Here is a text from the book Complex Variables and Applications by Churchill :

enter image description here

The proof to this theorem in the book is understandable except for that the proof and the theorem both doesn't say that each of the integrals on $C$ and $C_k$'s are separately zero also since $f(z)$ is analytic. Why is that so?

  • 2
    $\begingroup$ None of those integrals are zero, because $f$ is not analytic in the interior of any of those curves. Look again: $f$ is analytic in the region interior to $C$ and exterior to the $C_j$. $\endgroup$ – David C. Ullrich Mar 26 '16 at 15:01
  • 2
    $\begingroup$ @DavidC.Ullrich - but for example $f(z)=1/{z^2}$ is not analytic in the origin so why its integral around the origin is zero? $\endgroup$ – Liebe Mar 26 '16 at 15:04
  • $\begingroup$ When I said none of those integrals are zero of course I meant they're not necessarily zero. Think about $1/z$ instead of $1/z^2$. $\endgroup$ – David C. Ullrich Mar 26 '16 at 15:08
  • $\begingroup$ @DavidC.Ullrich - yes I got what you mean. But I really don't understand why integral of $f(z)=1/{z^2}$ is zero while $f(z)$ is not analytic? PS It's a bit different question but still relevant matter and I asked it prev and it marked duplicate but it wasn't at all. $\endgroup$ – Liebe Mar 26 '16 at 15:12
  • 1
    $\begingroup$ Cauchy's Theorem says if blah blah blah then the integral is zero. That's "if", not "if and only if". The integral of $1/z^2$ about the unit circle is zero because it is. (This may be clearer later, when you know the Residue Theorem.) $\endgroup$ – David C. Ullrich Mar 26 '16 at 15:25

Your Answer

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

Browse other questions tagged or ask your own question.