2660.11 – Imaginary Roots

If ω\omega is one of the imaginary roots of x3=1x^3 = 1, then the quantity (1−ω+ω2)(1+ω−ω2)(1 - \omega + \omega^2)(1 + \omega - \omega^2) equals

  1. 4
  2. ω2\omega^2
  3. 2
  4. −4ω2-4 \omega^2
  5. 1
Solution

Well, ω3=1\omega^3 = 1, so ω3−1=0\omega^3 - 1 = 0. Now,

0=ω3−1=(ω−1)(ω2+ω+1),0 = \omega^3 - 1 = (\omega - 1)(\omega^2 + \omega + 1),

and since ω\omega isn't real, ω−1\omega - 1 can't be 0. Therefore, ω2+ω+1=0\omega^2 + \omega + 1 = 0. This can be used, with some algebraic massaging, to finish the problem. First note,

1−ω+ω2=1+ω+ω2−2ω=0−2ω=−2ω,\begin{aligned} 1 - \omega + \omega^2 &=& 1 + \omega + \omega^2 - 2 \omega \\ &=& 0 - 2 \omega = - 2 \omega, \end{aligned}

and

1+ω−ω2=1+ω+ω2−2ω2=0−2ω2=−2ω2.\begin{aligned} 1 + \omega - \omega^2 &=& 1 +\omega + \omega^2 - 2 \omega^2 \\ &=& 0 - 2 \omega^2 = - 2 \omega^2. \end{aligned}

Therefore,

(1−ω+ω2)(1+ω−ω2)=(−2ω)(−2ω2)=4ω3=4,(1 - \omega + \omega^2)(1 + \omega - \omega^2) = (- 2 \omega) (-2 \omega^2) = 4 \omega^3 = 4,

since ω3=1\omega^3 = 1. The answer is a).