Irreducible polynomial in integer polynomial ring with complex root is a reciprocal polynomial I am working on this problem: 
Let $p(x) \in \mathbb{Z}[x]$ be an irreducible polynomial. Show that $p(x)$ is a reciprocal polynomial (i.e., that its coefficients equidistant from either end are equal) if one of its roots is a complex number $z$ with $|z|= 1$. 
My thinking so far is: 
Let $z  = a+bi$. Then $z = a-bi$ is also a root of $p(x)$, so both $(x-(a+bi))$ and $(x-(a-bi))$ are factors of $p(x)$ $\Rightarrow$ $(x-(a+bi))(x-(a-bi)) = x^2 - 2ax + a^2 + b^2 = x^2-2ax + 1$ is a factor of $p(x)$ (using that $|z| = 1$, and thus, $a^2 + b^2 = 1$). 
I suppose that, at this stage, I'm not sure how to use the fact that $p(x)$ is irreducible over $\mathbb{Z}[x]$ to obtain the result. It seems to me that $p(x)$ cannot have any other nontrivial polynomials over $\mathbb{Z}$ as factors, since $p(x) \in \mathbb{Z}[x]$. But then, how can I know about other factors of $p(x)$ ? 
Thanks! 

The following proof is due to a classmate of mine:
Let $K$ be the splitting field of $p(x)$ over $\Bbb Z$ and $\sigma\in Aut(K/\Bbb Z)$ such that for every root $\alpha$ of $p(x), \sigma(\alpha)=\dfrac1{\alpha}$. But $\sigma$ sends any root of $p$ to some root of $p$, which means that the set of all roots of $p$ is the same as the set of reciprocals of all roots of $p$. Hence for every root of $p$, its reciprocal is also a root of $p$, and thus $p(x)$ is reciprocal.
