# Suppose f(z) is analytic with [duplicate]

Let $$f$$ be an analytic function on $$D(0, 1)$$ and continuous on $$\overline{D(0, 1)}$$. We assume that $$f \equiv 0$$ on the $$arc$$ defined by {$$e^{i\theta}, 0\le \theta_1 \lt \theta \lt \theta_2 \le 2\pi$$}. $$\quad$$ prove that $$f= 0$$ on $$D(0, 1)$$.

Schwarz reflection principle across the arc. For $$|z|>1$$ define $$f(z) = \overline{f(1/\overline{z})}$$. This is analytic on $$D(0,1)$$ and $$|z|>1$$, and continuous on the union of these and the arc. Using Morera's theorem, it is analytic there. Then by the identity theorem, it is $$0$$ everywhere.
Another way: if you multiply together a bunch of functions of the form $$f(cz)$$, where $$c$$ is a complex number with $$|c|=1$$, you can get a function that is $$0$$ on the entire boundary. By the maximum principle, the function is $$0$$ everywhere. $$f$$ and the rotations can each only have countably many zeroes unless they are constant, but this product has uncountably many. Thus, $$f$$ is $$0$$ everywhere.