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If $f:\mathbb{C}\backslash S^1\rightarrow \mathbb{C}$ is holomorphic where $S^1=\{z\in \mathbb{C}:|z|=1\}$ and $f$ continuous in all $\mathbb{C}$ then $f$ is holomorphic in all $\mathbb{C}$.

All the results I know to prove that is holomorphic considers isolated singularities. I think that writing $$f(z)=\sum_{n=0}^\infty a_n z^n,|z|<1$$ might help. Because if I can prove that $$f(z_0)=\sum_{n=0}^\infty a_n z_0^n,|z_0|=1$$ then I finish. Right know the best I know is $f(z_0)=\lim_{z\rightarrow z_0}\sum_{n=0}^\infty a_n z^n$ taking the limit with $|z|<1$.

Luz
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  • you need some extra condition like continuity – Conrad Jul 26 '21 at 23:24
  • True, I forgot to write it, thank you @Conrad. – Luz Jul 26 '21 at 23:25
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    think of Morera theorem as the circle is nice and smooth so triangles have good intersection with it – Conrad Jul 26 '21 at 23:26
  • Okay, I understand know. Thank you @Conrad. – Luz Jul 26 '21 at 23:37
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    This is essentially the Schwarz reflection principle (remember that circles and lines are basically the same thing in the complex plane!)—but anyway that's usually proved using Morera's theorem as @Conrad suggests – Greg Martin Jul 26 '21 at 23:38
  • I tried showing the exact same statement by using Morera, but I'm not sure my approach is right. I'm interested in your proof. If you don't mind, I'd be very happy if you could sketch your proof. Here is my post. Thanks! – jasnee Jul 27 '21 at 10:58
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    See also https://math.stackexchange.com/q/3610683. – Paul Frost Aug 09 '21 at 22:45

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