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I have already shown that any polynomial $P\in\mathbb{R}[x]$ satisfies $\overline{P(\overline{x})}=P(x),\forall x\in\mathbb{C}$

My question is, given a polynnomial $P\in\mathbb{C}[x]$, how I can verify whether $\overline{P(\overline{x})}=P(x),\forall x\in\mathbb{C}\Rightarrow P\in\mathbb{R}[x]$ is always true?

So, does there exist a polynomial $P\in\mathbb{C}[x]\setminus\mathbb{R}[x]$ such that $\forall x\in\mathbb{C}:\overline{P(\overline{x})}=P(x)$?

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Let $P(x)=\sum_{k=0}^na_kx^k$. The hypothesis applied for $x$ real gives $$\overline{\sum_{k=0}^na_kx^k}=\sum_{k=0}^na_kx^k,$$ hence $$\sum_{k=0}^n(\overline{a_k}-a_k)x^k=0.$$ The LHS is a polynomial with complex entries, which vanishes on the real numbers: it's the null polynomial, hence $a_k\in\Bbb R$ for $0\leq k\leq n$ and $P\in\Bbb R[x]$.

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    Why is $\overline{\displaystyle\sum_{k=0}^na_kx^{k}}=\displaystyle\sum_{k=0}^na_kx^{k}$ for real numbers $x$ and complex numbers $a_k$?2012-10-14
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    It's given by the hypothesis $\overline{P(x)}=P(x)$ for $x$ real.2012-10-14
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    Did you mean $\overline{P(\overline{x})}=P(x)$? This is true for polynomials $P$ with real coefficients.2012-10-14
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    Yes, but when $x$ is real $\bar x=x$.2012-10-14
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    Right. $\overline{\displaystyle{\sum_{k=0}^{n}a_kx^{k}}}=\sum_{k=0}^{n}\overline{a_k}x^{k}$. I see no reason why that would be equal to $\displaystyle\sum_{k=0}^{n}a_kx^{k}$. (all $a_k$ can be complex, can they?)2012-10-14
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    I assume what we have on the left of the symbol $\Rightarrow$.2012-10-14
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    Ok. Of course, my fault. A nice proof, I finally understand :). You show that the polynomial $\displaystyle\sum_{k=0}^{n}(\overline{a_k}-a_k)x^{k}$ has infinitely many zero points (i don't know the word in English). Thus, it must be equal to the null-polynomial. Thus $a_k\in\mathbb{R},\forall k$. Isn't it?2012-10-14
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    Right (we can say 'zeros' of a polynomial, or 'roots'.2012-10-14