Sorry for a kind of overkill and for sure it's not an intended solution. But for sure it was the way it was invented. The claim is true for any holomorphic $f(z)$. Since $\text{Im} f(z)$ is harmonic and zero on $|z|=1$ it's also zero on $|z| \leq 1$, that's $f(z)\in \mathbb{R}$ when $|z|\leq 1$. Since any non constant holomorphic function is open mapping, the result follows.

propossed by Mohammad Ahmadi

Any other solution?

Suppose $P$ is not constant. Write $P(x) = \displaystyle\sum_{m=0}^n (a_m + b_m i )x^m$ for $a_i,b_i\in \mathbb R$ and suppose that $a_n,b_n$ are not both zero. Letting $x=\cos \theta + i\sin \theta$, we know that $\displaystyle\sum_{m=0}^n (a_m + b_m i )(\cos m\theta + i \sin m\theta) \in \mathbb R$. Taking the imaginary part yields that $\displaystyle\sum_{m=0}^n a_m\sin m\theta + b_m \cos m\theta = 0$. Case 1: $n=1$. We know $b_0 + a_1 \sin \theta + b_1 \cos \theta = 0$ for all $\theta$. Replacing $\theta$ with $-\theta$ yields $b_0 - a_1\sin \theta + b_1 \cos \theta = 0$, thus $a_1 \sin \theta = 0 \implies a_1=0$ since the equation holds for all $\theta$. Then we know $b_0 = -b_1 \cos \theta$ holds for all $\theta$, thus $b_1=0$ as well, a contradiction. Therefore assume $n>1$. For arbitrarily large integers $k$ we can take the $4k$th derivative of this constant expression to deduce that $\displaystyle\sum_{m=0}^n b_m m^{4k} \cos m\theta + a_m m^{4k} \sin m\theta = 0$. Since $a_n,b_n$ are not both $0$, it's easy to show we can find some $\theta $ so that $b_n \cos n\theta + a_n \sin n\theta = c \neq 0$. Then we must have $cn^{4k} = - \displaystyle\sum_{m=0}^{n-1} ( b_mm^{4k} \cos m\theta + a_mm^{4k} \sin m\theta)$, or $c = - \displaystyle\sum_{m=0}^{n-1} \dfrac{m^{4k}}{n^{4k}} (b_m\cos m\theta +a_m \sin m\theta)$. Now since $n>1$, we can make $\dfrac{m^{4k}}{n^{4k}} \le \left( \dfrac{n-1}{n} \right)^{4k}$ arbitrarily close to zero by taking an appropriately large $k$ (for a proof just take log of the expression). In particular, we can make $\dfrac{m^{4k}}{n^{4k}} < \dfrac{|c|}{n \text{max} (|a_0|+|b_0|, |a_1|+|b_1|, ... |a_{n-1}|+ |b_{n-1}|)}$. Then we will have $|c| = \left| \displaystyle\sum_{m=0}^{n-1} \dfrac{m^{4k}}{n^{4k}} (b_m\cos m\theta + a_m\sin m\theta) \right|$, which by the extended triangle inequality is $\le \displaystyle\sum_{m=0}^{n-1} \left| \dfrac{m^{4k}}{n^{4k}} (b_m\cos m\theta + a_m\sin m\theta) \right| \le \displaystyle\sum_{m=0}^{n-1} \dfrac{m^{4k}}{n^{4k}}( |a_m| + |b_m|) < \displaystyle\sum_{m=0}^{n-1} \dfrac{|c|}{n} = |c|$. Thus we have a contradiction, hence no such $P$ exist. $P$ must be constant.

Suppose that $p$ is not constant, $p(x)=a_nx^n+...+a_1x+a_0$ with $a_i\in \mathbb{C}$ and $a_n\ne 0$. $$p(x)=\overline{p(x)},\ \forall |x|=1\Leftrightarrow$$$$ a_nx^n+...+a_1x+a_0=\overline{a_n}\cdot \overline{x^n}+...\overline{a_1}\cdot \overline{x}+\overline{a_0},\ \forall |x|=1\Leftrightarrow$$$$x^n(a_nx^n+...a_0)=\overline{a_n}+..+\overline{a_1}x^{n-1}+\overline{a_0}x^n,\ \forall |x|=1$$This is a polynomial equality in an infinite number of points, so the two polynomials actually have the same degree. However, the degree of LHS is $2n$, while the degree of RHS is at most $n$, contradiction.

Let $\text{deg}(p(x))=n$ and let $x_1,x_2,\dots ,x_n,x_{n+1}$ be $n+1$ complex numbers such that $|x_i|=1$ then from Lagrange interpolation formula: $p(x)=\sum_{i=1}^{n+1}p(x_i)\times \frac{\prod(x-x_j)}{\prod_{i\neq j}(x_i-x_j)}$ now assume that $x$ is a number on the unit circle different from $x_i$ then because $|x|=1$ we have $p(x)\in \mathbb{R}$ hence $\overline{p(x)}=p(x)$ so because $\overline{x}=\frac{1}{x},\overline{x_i}=\frac{1}{x_i},p(x_i)=\overline{p(x_i)}$ we have $p(x)=\overline{p(x)}=\sum_{i=1}^{n+1}p(x_i)\times \frac{\prod(x-x_j)}{\prod_{i\neq j}(x_i-x_j)}\times\left(\frac{x_i}{x}\right)^n\Longrightarrow x^np(x)=\sum_{i=1}^{n+1}p(x_i)\times \frac{\prod(x-x_j)}{\prod_{i\neq j}(x_i-x_j)}\times x_i^n$ $(\bigstar)$ hence because the both sides are polynomial which are the same for infinite $x$ so they must be equiv so checking the degree of both sides we get $2n=n\Longrightarrow n=0$. DONE0

tastymath75025 wrote: ... $\displaystyle\sum_{m=0}^n a_m\sin m\theta + b_m \cos m\theta = 0$. ... The problem can be done quite easily from here; this is equivalent to the linear independence of the functions $1, \sin{x}, \cos{x}, \sin{2x}, \cos{2x}, \cdots $. This can be verified by taking the Wronskian of the functions and evaluating at zero, which is indeed nonzero.

Suppose that $P(x)=a_nx^n+...+a_1x+a_0$ and for $b_j , c_j \in \mathbb{R}$ , we have $a_j=b_j+ic_j$. So for each complex number on the unit circle with $arg(z)=\theta$ , from Moivre's theorem we can get : $$P(z)=\sum (b_j+ic_j)(\cos {j\theta}+i sin\ {j\theta}) \in \mathbb{R} \implies Im(P(z))=\sum b_j \sin{j\theta}+c_j \cos{j\theta}=0 (I)$$Now by induction on $n>0$ , we'll show that there aren't exist real numbers $b_0 , ... , b_n$ and $c_0 , ... , c_n$ , such that for each angle $\theta$ , the equality $I$ holds. So if we show this equality with $Q(\theta)$ and $0 \le \theta \le \frac{\pi}{2n}$ , then one can see that : $$Q(\theta+\pi)=\sum_{2|j}b_j \sin{j\theta}+c_j \cos{j\theta}-\sum_{\text{odd}k}b_k \sin{k\theta}+c_j \cos{k\theta}=0$$Now with changing variables from $2\theta$ to $\theta'$ , one can see that ( for each angle $\theta$ ) : $$\sum_{j \le [\frac{n}{2}]} b_j \sin{j\theta}+c_j \cos{j\theta}=0$$which is a contradiction by our induction case for $[\frac{n}{2}]$ , so we're done.

let $p(x) = a_nx^n + ... + a_1x + a_0$ so for every $|z|=1: p(z) = \bar{p(z)} $ and if $a_n \ge 0$ then : $a_nz^n + ... + a_0 = \frac{\bar{a_n}}{z^n} + ... + \bar{a_0}$ it means : $a_nz^{2n} + ... + a_0z^{n} = \bar{a_n} + ... + \bar{a_0}z^n$ so $Q(z) = a_nz^{2n} + ... + a_0z^{n} - \bar{a_n} - ... - \bar{a_0}z^n$ and we know $\forall |z|=1 : Q(z)=0 \Rightarrow \forall z\in \mathbb{C} : Q(z) = 0 \Rightarrow a_n=0, a_0\in \mathbb{R}$