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

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 $z$.
