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Let $f(z)$ be a complex valued function defined on an open subset $U$ of the complex plane. We say $f(z)$ is analytic if it can be expressed by a convergent power series on a neighborhood of every point of $U$. Suppose $U$ is simply connected. Then it is easy to prove that the contour integral of $f(z)$ on a piecewise continuously differentiable closed curve in $U$ is zero. It is easy because you can reduce it in the trivial case when $f(z) = z^n$. Let's call this the weak version of Cauchy's theorem.

I think we can get most of the important results of complex function theory of one variable by using this theorem. The reason is as follows. In most applications of complex analysis, it is easy to prove that the function is analytic in our sense. For example, it is easy to show that the Riemann zeta function is analytic in our sense.

Am I mistaken?

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  • $\begingroup$ I think this is basically correct and Morera's theorem. Or, are you aware of this result and want something more specific? $\endgroup$ – quid Jun 28 '16 at 22:15
  • $\begingroup$ "It is easy because you can reduce it to the trivial case when f(z) = z^n." Care to explain why? $\endgroup$ – zhw. Jun 29 '16 at 4:59
  • $\begingroup$ @zhw. Let p be any point of U. Let W be an open disc contained in U whose center is p such that f(z) can be expressed as a convergent power series on W. Consider a small triangle T contained in W. Then it is clear that the contour integral of f(z) on the boundary of T is zero. Since U is simply connected, the contour integral of f(z) along a closed polygon in U is zero. Since any piecewise continuously differential closed curve on U is arbitrarily approximated by a closed polygon, we are done. $\endgroup$ – Makoto Kato Jun 29 '16 at 5:36
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From your assumptions it only follows that $\int_\gamma f(z)\>dz=0$ for tiny closed curves $\gamma$, where "tiny" means that such a $\gamma$ has to lie in the domain of convergence of a single power series locally representing $f$.

For a full proof you would have to show that the interior of any (large) Jordan polygon $\pi$ can be triangulated into tiny $\triangle_k$ such that each $\triangle_k$ is contained in such a local domain of convergence, and then use that $\pi=\sum_k\partial\triangle_k$.

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  • $\begingroup$ That a Jordan polygon can be triangulated into sufficiently small triangles can be easily proved as follows. By induction on the number of edges of the polygon, we can assume that the polygon is a triangle. Then the assertion is clear. $\endgroup$ – Makoto Kato Jun 30 '16 at 1:04
  • $\begingroup$ @MakotoKato: Just to make a note here, we can't so easily do induction because the polygon might not be convex. And for the concave case it boils down to showing that it has an interior (Jordan curve theorem for polygons)... $\endgroup$ – user21820 Jul 28 '16 at 8:56
  • $\begingroup$ @user21820 I don't understand. Would you elaborate? $\endgroup$ – Makoto Kato Jul 30 '16 at 23:40
  • $\begingroup$ @MakotoKato: You claimed that you can triangulate a polygon. That's not at all obvious. Define "triangulation" rigorously and you'll see that you need to prove at least a large part of the Jordan curve theorem for polygons. Christian's answer uses the term "interior", which is not at all well-defined until you have proven it exists... $\endgroup$ – user21820 Jul 31 '16 at 1:53
  • $\begingroup$ @user21820 FYI, I have posted a question "Triangulation of a Jordan polygon". $\endgroup$ – Makoto Kato Jul 31 '16 at 22:30

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