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Given that $z$ is a non real cube root of 1. Find the exact values of $a = (1+2z+3z^2)$ and $b= (1+3z+2z^2)$. I ended up getting $a+b = =-3$ and $a*b=3$. Thus solving simultaneously I conceived $z=\frac{-3±i\sqrt3}{2}$. The problem is I am not exactly sure how to determine which result is paired off with a or b. If someone could help me that would be greatly appreciated.

allquiet1984
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  • If you start off with the fact that $z$ is a non-real cube root of unity that gives you only two possibilities as 1 is the purely real root so you just have the other two (which are conjugate). – Mandelbrot Sep 18 '18 at 10:16
  • Actually, $z=\frac{-1\pm i\sqrt 3}2$. – Bernard Sep 18 '18 at 10:22

2 Answers2

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Hint:

The non-real cube roots of $1$ have minimal polynomial $1+z+z^2$. So you may write, say: $$a=1+2z+3z^2=2(\underbrace{1+z+z^2}_{=0})+z^2-1=z^2-1=\bar z-1.$$

Similarly for $b$.

Bernard
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  • My textbook says $Im(z)=-Im(z^2)$ which I assume comes from $1+z+z^2=0$ please correct me if I am wrong. But then it proceeds to say, $Im(2z+3z^2) = Im(z^2) < 0$ so therefore $1+2z+3z^2 = \frac{-3-i\sqrt3}{2}$. I am trying to understand why this $Im(z^2) < 0$. – allquiet1984 Sep 18 '18 at 12:16
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    That is because they take $z=\mathrm e^{\tfrac{2i\pi}3}$ and so $z^2=\mathrm e^{\tfrac{4i\pi}3}$ (and it's also $\mathrm e^{\tfrac{-2i\pi}3}$). – Bernard Sep 18 '18 at 12:21
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As I read the question your obvious approach is to calculate the two values of a and the two values of b. (One for each value of z).

Peter
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