Let $f$ be analytic in $\mathbb D$ and continuous in $\overline {\mathbb D}$ . Let $A$ $=$ {$ e^{i \theta}| |\theta - \pi |< \epsilon$} , for $\epsilon > 0$ small enough such that $f|_{A} = 0$ . Show that $f$ is identically equal to $0$ on $\mathbb D$.
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One method to see it is the reflection principle. You can reflect $f$ across the arc $A$ obtaining a holomorphic extension to a connected open set containing the arc $A$. The identity theorem says that the extension vanishes identically since it vanishes on the arc $A$. Using
$$\tilde{f}(z) = \begin{cases} \overline{f(1/\overline{z})} &, \lvert z\rvert > 1 \\ f(z) &, \lvert z\rvert \leqslant 1,\end{cases}$$
you have the reflection across $A$ explicitly, the analyticity of $\tilde{f}$ is clear in the unit disk and the exterior of the unit disk, for the analyticity on the arc $A$, use Morera's theorem.
Another, lower-level way is to choose an $n$ such that $n\cdot \varepsilon > \pi$, and consider
$$g(z) = \prod_{k=0}^n f(e^{2\pi ik/n}z),$$
which is continuous on the closed unit disk, holomorphic on the open unit disk, and vanishes identically on the unit circle. The maximum modulus theorem implies $g\equiv 0$, and that in turn implies $f \equiv 0$.
Daniel Fischer
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