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Thread: Complex numbers and polynomials

  1. #1
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    Complex numbers and polynomials

    Having trouble with trying to prove this. It looks like it isn't too hard but I can't wrap my head around it.

    Let f be a polynomial with real coefficients. Prove: if a complex number z satisfies f(z)=0, then the conjugate of z (call it x) also satisfies f(x)=0
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  3. #3
    MHF Contributor chisigma's Avatar
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    Let the polynomial be of degree n so that...

    $\displaystyle P(z)= a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + \dots + a_{n}\cdot z^{n}$ (1)

    If $\displaystyle z_{0} = \rho\cdot e^{i \theta}$ satisfies the condizion $\displaystyle P(z_{0})=0$ that means that is...

    $\displaystyle a_{0} + a_{1}\cdot \rho\cdot e^{i \theta} + a_{2}\cdot \rho^{2}\cdot e^{2 i \theta} + \dots + a_{n}\cdot \rho^{n} \cdot e^{n i \theta} = 0$ (2)

    Of course (2) is valid if both the real and imaginary parts of $\displaystyle P(z_{0})$ vanish and that's true if we change in (2) $\displaystyle \theta$ with $\displaystyle - \theta$, so that $\displaystyle \overline{ z_{0}} = \rho\cdot e^{-i \theta}$ is also a zero of $\displaystyle P(z)$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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