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Suppose that $f:\mathbb{C}\rightarrow\mathbb{C}$ is analytic on the open unit disc and continuous on the closed unit disc. Assume that $f(z)=0$ on an arc of the circle $\{z\in\mathbb{C}:|z|=1\}$. Show that $f(z)=0$.

jkn
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Peyman Ghahremani
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Pedro May 30 '13 at 20:19

2 Answers2

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Another way to do it: First assume that the arc is $0 \le \theta \le \pi$, and define a new function $g(z) = f(z)f(-z)$. Then $g$ is analytic on $D$, continuous on $\bar D$ and equal to $0$ on the whole circle. What does that tell you about $g$, and consequently about $f$?

For the general case, put $g(z) = f(z)f(e^{i\alpha}z)f(e^{2i\alpha}z)\cdots$, where $\alpha$ is suitably chosen.

mrf
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Here is a rough outline.

Conformally map the closed disk to the closed upper half plane so that the arc on which $f(z)$ is zero is sent to the real axis. (This is a standard exercise you should be able to do.) This transforms $f$ into an analytic function on the upper half plane with a continuous extension to the real line. Using the Schwarz reflection principle, we can extend $f(z)$ to be analytic on a neighborhood of some point on this arc. Then the identity theorem shows that $f$ is zero on the entire region, which means $f$ was identically zero originally.

Potato
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    Or you can reflect across the arc directly, as in $f(z) = \overline{f(1/\overline{z})}$ and apply Morera afterwards. – A Blumenthal May 30 '13 at 21:03