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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 5.7

Exer 5.7 Prove $\overline{z}^2$ has no antiderivative on a non-empty region.

In the textbook, the discussion after Cor 5.9 (*) suggests the way to prove Exer 5.7 is to note that $\int_{\gamma} \overline{z}^2 \, dz$ is path dependent with the idea that existence of antiderivative implies path independent.

Now, I will attempt to do prove Exer 5.7 both along the suggestion (Way (1)) and alternatively (Way (2)). Please verify.

Proof of Exer 5.7: (Both ways start the same)

Let $G$ be a nonempty region. Observe that $\overline{z}^2$ is continuous in $G$, so by Morera's Thm 5.6, $$\exists \gamma \subset G: \int_\gamma \overline{z}^2 \, dz \ne 0$$

Now, I think we can proceed in 1 of 2 ways:

Way (1): Without noting path dependence of $\int_{\gamma} \overline{z}^2 \, dz$

Observe $G$ is open $\because G$ is a region.

Therefore, by a corollary (Cor 4.13) to the complex analogue of the Fundamental Theorem of Calculus Part II (Thm 4.11), $\overline{z}^2$ has no antiderivative on $G$.

QED via Way (1)

Way (2): Noting path dependence of $\int_{\gamma} \overline{z}^2 \, dz$

As in this proof for Cor 5.9, decompose $\gamma = \gamma_1 \wedge -\gamma_2$ where $\gamma_1$ and $\gamma_2$ have the same start and end points s.t. $$0 \ne \int_\gamma \overline{z}^2 \, dz = \int_{\gamma_1} \overline{z}^2 \, dz - \int_{\gamma_2} \overline{z}^2 \, dz \implies \int_{\gamma_1} \overline{z}^2 \, dz \ne \int_{\gamma_2} \overline{z}^2 \, dz$$

Recall that while the existence of an antiderivative of a holomorphic function on a simply-connected region implies path independence of integrals over said function over piecewise smooth paths in the region, we also have that

the existence of an antiderivative of a continuous function on an open subset implies path independence of integrals over said function over piecewise smooth paths in the subset. $\tag{**}$

Therefore, by $(**)$, $\overline{z}^2$ has no antiderivative on $G$.

QED via Way(2)

(*) (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.

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    First off: it's barely possibile to follow your thoughts. Please rethink the structure of the question. Second: How do you come up with those non-zero path integrals? You must give explicit calculations, otherwise how can you state that these exist? – b00n heT Aug 08 '18 at 12:55
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    It makes sense to write either $\displaystyle \int_\gamma f$ or $\displaystyle \int_\gamma f(z),dz,$ but writing $\displaystyle \int_\gamma f(z)$ is problematic. If, $z=g(w)$ then $\displaystyle \int_\gamma f(z), dw$ is different from $\displaystyle \int_\gamma f(z), dz,$ etc. $\qquad$ – Michael Hardy Aug 08 '18 at 12:56
  • @MichaelHardy Thanks! ^-^ – BCLC Aug 08 '18 at 12:57
  • @b00nheT 1. I guess you meant barely possible. I'll edit. I think I know where I was unclear. 2. Morera's Theorem? – BCLC Aug 08 '18 at 12:58
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    Anyway, what you are trying to achieve is rather nonsense: If you can find a closed path whose integral does not evaluate to $0$ then indirectly (as in the proof of the corollary) you have proved path dependence. Conversely if you first find two paths which evaluate to different results, then concatenating them leads to a non zero closed-path integral. The main part that is still missing is finding the explicit path. – b00n heT Aug 08 '18 at 13:01
  • @b00nheT Thanks. My desire to avoid path dependence is to see if I can prove Exer 5.7 using Cor 4.13 instead of a seemingly implicit assumption in Cor 5.9. This was in the original question: I'm actually not sure why Cor 5.9 is true. It might be that 'antiderivative implies path independent', a seemingly implicit assumption that I attempted to prove in the original question. I see how you might think this is nonsensical if you don't consider the context of the original question. – BCLC Aug 08 '18 at 13:11
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    By Corollary 5.5 of that text, I believe that it suffices to show that $f(z)=\bar{z}^2$ has no derivative on any non-empty region. – Mark McClure Aug 10 '18 at 23:30
  • @MarkMcClure That is so weird. The discussion after Cor 5.9 suggests to use path dependence. Am I understanding right? Anyhoo, thanks! – BCLC Aug 10 '18 at 23:37
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    In some ways, integration is more fundamental than differentiation so it might make sense to try to apply path independence. To do so, you might try integrating from $a+bi$ to $(a+bi) + r(1+i)$ along two paths - one diagonal across the square and the other along two lines parallel to the real and imaginary axes. I'm pretty sure you'll get different things. – Mark McClure Aug 11 '18 at 00:13
  • @MarkMcClure Yeah that's actually what Eg 2.8 does. Anyhoo, I posted an answer. How is it please? Thanks! – BCLC Aug 11 '18 at 00:15
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    @BCLC Well, it's been accepted! :) – Mark McClure Aug 11 '18 at 00:25
  • @MarkMcClure Um, i'm the OP and answerer.....anyhoo lol thanks – BCLC Aug 11 '18 at 00:26
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    Yes, hence the smiley joke indicator :) – Mark McClure Aug 11 '18 at 00:29
  • @MarkMcClure Had a feeling. Wasn't sure LOL – BCLC Aug 11 '18 at 00:30

1 Answers1

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Way (1): Right, but if you want to avoid noting path dependence, let's not use Morera's Thm (Thm 5.6). Instead, use Cor 5.5 in the textbook, as Mark McClure points out.

By Cor 5.5, if a function $f$ has an antiderivative $F$, then both the $F$ and $f$ are holomorphic. However, an example in the textbook, Eg 2.8, proves nowhere holomorphicity.

So actually, if a function is nowhere holomorphic, it has an antiderivative only in $\emptyset$.

Way (2): Right by Eric Wofsey's confirmation of weaker assumption.

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