0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.