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Proof that if a polynomial has a complex zero it's conjugate is also a zero

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    If [itex]P(x)[/itex] is a polynomial with real coefficients, then if [itex]z[/itex] is a complex zero of [itex]P(x)[/itex], then the complex conjugate [itex]\bar{z}[/itex] is also a zero of [itex]P(x)[/itex].

    2. Relevant equations

    Book provides a hint: assume that [itex]z[/itex] is a zero for [itex]P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}[/itex] and use the fact that [itex]\bar{z}=z[/itex] if [itex]z[/itex] is real (since every real can be written as a complex number with a zero imaginary part) and that [itex]\overline{z+w}=\bar{z}+\bar{w}[/itex] and [itex]\overline{z*w}=\bar{z}*\bar{w}[/itex] for all complex numbers.

    3. The attempt at a solution

    Well, it is proven for any real since the complex conjugate of a real number is that same number.

    I'm lost from here.
     
  2. jcsd
  3. Mar 28, 2012 #2

    Curious3141

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    Homework Helper

    Well, if z is a zero of P(x), then [itex]a_{n}z^{n}+a_{n-1}z^{n-1}+...+a_{1}z+a_{0} = 0[/itex].

    Now take the conjugate of both sides.

    For the LHS, use the rules they mentioned to systematically simplify the expression till you get it to: [itex]a_{n}{\bar{z}}^{n}+a_{n-1}{\bar{z}}^{n-1}+...+a_{1}{\bar{z}}+a_{0}[/itex]. You'll need to apply [itex]\overline{z+w}=\bar{z}+\bar{w}[/itex] first on the entire polynomial, followed by [itex]\overline{z*w}=\bar{z}*\bar{w}[/itex] on each term and finally, "[itex]\bar{z}=z[/itex] if [itex]z[/itex] is real" on the real coefficients.

    For the RHS, the conjugate of 0 is of course 0.

    You've now established [itex]P(\bar{z}) = 0[/itex] and you're done.
     
    Last edited: Mar 28, 2012
  4. Mar 28, 2012 #3
    I deleted a question that was here since I misunderstood the post.
     
    Last edited: Mar 28, 2012
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