$z$ is a root of $F$ iff $\bar z$ is root of $F$

Question

$F(x) \in R$ and $z \in C$. I need to prove that z is a root of $F$ iff $\bar z$ is root of $F$

I can't think of a way to prove that... will love some guidance.

Answer 1

Say, $F(x)=c_0+c_1x+c_2x^2+...$ where all these $c_i$ are real.

Then try to show that $\overline{F(z)}=F(\bar z)$ by using the nice properties of complex conjugation with respect to $+,-,\cdot$ operations of $\Bbb C$.

Answer 2

You only need to remember that $\overline{zw}=\bar z \bar w$ and $\overline{z+w}=\bar z + \bar w$, for any complex numbers. Also, for real numbers, its conjugate is the same.

Answer 3

Try this.

$z$ is a root of F iff $F(z)=0$ iff $\overline{F(z)}=\overline{0}$ iff $F(\overline{z})=0$ iff $\overline{z}$ is a root of $F$.

All you need to show is $\overline{F(z)}=F(\overline{z})$. This is pretty easy to show.:)