# The inverse of Cauchy's Integral Theorem

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

• 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 '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.