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$F(x) \in R$ 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.

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I guess you mean, the coefficients of the polynomial $F$ are real, i.e. $F\in\Bbb R[x]$. – Berci Dec 14 '12 at 1:12
up vote 8 down vote accepted

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 $\Bbb C$.

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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.

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still dont see how it helps me.. and I guess it so easy but... – baaa12 Dec 14 '12 at 1:24
$\overline{a_nz^n}=\overline{a_n}\overline{z^n}=a_n\overline{z}^n$. Take the sum and use the fact that $\sum a_iz^i=0$ and $\bar 0=0$. – Sigur Dec 14 '12 at 10:28

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.:)

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