= Solution
{c}
Suppose, for a contradiction, that $f$ has no zero in the open disk. The strict boundary inequality implies no boundary zero either. Thus $1/f$ is holomorphic on a neighborhood of the closed disk: compactness and continuity keep $f$ nonzero in a sufficiently small neighborhood. Apply the $k=0$ <Cauchy estimate> to $1/f$:
$$
\frac1{|f(z_0)|}\leq\max_{|z-z_0|=r}\frac1{|f(z)|}
=\frac1{\min_{|z-z_0|=r}|f(z)|}.
$$
This contradicts the assumed strict inequality. If $f(z_0)=0$, the conclusion is immediate without taking a reciprocal. Therefore \b[$f$ has a zero in the open disk]. This is a local form of the <minimum modulus principle>.
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