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This proof has two oddities in it. It is a proof of the Cauchy-Goursat theorem for triangle contours, i.e. the contour integral of a holomorphic function around the boundary of a triangle in $\Bbb C$ is zero.

In the following, $\triangle$ is the original triangle and $\triangle_n$ is a sequence of triangles, where $\triangle_n\subseteq\triangle_{n-1}$, and $\triangle_0$ is defined as $\triangle$.

$$\triangle\bigcap_{n=1}^\infty\triangle_n=\{z_0\}$$

For some unique $z_0$, and they reference Cantor - what is the name of this theorem?

The main question however is about this:

$$\int_{\partial\triangle_n}f(z)\,\mathrm{d}z=\int_{\partial\triangle_n}(f(z)-f(z_0)-f'(z_0)(z-z_0))\,\mathrm{d}z$$

They offer no explanation for this, other than that $f$ is complex differentiable at $z_0$. How can it be that the contour integral is equal to this expression? By definition of differentiability, $f(z)=f(z_0)+f'(z_0)(z-z_0)+\psi(z)(z-z_0)$, where $\psi$ is continuous and $\lim_{z\to z_0}\psi(z)=0$, so the RHS of that integral is infact just the remainder term, $\psi(z)(z-z_0)$, meaning that they have assumed that:

$$\int_{\partial\triangle_n}f(z)\,\mathrm{d}z=\int_{\partial\triangle_n}\psi(z)(z-z_0)\,\mathrm{d}z$$

Which is only explainable (as far as I can tell) if you presuppose the conclusion of that proof, which is that the contour integral around a triangle is $0$. What am I missing?

Many thanks.

FShrike
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1 Answers1

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Cantor here is a reference to Cantor's intersection theorem.

On the other hand, note that, since $f(z_0)$ is constant, $\int_{\partial\Delta_n}f(z_0)\,\mathrm dz=0$. And, since $f'(z_0)(z-z_0)$ has an antiderivative ($\frac12f'(z_0)(z-z_0)^2$) and $\partial\Delta_n$ is a closed path, $\int_{\partial\Delta_n}f'(z_0)(z-z_0)\,\mathrm dz=0$ too. Therefore, you have indeed$$\int_{\partial\Delta_n}f(z)\,\mathrm dz=\int_{\partial\Delta_n}f(z)-f(z_0)-f'(z_0)(z-z_0)\,\mathrm dz.$$

José Carlos Santos
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  • Thank you for taking the time with a trivial question like this - I am just learning these things, and I am missing the obvious. – FShrike Sep 17 '21 at 17:55
  • The expression $|z-z_0|$ is not analytic, but they claim $\int_{\partial\triangle_n}|z-z_0|,|dz|\le M^2$, where $M\ge|z-z_0|$. How does one arrive at this inequality? – FShrike Sep 17 '21 at 18:03
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    If the length of $\partial\Delta_n<M$ (which is true if $n\gg1$), then$$\int_{\partial\Delta_n}|z-z_0|,\mathrm dz\leqslant(\text{length of }\Delta_n)\times\sup|z-z_0|\leqslant M^2.$$ – José Carlos Santos Sep 17 '21 at 18:13