If $P$ is a polynomial with $|P(1)| = \max\limits_{|z| =1} |P(z)|$, then its root on the unit circle is separated away from 0 Let $P(z)$ be a nonzero polynomial of degree $n$ such that $$|P(1)| = \max\limits_{|z| =1} |P(z)|.$$
Furthermore let $z_0 = e^{i\varphi_0}$, $\varphi_0 \in [-\pi,\pi]$ be a root of $P$ on the unit circle. 
I want to prove that $|\varphi_0| \geq \pi /n$. Also, by my intuition, if this becomes an equality, $P(z)$ would be a multiple of $1+z^n$. Does anyone has an idea?
 A: EDIT. Mulpiplying on a constant, we may assume wlog that $P(1)=1$.
Consider the function $$f(t)=\frac{1}{2}\left( P(e^{it})+\overline{ P(e^{it})}\right)=\frac{1}{2}\left( P(e^{it})+P^*(e^{-it})\right),$$
where $P^*$ is the polynomial with complex conjugate coefficients.
This $f$ is an entire function of exponential type $n$, real on the real line, and and $|f(t)|\leq1$ on the real line, $f(0)=1$.
For such functions we have a generalization of Bernstein's inequality
due to Szego and Boas:
$$f'^2+n^2f^2\leq n^2,$$
which we rewrite as
$$|f'|\leq n\sqrt{1-|f|^2}.$$
Now define $g(t)=\arcsin f(t)$.
Suppose that $f(a)=0$ and $f(b)=1$, and $0<f(t)<1$ for $t\in(a,b)$. Then $g$ is well defined (with the principal branch of arcsin) 
and analytic on $(a,b)$, and $g(a)=0,\; g(b)=\pi$. 
Thus in view of our inequality above
$$\pi\leq\int_a^b|g'(t)|dt\leq\int_a^b \frac{|f'(t)|}{\sqrt{1-f^2(t)}}dt\leq n|b-a|.$$
So $|b-a|\geq\pi/n,$
as requested. The statement on the case of equality follows from the case of equality in Szego's inequality.
Refs. 
Szegö, Schriften der Königsberger gelehrten Gesellschaft, Naturwissenschaftliche
Klasse, Fünftes Jahr 4 (1928), p. 69. 
R. Duffin and A. Shaeffer, Some inequalities for functions of exponential type, Bull AMS, 43 (1937), 554. (Freely available online.)
EDIT. Further research shows that the result is not new: it was proved in
MR0018268
Turán, P.
On rational polynomials. 
Acta Univ. Szeged. Sect. Sci. Math. 11, (1946). 106–113. 
