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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 $D$. – hardmath Jan 15 '17 at 18:57
  • 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. – Daniel Fischer Jan 15 '17 at 19:10
  • @hardmath: If the set of zeroes of an analytic function on a connected domain has an accumulation point, the function is zero. –  Jan 15 '17 at 22:55
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    @Hurkyl. We are not assuming $f$ is analytic on $\ bar D.$... Analytic on $D$, continuous on $\bar D$. – DanielWainfleet Jan 15 '17 at 23:49
  • @DanielFischer I guess cauchy integral is not meant like cauchy sum as in 1, 2 or 3 but like 4, 5, 6, 7, 8, – BCLC Aug 10 '21 at 01:46
  • @DanielFischer 9 or 10? – BCLC Aug 10 '21 at 01:46

1 Answers1

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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.

Martin R
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