Determine the real numbers $a$, $b$, $c$ such that $1$, $\frac1{1+\omega}$ and $\frac1{1+\omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$

Question

I am stuck on this question:

Let $1$, $\omega$ and $\omega^*$ be the cube root of unity.

a. Show that $\dfrac1{1+\omega}=-\omega$ and $\dfrac1{1+\omega^*}=-\omega^*$.

b. Determine the real numbers $a$, $b$, $c$ such that $1$, $\dfrac1{1+\omega}$ and $\dfrac1{1+\omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$.

c. Hence, find $p(\omega)$ and $p(\omega^*)$.

So I was able to do part a by finding the roots in Cartesian forms, but I am not sure how to approach part b.

Answer 1

We use the familiar fact that $$ (x - \alpha) (x - \beta) = x^{2} - (\alpha + \beta) x + \alpha \beta. $$

For part a, note that $1, \omega, \omega^{*}$ are the three distinct roots of $$ x^{3} - 1 = (x -1) (x^{2} + x + 1). $$ Therefore $\omega, \omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that $$ \omega^{*} = \omega^{-1}, $$ and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that $$ \omega + \omega^{*} = -1, $$ and your formulas follow.

For b, you may note that because of the above formulas, $$ \frac{1}{1 + \omega} = - \omega, \quad\text{and}\quad \frac{1}{1 + \omega^{*}} = - \omega^{*} $$ are the roots of $x^{2} - x + 1$, and so $$ 1, - \omega, - \omega^{*} $$ will be the roots of $$ (x - 1) (x^{2} - x + 1) = x^{3} - 2 x^{2} + 2 x - 1. $$

Answer 2

For those three numbers to be roots of the cubic equation means that if you set $z$ equal to any one of them, then the cubic is zero. Therefore you can write $$z^3+az^2+bz+c\equiv(z-1)\left(z-\frac1{1+\omega}\right)\left(z-\frac1{1+\omega^*}\right)$$ To determine $a,b,c$, simply multiply this out and equate the coefficients of different powers of $z$.