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I am reviewing Stein's complex analysis , and there's one theorem I listed follows:

Suppose $ f$ is a holomorphic function on a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$ , then $f$ is identically $0$ . (Here region means both open and connect)

My question is : If we do not assume the limit point is in $\Omega$ , the limit point might in the boundary of $\Omega$ , can we get the similar conclusion ? Since the orginal proof using the power series of the limit point $z_0$ , I have no idea how to deal with this .

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No, if the limit point is outside the domain then we cannot conclude that the function is identically zero.

To be specific, the Blaschke product gives a very specific construction of functions with prescribed zeros within the unit disk (potentially infinitely many!) which are not identically zero. The zeros do need to satisfy some growth conditions, though - they have to be moving towards the boundary of the disk quickly enough.

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  • $\begingroup$ I see it ! Thanks for the answer . $\endgroup$ – J.Guo Dec 20 '18 at 17:20
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No, take $f(z)=\sin(1/z)$ on the punctured plane. Then $f=0$ on $z_k=\frac 1{k\pi}$ and $z_k\to 0$ but $f$ is not identically 0.

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