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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 $a_k$? –  barto Oct 14 '12 at 14:39
    
It's given by the hypothesis $\overline{P(x)}=P(x)$ for $x$ real. –  Davide Giraudo Oct 14 '12 at 14:42
    
Did you mean $\overline{P(\overline{x})}=P(x)$? This is true for polynomials $P$ with real coefficients. –  barto Oct 14 '12 at 14:45
    
Yes, but when $x$ is real $\bar x=x$. –  Davide Giraudo Oct 14 '12 at 14:45
    
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?) –  barto Oct 14 '12 at 14:51
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