I'm reading a book that defines and proves that there exists a series with negative and positive terms inside a disk. I'll have to write a bunch of things because of the context, but the main question is easy, just look at the image... Now, the book says:

Consider $f$ in the annulus $f:A(a,p_1,p_2)\to\mathbb{C}$, fix a point $z$ in $A(a,p_1,p_2)$ and let $r_1,r_2$ be two real numbers satisfying $p_1<r_1<|z|<r_2<p_2$. Look to the closed annulus $\overline{A}(a,r_1,r_2)$. Its boundary is formed by the circles $\gamma_1=\{|w-a|r_1\}$ and $\gamma_2=\{|w-a|=r_2\}$. Lets use the Cauchy Integral Formula to describe $f$ around $z$.

Initially, observe that $g(w) = \frac{f(w)}{w-z}$ is holomorph in the ring $A(a,p_1,p_2)$ except in the point $z$. Therefore, we'll isolate $z$ considering the disk with it in the center, $D(z,\tau)$ such that $\overline{D}(z,\tau)\subset A(a,r_1,r_2)$ and the boundary is the circle $\gamma$, which we oriented counterwise. By the Cauchy Integral Formula we have:

$$f(z) = \frac{1}{2i\pi}\int_{\lambda}\frac{f(w)}{w-z}\ dw$$

Consider now the region $V$ formed by the points interior to $\gamma_2$, exterior to $\gamma_1$ and $\lambda$. Its boundary $\partial V$ is formed by the three Jordan curves $\lambda, \gamma_1$ and $\gamma_2$. By considering the compatible orientation, we have: $\partial V = \gamma_1^{-}\cup \lambda^{-}\cup \gamma_2$. Since $g(w) = \frac{f(w)}{w-z}$ is holomorph in an open wich contains $V$ and $\partial V$, the Cauchy Theorem guarantees us that:

$$0 = \int_{\partial V}g(w)\ dw = \int_{\partial V}\frac{f(w)}{w-z}\ dw$$

Is the integral above $0$ because I can integrate like in the image I drawn?

If not, how should I break up the annulus in order for the integral to be $0$ and be represented as the sum of the integrals of each circle?

enter image description here

  • 1
    $\begingroup$ Yes, that is a standard consequence of Cauchy-Gousard, and is proven by connecting the two curves by segments as you did. $\endgroup$
    – N. S.
    Nov 27, 2016 at 17:28
  • $\begingroup$ @N.S. It's 'Goursat'. $\endgroup$
    – Pedro
    Nov 27, 2016 at 17:29
  • $\begingroup$ @N.S. could you explain better for me where is the closed path in a connected domain? I can't exactly see it. Is it one entire closed path? Or it's broken in more than one? $\endgroup$ Nov 27, 2016 at 17:32

1 Answer 1


To make your proof a bit more clear, connect the two circles by 2 segments as below. Apply Cauchy-Goursat on the left and right regions:

enter image description here

If you want to split it as you did, pick the two connecting pairs of segments to be $\epsilon$ apart. The orangeregion below is simply connected and your function is analytic inside and on the boundary, which is enough to apply Cauchy-Gousart.

enter image description here


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