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This is my first question in MSE:

Let $D$ denote the open ball of unit radius about origin in the complex plane $\Bbb C$.

Let $f$ be a continuous complex-valued function on its closure $D$ which is analytic on $D$. If $f(e^{it}) = 0$ for $0 < t <\frac{\pi}{2}$ , show that $f(z) = 0 $ for all $z$.

Here is what I tried:

$f$ is analytic on $D$. If I can find a sequence $(z_n)_n$ in $D$ such that $f(z_n)=0\forall n$ and additionally $(z_n)_n$ has a limit point in $D$ then we are done.

I think $f(e^{it}) = 0$ may help finding one sequence but I am not sure.

Will you kindly help?

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    Is there a basic property of analytic functions that this exercise might have been intended to illustrate? I'm not clear how finding a sequence in $D$, the open unit ball centered on the origin, with a limit point in $D$ will allow you to say "we are done". After all the sequence $\{1/n\}_{n=1}^\infty$ has a limit point in $D$. – 2017-01-15
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    Look at the Cauchy integral of $f$. If you look sharp, you will see that it defines a holomorphic function on a larger connected open set than just the unit disk. – 2017-01-15
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    @hardmath: If the set of zeroes of an analytic function on a connected domain has an accumulation point, the function is zero. – 2017-01-15
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    @Hurkyl. We are not assuming $f$ is analytic on $\ bar D.$... Analytic on $D$, continuous on $\bar D$. – 2017-01-15

2 Answers 2

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Let $\omega \ne 1$ be a 5-th root of unity ($\omega^5 = 1$) and define $$ g(z) = f(z)f(\omega z)f(\omega^2 z)f(\omega^3 z)f(\omega^4 z) \, . $$ $g$ is analytic in the unit disk $\Bbb D$ and continuous on it closure because $f$ is.

For every $z \in \partial \Bbb D$, at last one of the points $z, \omega, z, \omega^2 z, \omega^3 z, \omega^4 z$ has an argument between $0$ and $\pi/2$, so that $g(z) = 0$ on the boundary of the unit disk.

Now apply the maximum principle.

In the same way one can show that $f$ is identically zero if $f(z)=0$ on any segment of the unit circle.

Another option would be to use the Schwarz reflection principle (which is a more advanced topic however). It implies that $f$ can be continued analytically to a larger domain containing the arc $\{ e^{it} \mid 0 < t < \pi /2 \}$ in its interior. Then the identity theorem can be applied.

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$f(e^{it})$ is a continuous, and thus commutes with taking limits.