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Let $f$ be an analytic on a simply connected domain $D$, and $z_0\in D$

then $f(z_0)=\frac{1}{2\pi i}\int_C{\frac{f(z)}{z-z_0}}$ for a closed path C which is all in C, and surrounds $z_0$...

Is it really necessary for $f$ to be analytic on every point in $C$? or continuity of $f$ on $C$ is sufficient?

Please explain why :)

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  • $\begingroup$ Shouldn't it be $\frac{f(z)}{z-z_0} dz$ ? $\endgroup$ – C. Dubussy Apr 13 '16 at 19:32
  • $\begingroup$ yes it is really necessary for $f$ to be analytic (holomorphic) on an open connected set containing the contour $C$ and $z_0$ $\endgroup$ – reuns Apr 13 '16 at 19:34
  • $\begingroup$ You are right, edited :) $\endgroup$ – Daniel Apr 13 '16 at 19:34
  • $\begingroup$ read a complex analysis course for proving the Cauchy integral theorem that $\int_C g(z) dz = 0$ whenever $g(z)$ is holomorphic on $U$ containing $C$. then use that if $f(z)$ is holomorphic on $U$ then $g(z) = \frac{f(z)-f(z_0)}{z-z_0}$ is holomorphic on $U$ hence $\int_C \frac{f(z)-f(z_0)}{z-z_0} dz = 0$ and $\int_C \frac{f(z)}{z-z_0} dz = \int_C \frac{f(z_0)}{z-z_0} dz = 2 i \pi f(z_0)$ (if $z_0$ is in the interior of the contour) $\endgroup$ – reuns Apr 13 '16 at 19:45
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Based on your question, let's assume that $f$ is continuous on $C$. If $f$ is not analytic at some point $z_1 \in C$, but continuous at $z_1$, we may take the analytic continuation of $f$ to the entire boundary $C$, and then it reduces to the case when $f$ is analytic on $C$.

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  • $\begingroup$ In general, there is no reason to believe that a holomorphic function continuous up to boundary admits an analytic continuation (not even to a single point on $C$). $\endgroup$ – mrf Apr 13 '16 at 20:24
  • $\begingroup$ We already know that $C$ is contained in the $D^\circ$, so if $f$ is holomorphic in $D-\{z_1\}$, and continuous on $D$ (or bounded on $D$), then we may take the analytic continuation. $\endgroup$ – Yuxin Wang Apr 13 '16 at 20:38
  • $\begingroup$ It's not clear from the statement that $C$ is contained in $D$. If it is, the OP is already assuming that $f$ is analytic on $D$. I suspect he wants to ask about when $C = \partial D$ (but your intepretation may very well be the intended one). $\endgroup$ – mrf Apr 13 '16 at 20:41
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Once you have Cauchy's integral theorem, it's fairly straight-forward to prove Cauchy's integral formula under the same assumptions.

It's known (but not easy to prove in full generality) that if $\Omega$ is a simply connected domain such that $\partial\Omega$ is a rectifiable Jordan curve, and $f$ is a continuous function on $\bar\Omega$, holomorphic on $\Omega$, it is true that $$ \int_{\partial\Omega} f(z)\,dz = 0. $$ See this question for details. Consequently, under the same assumptions, $$ \int_{\partial\Omega} \frac{f(z)}{z-z_0}\,dz = 2\pi i f(z_0) $$ for $z_0 \in \Omega$.

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