ω^{2}}{r + p \cdot ω + q \cdot ω^{2}}$ $(p+qomega+romega^2)/(r+pomega+qomega^2)$ ?