Is a meromorphic function determined by its boundary values?

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Let $f: \mathbb D \to \widehat {\mathbb{C}}$ be a meromorphic function inside the unit disk.

Assume that $f$ is zero on the boundary and continuous in the closed disk (as a function into $\widehat {\mathbb{C}}$).

Is $f$ necessarily identically zero?

If $f$ is not surjective then we can take $a\not \in \text{Image}(f)$ and reduce to the holomorphic case via $\frac 1 {f(z)-a}$, where the maximum modulus finishes.

What if $f$ is not surjective?

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User Ted Shifrin proved it:

By the continuity on the boundary, there must be finitely many poles, and we may multiply by suitably many factors of the form $z-a_j$ to cancel them out and reduce to the holomorphic case again.