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

  • $\begingroup$ And did they satisfy the criterium? $\endgroup$ – Berci Jan 28 '13 at 17:28
  • $\begingroup$ Yes, they satisfy . $\endgroup$ – Miss Independent Jan 28 '13 at 17:37
  • $\begingroup$ So, you already answered your own question. $z\mapsto \bar z$ is thus a counterexample, so the answer on your question is no. $\endgroup$ – Berci Jan 28 '13 at 18:28
  • $\begingroup$ Yes, but i want to be sure and if any one has any other counterexample or if there is a proof. $\endgroup$ – Miss Independent Jan 28 '13 at 21:11
  • $\begingroup$ A counterexample is proof enough. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 25 '14 at 8:27

In the formula: $$ f(z)=\frac{1}{2\pi i} \int_C\frac {f(\zeta) \,d\zeta}{\zeta-z} , $$ the right hand side is an analytic function, with respect to $z$.

Not too hard to show: One can differentiate inside the integral, since $z$ has a positive distance from the contour $C$.

Thus, the left hand side is analytic as well.


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