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In his Complex Variables text, Conway defines a triangular path as follows:

A closed path $T$ is said to be triangular if it is polygonal and has three sides.

I am having some confusions regarding his definition. First of all, why does he specify that the path is polygonal ? Does this mean some general version of a triangular path? My point is, if the path has three sides and is closed, then why does he need to specify that it is polygonal? What else could it be other than a triangle?

Additionally, I have another confusion regarding his proof of Morera's Theorem.

Morera's Theorem. Let $G$ be a region and let $f:G\to\mathbb C$ be a continuous function such that $\int_{T}f=0$ for every triangular path $T$ in $G$; then $f$ is analytic in $G$.

In his proof (which according to this question is wrong?), where does Conway make use of the hypothesis that $\int_{T}f=0$ for every triangular path $T$? TIA.

math-physicist
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  • He may mean, to make precise definition - "3-sided polygona = triangle"; but this understanding in geometry involves non-orientation. The triangular path as per Conway's convention may be oriented triangular path. – Maths Rahul Aug 03 '22 at 04:23
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    $F(z)=\int_{[a,z_0]}f+\int_{[z_0,z]}f$ This requires the fact that the integral of $f$ over the triangle determined by $a,z_0,z$ is $0$. – geetha290krm Aug 03 '22 at 05:04
  • @MathsRahul : what do you actually mean by non-orientation? What's the difference between an oriented triangle path vs. one that's not?? – math-physicist Aug 03 '22 at 05:52
  • @geetha290krm : I still do not see why the path $[a,z_0]\cup[z_0,z]$ is a triangular path. To me this is just the connected path of two line segments $[a,z_0]$ and $[z_0,z]$. Additionally, why does your mentioned equality require $\int_{T}f=0$? – math-physicist Aug 03 '22 at 05:59
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  • Polygonal refers to the fact the sides are straight and not curved. It is also oriented. Another term often used for this in maths is PL, meaning piecewise-linear.

  • Regarding the integration, let $T$ be the oriented triangle formed by $a, z, z_0$, so $T = [z_0,z] + [z,a] + [a,z_0]$, and note $0 = \int_T f = \int_{[z_0,z]} f + \int_{[z,a]} f + \int_{[a,z_0]} f = \int_{[z_0,z]} f - \int_{[a,z]} f + \int_{[a,z_0]} f$, where we have reversed the orientation on $[z,a]$ and thus the sign of the integral. We rearrange to get $F(z) := \int_{[a,z]}f = \int_{[z_0,z]} f + \int_{[a,z_0]} f$.

    • – Brevan Ellefsen Aug 03 '22 at 09:53
  • @BrevanEllefsen : Thanks for the detailed explanations. Totally understood now. If you want to post the above as an answer, I'd be happy to accept it. – math-physicist Aug 03 '22 at 15:49