$F(x) \in R$ 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.
$F(x) \in R$ 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.
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 $\Bbb C$.
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.
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.:)