Roots of polynomial with real coefficients appear in conjugate pairs.

How to prove most simply that if a polyonmial $f$, has only real coefficients and $f(c)=0$, and $k$ is the complex conjugate of $c$, then $f(k)=0$?

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Did you try and conjugate $f(c)=0$? You get exactly $f(k)=0$, by using basic properties of the conjugation. –  Beni Bogosel Mar 18 '12 at 12:15

Look at $\overline{f(c)}$ and use that conjugation is a homomorphism of $\mathbb{C}$. That is, $\overline{a+b} = \overline{a}+\overline{b}$ and $\overline{a\cdot b} = \overline{a} \cdot \overline{b}$.

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You use the fact that the coefficients of $f$ are real to show that $$f(\overline c)=\overline{f(c)}.$$

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I've long forgotten the details so maybe someone else can complete this. I also don't see how folks have formatted the conjugate with the overhead bar so I will use ~ for conjugate.

When I took complex variable, the proof was along the lines of:

Suppose z1 is indeed a zero of f(z). Assume ~z1 is not a zero of f(z). Then 1/(fz) does not have a singularity at ~z1 i.e. 1/f(z1) is a permitted operation.

And that's where my memory fails me: Somehow, the professor (and the book) show that this leads to a contradiction.

Just a starter..

-- JS

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Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. –  Zev Chonoles Jul 17 '13 at 5:19