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I have two questions regarding complex numbers:

  1. Assume $\mathbb{\lambda, \overline\lambda}$ are solutions to the equation $\mathbb{z^2+az+b=0}$ and $\mathbb{\lambda}$ isn't real. are $\mathbb{a,b}$ real?

  2. Assume $\mathbb{\lambda, \overline\lambda}$ are solutions to the equation $\mathbb{z^3+az^2+bz+c=0}$ and $\mathbb{\lambda}$ isn't real. are $\mathbb{a,b,c}$ real?

My attempts:

  1. I managed to prove a is real (by creating two equations, once setting z to be $\mathbb{\lambda}$ and once $\mathbb{\overline\lambda}$ and substracting them apart, I got to $\mathbb{a = 2 * real(z)}$, afterwards, I know $\mathbb{-b=\lambda^2+a*\lambda=\lambda(\lambda+a)}$, both $\mathbb{\lambda}$ and $\mathbb{\lambda+a}$ aren't real, so after multiplying them we will get b isn't real, but I don't think it is correct.

  2. I couldn't get to anything meaningful because substracting the equations after setting the solutions only eliminates $\mathbb{c}$ but you have both $\mathbb{a,b}$ . I only got to $\mathbb{(\lambda-\overline\lambda)(\lambda^2+\lambda\overline\lambda+\overline\lambda^2)+a(\lambda-\overline\lambda)(\lambda+\overline\lambda)+b(\lambda-\overline\lambda)=0}$.

Thanks.

user3731180
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  • If $λ=x+iy$ where $x$ and $y$ are real then $\mathbb{\lambda\cdot\overline\lambda}=x^2+y^2$ which is real. So $a,b$ in the first case need to be real but $a,b,c$ in the second case is not real if third root is not real. – Asher2211 May 25 '21 at 11:45
  • We have $\mathbb{\lambda\cdot\overline\lambda}=x^2+y^2$ and not$=x^2-y^2.$ – Fred May 25 '21 at 11:49

1 Answers1

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  1. Yes. You always have $$z^2+az+b=(z-\lambda)(z-\overline\lambda)=z^2-2\operatorname{Re}\lambda+|\lambda|^2,$$so $a,b$ are real.

  2. No. It depends on whether the third root $\mu$ is real or not. Based on part 1, you have $$ z^3+az^2+ba+c=(z^2+a'z+b')(z-\mu), $$ with $a',b'$ real.

Martin Argerami
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