Complex polynomial that has no roots in the closed unit disc

Question

Let $c_0 > c_1 >... > c_n > 0$. Show that the polynomial $p(z) = c_0 + c_1z+...+c_nz^n$ has no zeros in the closed unit disc. My proof is not self-contained at all, and I've no idea how to approach the problem.

Answer 1

Write $$(1-z) p(z) = c_0 + (c_1 - c_0) z + \ldots + (c_n - c_{n-1}) z^{n-1} - c_n z^n = c_0 - g(z)$$ where $ |g(z)| \le c_0$ for $|z|\le 1$ using the triangle inequality. Equality can hold only when $|z|=1$ and all terms in $g(z)$ have the same argument, i.e. $z=1$. Thus $(1-z)p(z)$ has no zeros in the closed unit disk except $z=1$, which is not a zero of $p(z)$.