Zeros of a holomorphic function on the boundary of a closed region Is there a holomorphic nonzero function on a closed bounded connected subset of $\mathbb{C}$ which has infinitely many zeros on the boundary and at most a finitely many zeros in the interior (open connected subset of the closed set)?
If there are infinitely many zeros in the interior, I am not certain whether the function reduces identically to zero by using limit point compactness and constructing a sequence of points with an accumulation point which is certainly in the closure but may not be in the interior.
For clarity: By "holomorphic on a closed bounded connected subset", I mean holomorphic in the interior and continuous on the boundary since holomorphicity is defined only over open sets. For e.g., consider a nonzero function continuous on the closed unit disc and holomorphic on the open unit disc. Can it have only finitely many zeros on the open disc and infinitely many zeros on the boundary? 
Thanks.
 A: Consider the union, over all rationals $p/q\in [0,1)$, written in lowest terms, of the  line segments starting at $0$, having angle $2\pi p/q$ with the positive real axis, and having  length $1/q$. Consider the union of this set with the unit circle and call this set $K$. Then $K$ is a compact, locally connected subset of the Riemann sphere (consisting of the unit circle, the interval [0,1] and countably many "spikes" protruding from zero. Let $U$ be the bounded complementary component of $K$, so $U$ is a simply-connected bounded subset of the plane, with locally connected boundary.
Let $\phi:\mathbb{D}\to U$ be a Riemann map, i.e. a conformal isomorphism between the unit disk and the domain $U$. By the Carathéodory-Torhorst theorem, the map $\phi$ extends continuously to the unit circle. This extension will have uncountably many zeros (one for each "access" to zero from the complement of $K$), but of course the map $\phi$ itself has no zeros.
To get an example of a holomorphic function that has infinitely many zeros, extends continuously to the boundary but has only one zero there (the minimum possible due to continuity) is very easy. For example, restrict the function \sin(z)/z to a horizontal half-strip surrounding the positive real axis, and whose boundary does not pass through any zeros. Precompose with a Riemann map taking the disk to this strip to get the desired map. (This is similar to J.J.'s example as above, but I've divided by z to ensure a continuous extension.)
Edit. It is worth noting that by the F. And M. Riesz theorem, the set of zeros on the boundary has zero one-dimensional Lebesgue measure.
