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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 $b,c$?

newhere
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  • Are $z_1$ and $z_2$ supposed to be roots? – Cornman Aug 19 '18 at 19:41
  • @Cornman Yes it is – newhere Aug 19 '18 at 19:42
  • Can you be more specific about what you mean with "straight away". Are you searching for non-trivial properties of $b, c$ only? One could say straight away, that $c\neq 0$, since there are two complex solutions different from $0$. – Cornman Aug 19 '18 at 19:52
  • Alt. hint: $,\require{cancel}(1+i)^2=\bcancel{1}+2i+\bcancel{i^2}=2i,$ so $,(1+i)^4=-4,$ so $,z_1=-\big(\sqrt{2}(1+i)\big)^5=\ldots,$ – dxiv Aug 19 '18 at 20:58

3 Answers3

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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.

MrYouMath
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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 $(1)$.

mathcounterexamples.net
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  • At first I was confused as to why $z_3$ was constrained at all, but then I realized that $z$'s coefficient is 0. – Striker Aug 19 '18 at 20:46
0

Find $b,c\in \mathbb{R}$ where $z^3+bz^2+c=0$

And $z_1=(-\sqrt{2}-\sqrt{2}i)^5$ and $z_2=(-\sqrt{2}+\sqrt{2}i)^5=z_1^*$

$$z^3+bz^2+c=(z-z_1)(z-z_1^*)(z-a)$$ 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 $b,c$.

Ahmad Bazzi
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