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I was given the definition of anlytic continuation as in Stein's book- given two regions $\Omega\subset \Omega'$ for analytic functions $F:\Omega'\to \mathbb{C}$ is an analytic continuation of $f:\Omega \to \mathbb{C}$ if $F$ agrees with $f$ on $\Omega$. I know that $\sin:\mathbb{C}\to \mathbb{C}$ is an analytic continuation of $\sin:\mathbb{R}\to \mathbb{R}$ because they agree on $\mathbb{R}$, but this is not a region in $\mathbb{C}$, so how does that add up with the defintion?

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There is some ambiguity here (not in what you wrote, which is perfectly clear, but in Stein's textbook). Sometimes, while writing about analytic continuation, Stein indeed assumes that $\Omega$ and $\Omega'$ are regions. But when finally Stein states a theorem about analytic continuation, the statement is:

Theorem: Suppose $f$ is a holomorphic function in a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$. Then $f$ is identically $0$.

As you can see, when it is stated like this, the theorem on analytic continuation applies indeed to the situation that you described.

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  • $\begingroup$ I think it answered my question; We know that $\sin(z)$ agrees with $\sin(x)$ on $\mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer! $\endgroup$ – Simon Green Jan 18 at 11:10
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    $\begingroup$ Yes, that is correct. $\endgroup$ – José Carlos Santos Jan 18 at 11:20

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