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 $b,c$?
 A: Hint: Expand $$(z-z_1)(z-z_2)(z-z_3)$$ and compare it to your expression. I am assuming that $z_3$ is the third root of the polynomial.
A: Using Vieta's formulas, we get that the product $z_1 z_2 z_3$ of the roots is equal to $-c$. Hence $z_3 = -2^{-10}c$. Plugging this value into the equation you get
$$-2^{-30} c^3 + 2^{-20}bc^2 +c=0$$ As $c \neq 0$ we get the equation $-c^2+2^{10}bc + 2^{30}=0 \tag{1}.$
Also $$z_1z_2+z_1z_3+z_2z_3=2^{10}-2^{10}c(z_1+z_2)=2^{10}-2^{15}\sqrt{2}c=0.$$
Hence $c=\dfrac{2^{-5}}{\sqrt{2}}$ and you get the value of $b$ by plugging in this value in equation $(1)$.
A: 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 $b,c$. 
