
> 
> Let $f$ be a monic polynomial with bounded integer coefficients and such that all zeros are (in absolute value) greater than $1$. How close can the zeros of $f$ reach $1$ (in absolute value)?
> 
> 
> 


More precisely, let
$$
f = a\_0 + a\_1 X + \cdots + a\_{n-1}X^{n-1} + X^n
$$
with $a\_i \in \mathbb{Z}$ and $\lvert a\_i \rvert < M$. Suppose $f(z)=0$ implies $\lvert z \rvert > 1$ (so all zeros have absolute value greater than $1$). 



> 
> What is an (explicit) positive function $B(n,M)$ with 
>  $$\lvert z \rvert - 1 \geq B(n,M)$$
>  for any zero $z$ of $f$?
> 
> 
> 


If it is easier, then $M=2^n$ can be assumed. Further, I am only interested in the behavior for large $n$, i.e., I want something like
$$
\frac{1}{\lvert z \rvert - 1} = O(B(n,2^n))
$$
for $n\to\infty$ and for each zero $z$ of any monic polynomial $f$ with degree at most $n$ and integer coefficients bounded by $2^n$.



bound for zeros of a polynomial with bounded integer coefficients
$\def\conj#1{\overline{#1}}\DeclareMathOperator\Res{Res}$If $z$ is a zero of $f$, then $|z|^2-1=z\conj z-1$ is a zero of the resolvent $g(w)=\Res\_z(\conj f(z),z^nf((w+1)/z))$. You can extract a bound on the (integer) coefficients of $g(w)$ from the definition, and then e.g. Cauchy's bound will give you a lower bound on $|z|^2-1$, which in turn implies a lower bound on $|z|-1$. I don’t feel like working it out myself now, but you should get something of the form $B(n,M)\ge M^{-O(n)}$.


