I have been re-reading through my complex analysis text and wanted to try something different. Cauchy's Integral Theorem is typically proved using an application of Green's Theorem and then by virtue of the Cauchy-Riemann Equations the integral vanishes. I have been trying to do the same to Cauchy's Integral Formula. That is, starting with $\int_{\gamma}\frac{f(z)}{z-a}dz$, with $a$ inside the region defined by the curve $\gamma$ I want to get back $f(a) 2 \pi i$. However, whenever I try Green's theorem on the integral, I get that it vanishes. Here is my work so far (mostly based on the proof of Cauchy's Integral Theorem):

$$\begin{align} & \int_\gamma \frac{f(z)}{z-a}dz \\[10pt] & = \int_\gamma \frac{u(x,y)+iv(x,y)}{z-a}(dx+i\,dy) \end{align} $$

Now let $l(x,y)=\dfrac1{x+iy+a}$

So we have $$ \begin{align} & = \int_\gamma \frac{u(x,y)+iv(x,y)}{z-a}(dx+i\,dy) \\ & = \int_\gamma ul\;dx -vl\; dy + i \int_{\gamma}vl\; dx+ ul\; dy \\ & = \int\int_D -\frac{\partial vl }{\partial x} -\frac{\partial ul }{\partial y} dx\,dy + i \int\int_D \frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \end{align} $$

$\dfrac{\partial ul }{\partial x}=\dfrac{\partial l }{\partial x}u+\dfrac{\partial u }{\partial x}l$ and we know $$\frac{\partial l }{\partial x}=\frac{\partial }{\partial x} \frac1{x+iy+a}=\frac{-1}{(x+iy+a)^2} $$

and similarly

$$\frac{\partial }{\partial x}\frac1{x+iy+a}=\frac{-i}{(x+iy+a)^2} $$

Thus we have $$ \begin{align} & {} \quad \int\int_{D}-\frac{\partial vl }{\partial x} -\frac{\partial ul }{\partial y} dx\,dy + i \int\int_{D}\frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \\[8pt] & =-\int\int_{D}\frac{\partial vl }{\partial x} + \frac{\partial ul }{\partial y} dx\,dy + i \int\int_{D}\frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \\[8pt] & =-\int\int_{D}\frac{\partial l }{\partial x}v+\frac{\partial v }{\partial x}l + \frac{\partial l }{\partial y}u+\frac{\partial u }{\partial y}l \, dx\,dy + i \int\int_{D}\frac{\partial l }{\partial x}u+\frac{\partial u }{\partial x}l-\frac{\partial l }{\partial y}v-\frac{\partial v }{\partial y}l \, dx\,dy \\[8pt] & =-\int\int_{D}\frac{-1}{(x+iy+a)^2}v+\frac{\partial v }{\partial x}l + \frac{-i}{(x+iy+a)^2}u+\frac{\partial u }{\partial y}l\; dx\,dy + i \int\int_{D}\frac{-1}{(x+iy+a)^2}u+\frac{\partial u }{\partial x}l-\frac{-i}{(x+iy+a)^2}v-\frac{\partial v }{\partial y}l\;dx\,dy \end{align} $$

edit: Taking out a small area around the singularity gives the correct answer. Thank you to froggie for pointing this out!

This also gives:

If $g(z)$ is a holomorphic function that is bounded for $| z| < 1$ then for all $|a|< 1$ we have $\displaystyle\pi g(a)=\int\int_{|z|<1} \dfrac{g(z)}{(1-a\bar z)^2}\,dx\,dy$ where $z=x+iy$.

  • $\begingroup$ I don't have time to look at this right now, but I wanted to say that I'm very impressed by your proactiveness so far. Keep up the good work. $\endgroup$
    – davidlowryduda
    Commented May 8, 2012 at 23:29
  • $\begingroup$ I'll have to set aside time to look at this, but it seems like you have to be careful in the sense that $\gamma$ encloses a singularity of the integrand. There is a version of Stokes that deals with this, but maybe you can use a limit argument? $\endgroup$ Commented May 8, 2012 at 23:34
  • 3
    $\begingroup$ This is a great exercise! It's important to remember that Green's theorem doesn't work as usual when the integrand has a singularity in $D$. Here there is a singularity at $a$. I recommend removing a small disk $D(a,\varepsilon)$ around $a$ from the region $D$ before applying Green's theorem. Then take the limit as $\varepsilon\to 0$. $\endgroup$
    – froggie
    Commented May 8, 2012 at 23:37
  • $\begingroup$ So I need to subtract the integral of the line curve around this small disk (with opposite orientation)? $\endgroup$ Commented May 8, 2012 at 23:43
  • $\begingroup$ Yes, that should work. $\endgroup$
    – froggie
    Commented May 8, 2012 at 23:44

1 Answer 1


We may safely assume $a=0$ since translation will not influence the integral in any ways. We want to prove that $\oint_{C}\frac{f(z)}{z}dz=2\pi if(0)$. Consider $z=re^{i\theta}$ we have $$\int_{|r|=R}\frac{f(re^{i\theta})}{re^{i\theta}}(dre^{i\theta}+re^{i\theta}id\theta)=\int_{|r|=R}\frac{f(re^{i\theta})}{r}dr+if(re^{i\theta})d\theta$$

We cannot apply Green's theorem directly because it assumes the area is simply connected. But we can calculate the integral directly via taking limit $R\rightarrow 0$. This can be done by noticing that since $r=R$ the left term actually vanishes; Thus we are left with $$\oint_{|r|=R}if(Re^{i\theta})d\theta=i\int^{2\pi}_{\theta=0}f(Re^{i\theta})d\theta$$

Taking the limit of this integral when $R\rightarrow 0$ should give you $2\pi if(0)$.

This small trivial calculation is associated with Poincare's lemma and DeRham Cohomology. You may venture to read some reference books if you are interested.

  • $\begingroup$ Thanks Chengwei, but I was hoping to get an answer to my second question since froggie helped me figure out my mistake. $\endgroup$ Commented May 9, 2012 at 1:21
  • $\begingroup$ @ricky: I recommend you to work through Stein's book before raising any more such questions. It should be suffice for questions at this level. $\endgroup$
    – Kerry
    Commented May 9, 2012 at 1:23
  • $\begingroup$ I am working through his book, it's a little hard and I have been trying to supplement with Marsden and Ullrich, but I wanted to see if I could this. $\endgroup$ Commented May 9, 2012 at 1:34
  • $\begingroup$ @ricky: His book is not the hardest. Maybe you will feel a little frustrated but you should find it becoming easier after you get used to the material. I worked over his book three years ago and I used to feel the same way. $\endgroup$
    – Kerry
    Commented May 9, 2012 at 1:36
  • $\begingroup$ Thank you. I actually just got the problem! It's cliche, but this stuff is challenging but rewarding. $\endgroup$ Commented May 9, 2012 at 1:57

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