Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Cauchy-Goursat theorem for a triangle contour states the following:

Let $\triangle=\triangle(a,b,c)$ be a triangle in an open set $\Omega \subseteq \mathbb{C},p\in \Omega,f:\Omega\rightarrow \mathbb{C}$ continuous and f analytical on $\Omega \setminus \{p\}$. Then:

$\int\limits_{\partial\triangle}f(z)dz=0$ , where $\partial\triangle$ means on and inside the triangle $\triangle$.

The proof then distinguishes a few cases around $p$... but why?:

1) $p \notin \triangle$

2) $p=a$

3) $p \in \triangle$ randomly.

I do understand the proof, but I don't understand what's the purposes of $p$. Why do we even mention it? What's the thought behind this? Some books don't mention it.



Case 1: $ p\not\subset\triangle$. Let $a',\ b',\ c'$ be the midpoints of the segments $[b,\ c],\ [c,\ a],\ [a,\ b]$.

enter image description here

We denote with $\triangle^{j}(j=1,2,3,4)$ the triangles $(a,\ c',\ b'),\ (b,\ a',\ c'),\ (c,\ b',\ a'),\ (a',\ b',\ c')$. Let $I =\displaystyle \int_{\partial\triangle}f(z)dz$. We now have: $$ I\ =\sum_{j=1}^{4}\int_{\partial\triangle^{j}}f(z)\mathrm{d}z $$ There is a $j_{0}\in\{1,2,3,4\}$ so that $$ \left|\int_{\partial\triangle^{j_{0}}}f(z)\mathrm{d}z\right|\geq\frac{|I|}{4} $$ Let $\triangle_{1}:=\triangle^{j_{0}}$.

The same construction with $\triangle_{1}$ instead of $\triangle$ yields a triangle $\triangle_{2}$, etc. By induction we get a sequence $(\triangle_{n})$ of triangles with $\triangle\supseteq\triangle_{1}\supseteq\triangle_{2}\supseteq\ldots$, so that

$|I| \displaystyle \leq 4^{n}\left|\int_{\partial\triangle_{n}}f(z)\mathrm{d}z\right|,\ (n\in \mathbb{N})$

Further we have the length $L(\partial\triangle_{n})=2^{-n}L(\partial\triangle)$ and $\displaystyle \lim_{n\rightarrow\infty}$ diam $(\triangle_{n})=0$ because diam $(\triangle_{n})\leq L(\partial\triangle_{n})$. It follows that (Cantor) $$ \triangle \bigcap_{n=1}\triangle_{n}=\{z_{0}\}. $$ Because $f$ is in $z_{0}$ complex differentiable, there exists for $\varepsilon>0$ a $r>0$ with $$ |f(z)-f(z_{0})-f'(z_{0})(z-z_{0})|\leq \varepsilon|z-z_{0}|\ (z\in C(z_{0},\ r)\subseteq\Omega) $$ and there is a $n\in \mathbb{N}$ mit $\triangle_{n}\subseteq C(z_{0},\ r)$. For this $n$ we have: $$ |z-z_{0}|\leq 2^{-n}L(\partial\triangle)\ (z\in\triangle_{n}) $$ Furthermore: $$ \int_{\partial\triangle_{n}}f(z)\mathrm{d}z=\int_{\partial\triangle_{n}}f(z)-f(z_{0})-f'(z_{0})(z-z_{0})\mathrm{d}z $$

We have: $$ \left|\int_{\partial\triangle_{n}}f(z)\mathrm{d}z\right|\leq \varepsilon\cdot(2^{-n}L(\partial\triangle))^{2} $$ and thus $$ |I|\leq 4^{n}\left|\int_{\partial\triangle_{n}}f(z)\mathrm{d}z\right|\leq \varepsilon(L(\partial\triangle))^{2} $$ Because $\varepsilon>0$ was chosen arbitrarly, $I =0$ follows.

$\color{red}{\text{In my point of view, we are now completely done with the proof. But the proof continues, see below}}$

Case 2:

Now, let $p$ be a vertex of $\triangle$, for instance let $p=a$. If $a,\ b,\ c$ lie on the same line, then we have trivially $I=0$. If not, we choose $x\in[a,\ b],y\in[a,\ c]$ near $a$.

enter image description here

Then we have: $$ \displaystyle \int_{\partial\triangle}f(z)\mathrm{d}z=\int_{\partial\triangle(a,x,y)}f(z)\mathrm{d}z+\int_{\partial\triangle(x,b,y)}f(z)\mathrm{d}z+\int_{\partial\triangle(b,c,y)}f(z)\mathrm{d}z $$ $$ =\int_{\partial\triangle(a,x,y)}f(z)dz. $$ Because we can make $L(\partial\triangle(a,x,\ y))$ arbitrarily small, $I =0$ follows.

Case 3: If $ p\in\triangle$ arbitrarily, $I =0$ follows from case 2:

$$ \displaystyle I\ =\int_{\partial\triangle(a,b,p)}f(z)\mathrm{d}z+\int_{\partial\triangle(b,c,p)}f(z)\mathrm{d}z+\int_{\partial\triangle(c,a,p)}f(z)\mathrm{d}z=0 $$

enter image description here

$\color{red}{\text{Why the hassle of introducing } p?}$

share|cite|improve this question
These are not the usual assumptions in Goursat's theorem. I don't think I've ever seen it formulated like that. Normally you'd require that $f$ is analytic (in the sense that $f'$ exists everywhere) on $\Delta$. Continuity is then implied. Without knowing where your version comes from, I can't really guess why the author fomulate the theorem like this? Perhaps they want a quick proof of Riemann's theorem of removable singuliarities without first developing power series expansions of holomorphic functions? – mrf Jun 5 '12 at 22:32
It's from my complex analysis course at my university. I will soon provide the complete proof. Maybe this will shed some light on the problem. – Chris Jun 5 '12 at 22:37
haha okay yeah the point is basically $f$'s gonna be analytic at $p$ too - if it weren't, it'd blow up to $\infty$ at least as fast as $\frac{1}{z}$, so not continuous – uncookedfalcon Jun 5 '12 at 23:06
I've added the proof! – Chris Jun 6 '12 at 1:04

The introduction of $p$ weakens the hypotheses of the theorem: $f$ is not assumed to be differentiable at $p$. In effect, along with the usual Goursat's theorem (integral of complex differentiable function over a closed contour is zero) we are dealing with the removability of a point for continuous holomorphic functions. (A set $E$ is removable for continuous holomorphic functions in $\Omega$ if every continuous function on $\Omega$ that is holomorphic on $\Omega\setminus E$ is actually holomorphic in $\Omega$. See Which sets are removable for holomorphic functions?)

I find that combining these two issues within the same theorem makes its meaning less clear. You can restore some clarity by dividing it into two theorems. The case $p\notin \triangle$ does not involve $p$ at all: this is a standard theorem you normally find in a textbook. Parts 2 and 3 introduces the aforementioned aspect of removability: we obtain the same conclusion without assuming differentiability at a point $p$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.