Let the polynomial be $f(x)$ . If $f(x)=x^{n}$ , then the result is obvious, otherwise, if $a_{0}=\cdots=a_{k-1}=0$ , we can divide the polynomial by $x^{k}$ ($k$ zero roots), so that the resulting polynomial of degree $n-k$ is such that $a_{0}\neq0$ and it has the remaining roots of $f(x)$ . So, without loss of generality we assume $a_{0}\neq0$ and there are no zero roots. Then for $x\neq0$ , $f(1/x)=\frac{1}{x^{n}}g(x)$ , and $f$ has all roots in the interior of $B^{2}$ iff all roots of $g$ are outside of $B^{2}$ . But since $1=g(x)-\sum a_{i}x^{n-i}$ , for $|x|\le1$ , $|g(x)|\ge1-\sum_{i}|a_{i}|>0$ .