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

Question

Eli Assouline · Accepted Answer

See explanation...

#omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#. So #omega^3 = 1# and #omega^4 = omega# #(p+q*omega+r*omega^2)/(r+p*omega+q*omega^2)# #=(p*omega^3+q*omega^4+r*omega^2)/(r+p*omega+q*omega^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#

Kennedy Adams · Answer

undefined To simplify the expression (p + qω + rω^2) / (r + pω + qω^2), where ω is the complex cube root of unity, we can use the fact that ω^3 = 1. By multiplying the numerator and denominator by the conjugate of the denominator, we can eliminate the complex terms. The simplified expression is:

(p + qω + rω^2) / (r + pω + qω^2) * (r^2 - rqω + pqrω^2) / (r^2 - rqω + pqrω^2)

After multiplying and simplifying, the expression becomes:

(p^2 r - pqr + q^2 ω^2 r + pq rω + qr^2 - pq rω^2 + p^2 qω + pq^2 ω^2 + pr^2 ω^2) / (r^2 + p^2 ω + q^2 ω^2 - rpω - rqω^2 - pqrω)

Since ω^3 = 1, we can simplify further:

(p^2 r - pqr + q^2 r + pq rω + qr^2 - pq rω + p^2 qω + pq^2 + pr^2) / (r^2 + p^2 ω + q^2 ω^2 - rpω - rqω - pqrω)

Finally, rearrange the terms:

(p^2 r + q^2 r + pr^2 + pq(r + q) + r^2(p + q)) / (r^2 + p^2 ω + q^2 ω^2 - rpω - rqω - pqrω)

Eli Assouline · Answer

undefined To simplify the expression $\frac{{p+q\omega+r\omega^2}}{{r+p\omega+q\omega^2}}$, where $\omega$ is a complex cube root of unity, you can use the properties of $\omega$ to simplify it further.

Since $\omega^3 = 1$, you can rewrite $\omega^2$ as $\omega^{-1}$, where $\omega^{-1}$ is the complex conjugate of $\omega$, given by $\omega^{-1} = \omega^2$.

Now, substituting $\omega^2$ with $\omega^{-1}$ in the expression, we get:

$\frac{{p+q\omega+r\omega^{-1}}}{{r+p\omega+q\omega^{-1}}}$

This can be further simplified by multiplying the numerator and denominator by $\omega^{-1}$ to rationalize the denominator:

$\frac{{p\omega^{-1}+q+r\omega^{-2}}}{{r\omega^{-1}+p\omega^{-1}\omega+q}}$

Since $\omega^{-2} = \omega$, the expression simplifies to:

$\frac{{p\omega^{-1}+q+r\omega}}}{{r\omega^{-1}+p\omega+q}}$

Now, you can reorder the terms to simplify it further:

$\frac{{q+p\omega+r\omega}}}{{q+r\omega+p\omega}}$

Finally, since addition is commutative, you can rearrange the terms to get:

$\frac{{q+(p+r)\omega}}}{{q+(p+r)\omega}}$

Therefore, the simplified form of the expression is $\frac{{q+(p+r)\omega}}{{q+(p+r)\omega}}$.