What if in $ aw^2 +c +bw =0 $ $ a,b,c $ are not real ? $ w $ is the cube root of unity.

Question

I was just solving a problem in which i was stuck that if $ a,b,c $ are real then Is it true that if $ aw^2 +c +bw =0 $ then $ a=b=c $ ? What in case if $ a,b,c $ are not real?

Answer 1

You know that $w^2=-w-1$. Suppose that $a+bw+cw^2=0$. Replacing $w^2$ we find that $(a-c)+(b-c)w=0$.

If $b-c\neq0$, then we have that $w=-\frac{a-c}{b-c}$, and the right hand side is real: as $w$ is not real this is absurd. It follows that $b=c$ and that our equation is in fact that $a-c=0$. We thus conclude that $a=b=c$.

Answer 2

Yes, it is true. Note that $\omega^2=-1-\omega$. Therefore$$a\omega^2+b+c\omega=0\iff-a+b+(-a+c)\omega=0.$$Since $1$ and $\omega$ are linearly independent, it follows from this that $-a+b=-a+c=0$. In other words, $a=b=c$.

Answer 3

For real $a,b,c$ it is true. Note $w^2=\overline{w}$ so $0=aw^2+bw+c=a\overline{w}+bw+c=(a+b)\Re(w)+(b-a)\Im(w)+c$, which then implies $a=b$, and then, because $\Re(w)=-\frac{1}{2}$: $-a+c=0$, i.e. $a=b=c$.

For complex $a,b,c$ it is not true. Take $a=w, b=w^2, c=-2$.

Answer 4

Complex roots of real polynomials are found in conjugate pairs.

If $\omega$ is a root of the polynomial $ax^2 + bx + c$ then $\bar \omega$ is also a root.
$ax^2 + bx + c = a(x-\omega)(x-\bar\omega) = a(x^2+x +1)$