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

1. Mar 28, 2012

### mindauggas

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Book provides a hint: assume that $z$ is a zero for $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$ and use the fact that $\bar{z}=z$ if $z$ is real (since every real can be written as a complex number with a zero imaginary part) and that $\overline{z+w}=\bar{z}+\bar{w}$ and $\overline{z*w}=\bar{z}*\bar{w}$ 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. Mar 28, 2012

### Curious3141

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

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: $a_{n}{\bar{z}}^{n}+a_{n-1}{\bar{z}}^{n-1}+...+a_{1}{\bar{z}}+a_{0}$. You'll need to apply $\overline{z+w}=\bar{z}+\bar{w}$ first on the entire polynomial, followed by $\overline{z*w}=\bar{z}*\bar{w}$ on each term and finally, "$\bar{z}=z$ if $z$ is real" on the real coefficients.

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

You've now established $P(\bar{z}) = 0$ and you're done.

Last edited: Mar 28, 2012
3. Mar 28, 2012

### mindauggas

I deleted a question that was here since I misunderstood the post.

Last edited: Mar 28, 2012