Roots of real polynomial $f(x)$ is a real polynomial.
Show that $z=a+bi$ and $\bar z=a-bi$ have the same algebric multiplicity.
I know that if $z=a+bi$ is a root of $f$ then $\bar z=a-bi$ is too, but don't know how to use it...
Thanks
 A: If the root is real, the result is obvious, so we may assume that $b\ne 0$. Let us suppose the result is true for all polynomials of degree $\lt n$. We show it is true for polynomials of degree $n$.
So let $w$ be a non-real root of the polynomial $P(x)$ of degree $n$. Then $\bar{w}$ is a root. It follows that the real polynomial $(x-w)(x-\bar{w})$ divides $P(x)$, say $P(x)=(x-w)(x-\bar{w})Q(x)$. By induction hypothesis, $w$ and $\bar{w}$ have the same multiplicity (possibly $0$) as roots of $Q(x)$. It follows they have the same multiplicity as roots of $P(x)$.   
A: Suppose that $z=a+bi$ and $z=a-bi$ have different algebraic multiplicities, say $m$ and $n$ respectively. Without loss of generality, suppose $m>n$. Then in the complex conjugate of the polynomial $f(x)$, written $\overline{f(x)}$, the algebraic multiplicities of $z=a+bi$ and $z=a-bi$ will be reversed. But since for real polynomials $f(x)=\overline{f(x)}$, this would require that both $m>n$ and $m<n$. Contradiction.
A: You need to show that if $f(x) = (x-z)^kq(x)$ then $f(x) = (x-\overline{z})^kp(x)$. Hint - you'll find that $p(x) = \overline {q(x)}$. Remember that $x \mapsto \overline x$ is an automorphism, fixed on real numbers.
