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I do not know if the following is true: A bounded region $U$ is simply connected if and only if for any holomorphic function $f$ on $U$ and any closed curve $\gamma\subseteq{U}$ we have $\int_{\gamma}f(z)dz=0$. The "only if" part follows from the homotopy of curves theorem, but I don't know how to begin with the "if" part, intuitively I think it is because the integral condition implies that $U$ has no "holes", and this implies $U$ is biholomorphic to an open ball.

Or to prove that the interior of a closed curve is simply connected.

Thanks

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It $U$ is not simply connected, then there exists a simple closed curve $\gamma:S^1\to U$, such that $U$ does not contain the bounded connected component $C$ of $\mathbb{C}\setminus{\gamma(S^1)}$. Choose $z_0\in C\setminus U$ and let $f(z)=\frac{1}{z-z_0}$. Then $f$ is holomorphic on $U$ but $\int_{\gamma}f(z)dz=2\pi i$.

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    @CamiloArosemena: There may not be complex analysis involved in the proof of Jordan–Schönflies theorem, but it has a strengthened version in complex analysis, [Carathéodory's theorem](http://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(conformal_mapping)), which also implies your result.2012-11-11