Image of integer coefficient polynomial on unit circle. Let T be an unit circle and f(x) be an integer coefficient polynomial.

Then What is $f(T)$?

Using complex analysis,
I think there is a point z in T such that |f(z)|>=1. But I do not know why this happend.
 A: Unless $f(z)\equiv0$, since $\max\limits_{|z|=1}\left|\frac{f(z)}{z^n}\right|=\max\limits_{|z|=1}\left|\,f(z)\,\right|$, we can assume $f(0)\ne0$. Since $f(0)\in\mathbb{Z}$, we must have $\left|\,f(0)\,\right|\ge1$.
Polynomials are holomorphic, so they have the Mean Value Property:
$$
f(0)=\frac1{2\pi}\int_0^{2\pi}f\!\left(e^{it}\right)\mathrm{d}t\tag{1}
$$
Suppose
$$
\frac1{2\pi}\int_0^{2\pi}\left|\,f\!\left(e^{it}\right)-f(0)\,\right|^2\,\mathrm{d}t=\sigma^2\tag{2}
$$
Then
$$
\begin{align}
\frac1{2\pi}\int_0^{2\pi}\left|\,f\!\left(e^{it}\right)\,\right|^2\,\mathrm{d}t
&=\frac1{2\pi}\int_0^{2\pi}\left(2\mathrm{Re}\!\left(\,f\!\left(e^{it}\right)\overline{f(0)}\right)-\left|\,f(0)\,\right|^2\right)\,\mathrm{d}t+\sigma^2\\[6pt]
&=\left|\,f(0)\,\right|^2+\sigma^2\tag{3}
\end{align}
$$
Thus, if $\left|\,f(0)\,\right|\gt1$ or $\sigma^2\gt0$, then $(3)$ says that for some $t\in[0,2\pi]$, we have $\left|\,f\!\left(e^{it}\right)\,\right|\gt1$.
If $\left|\,f(0)\,\right|=1$ and $\sigma^2=0$, then $f(z)=f(0)=\pm1$ for all $z$.

Therefore, except for $f(z)=\pm z^n$ and $f(z)=0$, all polynomials with integer coefficients satisfy
  $$
\max_{|z|=1}\left|\,f(z)\,\right|\gt1\tag{4}
$$

