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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 ?

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  • $\begingroup$ Just to make sure. Do you require that the zeros of $Q$ and $R$ have absolute values exactly $1$, or will $\le1$ suffice? To me on the unit circle suggests the former, but Robert read it differently. Furthermore, in the question body you use in the unit circle which may mean $<1$. The unit disk is commonly used when you want $\le 1$ or $<1$. The unit circle is just the perimeter. $\endgroup$ – Jyrki Lahtonen Oct 3 '16 at 7:13
  • $\begingroup$ The idea (of Robert's answer) is that the roots of $R+ b P$ move continuously with $b$, so with $b$ small enough and $R$ having all its roots inside the unit disk, the roots of $R+ b P$ will be inside the disk, and you can write $P = \frac{1}{b}((R+ b P)+ (-R))$. $\endgroup$ – reuns Oct 3 '16 at 7:15
  • $\begingroup$ @JyrkiLahtonen on the unit circle, so exactly $1$ $\endgroup$ – user374692 Oct 3 '16 at 7:36
  • $\begingroup$ @user374692 it changes everything. $\endgroup$ – reuns Oct 3 '16 at 7:38
  • $\begingroup$ If $P(z) = z-p, Q(z) = a(z-b),R(z) = c(z-d)$, it reduces to $|b| = |d| = 1, a+c = 1, ab+cd = p = ab+(1-a)d = a(b-d)+d$. So once you have fixed $d\ne b$ there is always a solution for $a$ $\endgroup$ – reuns Oct 3 '16 at 7:45
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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 $$P(z)=Q(z)+R(z).$$

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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.

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