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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

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

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

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

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 '13 at 17:28
    
Yes, they satisfy . –  Miss Independent Jan 28 '13 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 '13 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 '13 at 21:11
    
A counterexample is proof enough. –  Martín-Blas Pérez Pinilla Jan 25 at 8:27

1 Answer 1

In the formula: $$ f(z_0)=\frac{1}{2\pi i} \int_C\frac {f(z)}{z-z_0} dz, $$ the right hand side is an analytic function, with respect to $z_0$. (Not too hard to show.) Thus so is the left hand side.

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