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Cauchy's 1st integral formula : let f(z) be analytic in simply connected domain D containing a simple closed contour C . If $z_0$ is inside C then

$ f(z_0)=\frac{1}{2\Pi i} \int\frac {f(z)}{z-z_0} dz $

my question is :suppose that C is simple closed contour such that for each $z_0$ inside C we have :

$ f(z_0)=\frac{1}{2\Pi i} \int\frac {f(z)}{z-z_0} dz $

Does it follow that f is analytic inside C??

i tried $\overline z $ and $|z|^2$ they are not analytic

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And did they satisfy the criterium? – Berci Jan 28 at 17:28
Yes, they satisfy . – Miss Independent Jan 28 at 17:37
So, you already answered your own question. $z\mapsto \bar z$ is thus a counterexample, so the answer on your question is no. – Berci Jan 28 at 18:28
Yes, but i want to be sure and if any one has any other counterexample or if there is a proof. – Miss Independent Jan 28 at 21:11

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