I'm using Donald Sarason's book as a reference.

(One version of) Cauchy's theorem states that for a holomorphic function defined on an open, convex set, $\int_\gamma f(z)\mathrm{d}z$ = 0 for any piecewise-$C^1$ closed curve $\gamma$ in this set.

On the other hand, Cauchy's formula for a circle gives one a method to compute the value of a holomorphic $f$ for any point $z$ in the interior of the circle: $$ f(z) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi - z} \mathrm{d}\xi\ . $$

My question is the following: why doesn't the integral on the right hand side of the above question vanish by Cauchy's theorem? To be painfully explicit, if we consider a circle centered around the origin, why is it not true that $$ f(0) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi} \mathrm{d}\xi\ = 0 $$ for all holomorphic functions $f$?

I know I'm missing something elementary; I just can't see what it is.


2 Answers 2


Because the denominator of the integrand gives rise to a singularity, so the integrand is not holomorphic on any open convex set containing the circle you are integrating over.

  • $\begingroup$ But the denominator is not singular on the circle itself, which is the contour along which we're integrating, correct? $\endgroup$
    – Anthony
    Aug 27, 2018 at 16:57
  • 3
    $\begingroup$ Yes, but the Cauchy theorem requires the function to be holomorphic on an open convex set containing the contour, so the point $z = \xi$ will always be contained there. More particularly, in your example, if you integrate around any circle centered at the origin, the open convex set that is required to apply Cauchy's theorem will always contain the origin, since it would not be convex otherwise. I can try to elaborate this further in my answer, if you need me to. $\endgroup$
    – MSDG
    Aug 27, 2018 at 16:58
  • $\begingroup$ No, that's clear now, thanks! I just got hung up thinking about the contour itself and forgot that the region in which it's contained matters. $\endgroup$
    – Anthony
    Aug 27, 2018 at 17:05
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    $\begingroup$ @Anthony If I remember correctly, I had a similar problem when I was learning this. :) $\endgroup$
    – MSDG
    Aug 27, 2018 at 17:07

the integrand in $$f(z) = \frac{1}{2 \pi i} \int_C \frac{f(\xi)}{\xi - z} \mathrm{d}\xi\ $$

is $$\frac{f(\xi)}{\xi - z}$$ which is not holomorphic within the region bounded by the boundary.

Consider the point $z=\xi $ where your function is not defined.


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