Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $f(k)=0$?

share|improve this question
    
    
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
add comment

3 Answers

up vote 6 down vote accepted

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 $\overline{a\cdot b} = \overline{a} \cdot \overline{b}$.

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.