A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Cor 5.9

(Cor 5.9) If $f$ is a holomorphic function on a simply-connected region $G \subseteq \mathbb C$, then $\forall \gamma \subset G$ piecewise smooth, $\int_{\gamma} f$ is path independent.


  1. Could we replace '$f$ is holomorphic on a simply connected-region' with the weaker '$f$ has an antiderivative and is continuous on an open subset'?

This seems to still satisfy the conditions of a corollary (Cor 4.13) to the complex analogue of the Fundamental Theorem of Calculus Part II (Thm 4.11)

  1. Actually, even if $f$ is continuous on an open subset that is not a region, $G$, can $f$ have an antiderivative on $G$?

In the book, Cor 4.13 allows for antiderivatives on open subsets yet antiderivatives are defined on regions, defined as open connected subsets.

This might be an error in the text, in which case I would like to know if antiderivatives on disconnected open subsets are possible, and if it's not an error, I would like to know, well, how it's not an error.

  • 2
    $\begingroup$ "Open set that is not a region" ... what is the definition of "region" in that book? $\endgroup$ – GEdgar Aug 10 '18 at 10:26
  • $\begingroup$ @GEdgar Edited (again): Open connected subsets. Thanks! $\endgroup$ – BCLC Aug 10 '18 at 10:33

Yes, it is enough for $f$ to have an antiderivative to conclude that its integrals are path-independent. You can certainly have a function on a disconnected open set with an antiderivative. This is really quite obvious: given a disconnected open set $G$, let $g$ be any holomorphic function on $G$, and consider $f=g'$.

In practice though there is no loss of generality in restricting to connected open subsets. If $G$ is open and $f$ has an antiderivative on each connected component of $G$, then those antiderivatives combine to form an antiderivative of $f$ on all of $G$. So $f$ has an antiderivative iff its restriction to each connected component has an antiderivative.

  • $\begingroup$ Thanks Eric Wofsey! What about question 1 please? $\endgroup$ – BCLC Aug 10 '18 at 23:00
  • $\begingroup$ I answered question 1 in the affirmative in the first sentence. $\endgroup$ – Eric Wofsey Aug 10 '18 at 23:01
  • $\begingroup$ Eric Wofsey oh wait so your 1st sentence means 'f's having an antiderivative, and G does not have to be connected is enough to conclude path independence' rather than 'f's continuity on an open disconnected subset is enough for f to have an antiderivative, and G does not have to be connected' ? Could be wrong, but I think there's some ambiguity $\endgroup$ – BCLC Aug 10 '18 at 23:06
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    $\begingroup$ That is correct. It is certainly not true that an arbitrary continuous function has an antiderivative! $\endgroup$ – Eric Wofsey Aug 10 '18 at 23:07
  • $\begingroup$ Ayt thanks! ^-^ $\endgroup$ – BCLC Aug 10 '18 at 23:08

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