Find $z^3+bz^2+c=0$ Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$ 
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5$
We have $z_1=(-\sqrt{2}-\sqrt{2}i)^5=32e^{i\frac{15\pi}{4}}$
and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=32e^{-i\frac{15\pi}{4}}$
Can we conclude straight  away something about $b,c$?

Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$ 
And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=z_1^*$
$$z^3+bz^2+c=(z-z_1)(z-z_1^*)(z-a)$$ where $a$ has to be real, right?
$$z^3+bz^2+c=z^3-(z_1+z_1^*+a)z^2+(a(z_1+z_1^*)+|z_1|^2)z-a|z_1|^2$$
You can conclude right away anything you want here. You could quickly compute what $a$ is and find any relation between $b,c$. You could also quickly solve for $b,c$. 
