Roots of a polynomial on the unit circle

Question

Let $p(z)=z^n+a_1z^{n-1}+\cdots+a_{n-1}z+a_n$ and suppose that $|p(z)|<1$ for $|z|=1$. Show that $p$ has a zero on $|z|=1$.

I am having trouble to solve this problem. My attempt was to suppose that $p$ has no roots on the unit circle, then apply Rouché's Theorem and arrive at a contradiction. I tried to compare to $z^{n+1}$ and conclude that $p$ has $n+1$ roots, but I was unsuccessful.

Any hints?

Thanks.

Answer 1

If $g(z) = -p(z)$, and $f(z) = z^n$, then $|f(z)| = 1 > |p(z)| = |g(z)|$ on the unit circle. It would seem that Rouche's theorem implies that $f(z) = z^n$ and $$f(z) + g(z) = z^n - p(z) = -a_{1} z^{n-1} + \ldots $$ have the same number of zeros inside the unit disc, which is a contradiction, because the first polynomial has $n$ such roots (with multiplicity) and the second has at most $n-1$ roots. So no such $p(z)$ exists with these properties. Hence(!?) $p(z)$ has a zero on $|z| = 1$.

Answer 2

Let $q$ be a complex primitive unit root of degree $d>n$. Then on the one hand $$ \left|\sum_{k=0}^{d-1}q^{-kn}p(q^k)\right|\le \sum_{k=0}^{d-1}|p(q^k)|<d $$ on the other hand, we can compute that sum explicitly using the geometric sum formula $$ \sum_{k=0}^{d-1}q^{-kn}p(q^k)=\sum_{k=0}^{d-1}\sum_{j=0}^n a_jq^{-kn+k(n-j)} =d+\sum_{j=1}^n a_j\sum_{k=0}^{d-1}q^{-kj} =d $$ This is a contradiction.