Find all $p\in \mathbb{C}[x]$ with $p(0)=1$ such that $p(z)\leq 1 \forall z$ with $|z|=1$. I'm asked to find all complex-valued polynomials $p\in \mathbb{C}[x]$ with $p(0)=1$ such that $p(z)\leq 1$ for any $z$ on the unit circle.
My progress is as follows: Let deg $p = n-1$. Consider the $n$th roots of unity $$U_n = \left\{ 1,\epsilon,\epsilon^2,...,\epsilon^{n-1}\right\}$$
where $\epsilon = e^\frac{2\pi i}{n}.$
Observe that $\sum_{j=1}^n \epsilon^{kj} = \frac{\epsilon^{nk}-1}{\epsilon-1} = 0$ for any $k$ in $\overline{1,n-1}$ so by summing $p$ over the $n$th roots of unity we get 
$$\sum_{k=0}^{n-1} p(\epsilon^k) =n\cdot p(0)= n.$$
Thus 
$$n = \left| \sum_{k=0}^{n-1} p(\epsilon^k) \right|  \leq \sum_{k=0}^{n-1} |p(\epsilon^k)| \leq n $$
where the last inequality follows from $|p(z)| \leq 1$ for $z$ on the unit circle.
It follows that 
$$|p(\epsilon^k)| = 1$$
for any k. 
This is where I got stuck. Any ideas?
 A: Your argument extends to show that $|p(\omega)| = 1$ for any root of unity $\omega$. Now, write $p(z) = \sum_{i=0}^{n-1} c_i z^i$ and denote $$\overline{p}(z) = \sum_{i=0}^{n-1} \overline{c_i} z^{i}$$ Then, the fact $|p(\omega)| = 1$ for $\omega$ a root of unity translates to $$1 = p(\omega) \overline{p(\omega)} = p(\omega) \overline{p}(1/\omega)$$ Since there are infinitely many roots of unity, this is in fact a polynomial identity, i.e. $1 = p(z) \overline{p}(1/z)$ as rational functions. If $p$ has a root away from $0$, this equation yields a contradiction. Thus, $p(z) = cz^k$ for some $k$ and $|c| = 1$. But we are told $p(0)=1$, so accordingly we must conclude $p(z)=1$ identically.
A: Yet another idea is to use the maximum modulus principle (if allowed). It states, for a connected open subset $D$ of $\mathbb{C}$, and a holomorphic function $f:D\to\mathbb{C}$, that if $z_0\in D$ is a local maximum of $|f(z)|$, then $f$ is constant (equal to $f(z_0)$).
Applied to $|z|=1\implies|p(z)|\leqslant 1$, it produces $|z|\leqslant 1\implies|p(z)|\leqslant 1$ (for any holomorphic $p(z)$, not only for a polynomial); and now, applied to $p(0)=1$, it yields $p(z)=1$ identically.
