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

1 Answer
Mar 20, 2016

Answer:

See explanation...

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#