The identity theorem at the boundary (complex analysis) Let $\mathbb{D}^2$ be the closed unit disk, and let $f:\mathbb{D}^2 \to \mathbb{C}$ be a smooth map, which is holomorphic on the open unit disk $\text{int}(\mathbb{D}^2)$.

Suppose that there exists a sequence $z_ n \in \text{int}(\mathbb{D}^2)$, $z_n \to z_0 \in \partial \mathbb{D}^2$ such that $f(z_n)=0$. Is $f$ identically zero on $ \mathbb{D}^2$?

The usual formulation of the identity theorem is for open connected domains; it states that a holomorphic function whose zero set has an accumulation point (inside the open domain) is identically zero.
Note that I assumed that $f$ is smooth on the closed disk. (In a sense it is "holomorphic" at the boundary too, as the condition of being conformal is a closed one).
Edit:
If $f$ could be extend $f$ holomorphically to an open neighbourhood of $\mathbb D^2$, then the answer would be positive, by the usual identity theorem (as the accumulation point would now be in the interior of the new extended domain).
I am not sure if such an extension is always possible. There are certainly continuous examples that cannot be extended: e.g. $ f(z) = \sum_{n=1}^\infty \frac{z^{n!}}{n!}$. (See here for details). However, I don't know any smooth example which cannot be extended.
 A: This is not a complete answer, but it was too long for a comment. As such, I am making it community wiki, if anyone wants to fill the missing steps.
An easy example (provided you know the Ostrowski-Hadamard gap theorem) of a $C^{\infty}$ map on the closed disk, holomorphic in the interior which cannot be extended (i.e. the circle is its natural boundary):
$$f(z)=\sum_{n=0}^{\infty} \frac{z^{2^n}}{n!}$$
Back to the original problem:
Consider the orizontal strip $\Omega=\{-\frac\pi2<\Im(z)<\frac\pi2\}$, and the conformal mapping $\varphi:\Omega\to \text{int}(\mathbb{D}^2)$, which is easily seen to be $\varphi(z)=\frac{e^z-1}{e^z+1}=\text{tanh}\left(\frac z2\right)$.
On the horizontal strip, we can consider the function $g(z):=\sin(z)h(z)$, where $h$ is a map that goes to $0$ suitably fast as $z\in \Omega\to \infty$. This function has infinite zeros, and thus precomposing with $\varphi^{-1}$ gives a holomorphic map on the unit disc with infinite zeros. It remains only to prove that it is smooth on the boundary: if $z\neq \pm 1$ this is trivial, and if we choose an $h$ that goes to $0$ fast enough, this should suffice to ensure smoothness. It surely is enough to ensure $n$-th differentiability
