Polynomial as sum of two polynomials with roots on the unit circle 
Can any polynomial $P\in \mathbb C[X]$ be written as $P=Q+R$ where $Q,R\in \mathbb C[X]$ have all their roots on the unit circle (that is to say with magnitude exactly $1$) ? 

I don't think it's even trivial with degree-1 polynomials... In this supposedly simple case, with $P(X)=\alpha X + \beta$,  this boils down to finding $\alpha_1$ and $\beta_1$ such that $|\alpha_1|=|\beta_1|$ and $|\alpha-\alpha_1|=|\beta-\beta_1|$. I can't prove that geometrically, let alone analytically...
Furthermore I don't think anything can be said about the sum of two polynomials with known roots...
Can someone give me some hints ? 
 A: The original post was.

Can any polynomial $P\in \mathbb C[X]$ be written as $P=Q+R$ where $Q,R\in \mathbb C[X]$ have all their roots in the unit circle? 

I read "in the unit circle" as "inside the unit disc".
Let $P(z)=a_nz^n+\dots+a_0$ with $a_n\not=0$ and take
$$a>|a_n|+\dots +|a_0|.$$
Then, for any $z$ on the unit circle, we have that 
$$|az^{n+1}|=|a|>|a_n|+\dots +|a_0|=|a_n||z^n|+\dots +|a_0|\geq |P(z)|.$$
Hence, by Rouché's theorem,
$$Q(z):=az^{n+1}+P(z)$$
has all its $n+1$ roots inside the unit disc. 
Let $R(z):=-az^{n+1}$ which has (trivially) all the zeros inside inside the unit disc too.
Finally
$$P(z)=Q(z)+R(z).$$
A: What you can also do is consider, if the degree of $P$ is $d$ and if all its roots are in the (open) unit disk, is $$Q_\omega=P-\omega X^d P(1/X)$$ for any $\omega$ on the unit circle.
Indeed, if you write $P=\prod_k(X-r_k)$ then a complex $z$ of modulus $1$ is a root of $Q_\omega$ iff $$\prod_k \frac{z-r_k}{1-\bar{r_k}z} = \omega.$$
But the homography $f_k:z\mapsto \frac{z-r_k}{1-\bar{r_k}z}$ preserves the unit circle, actually $t\in\left[0,2\pi\right[\mapsto f_k(e^{it})$ makes a full turn counterclockwise. Hence $t\in\left[0,2\pi\right[\mapsto \prod_{k=1}^d f_k(e^{it})$ makes $d$ full turns, thus it is equal to $\omega$ $d$ times. It proves that all the roots of $Q$ are on the unit circle.
Now if you write $P=\frac12Q_\omega+\frac12Q_{-\omega}$ you may conclude in this particular case.
This does not prove the result in the general case, but combined with Robert's answer it shows that any polynomial can be written as the sum of $4$ polynomials with all their roots on the unit circle.
