I've been trying to solve this question for a few hours but I'm stumped, any guidance would be appreciated
If $\gamma$ and $\bar{\gamma}$ are solutions to $z^2+az+b=0$ And $\gamma$ is not real then a,b are real
Another question of the same vein
If $\gamma$ and $\bar{\gamma}$ are solutions to $z^3+az^2+bz +c=0$ And $\gamma$ is not real then a,b and c are real
If $\gamma$ and $\overline{\gamma}$ are solutions of $z^2+az+b=0$, then the polynomial factorizes in $z^2+az+b=(z-\gamma)\cdot(z-\overline\gamma)=z^2-(\gamma+\overline\gamma)z+\gamma\overline\gamma=z^2-2Re(\gamma)z+|\gamma|^2$.
So $a$ and $b$ are reals.
The other question works the same.
HINT: You know that the third root of the polynomial must be real, factorize and apply the previous result.
