So I am reading through the book Complex Analysis by Lars Ahlfors, and there is one point that is causing some confusion for me:

In chapter 4.2 - Cauchy's integral formula, we first encounter the following theorem:

$\textbf{Theorem 1.}$ The line integral $\int_\gamma p\ dx + q\ dy$, defined in $\Omega$, depends only on the end points of $\gamma$ if and only if there exists a function $u(x,y)$ in $\Omega$ with the partial derivatives $\partial U/\partial x = p$ and $\partial U/\partial y = q$.

And, as it is stated on the following page:

$\textbf{Theorem 2.}$ The integral $\int_\gamma f\ dz$, with continuous $f$, depends only on the end points of $\gamma$ if and only if $f$ is the derivative of an analytic function in $\Omega$.

A little bit later, however, we are presented with the following result:

$\textbf{Theorem 5.}$ Let $f(z)$ be analytic in the region $\Delta'$ obtained by omitting a finite number of points $\zeta_i$ from an open disk $\Delta$. If $f(z)$ satisfies the condition $$ \lim_{z \to \zeta_j} (z - \zeta_j)f(x) = 0 \quad \forall j $$ then the $\int_\gamma f(z)\ dz = 0$ for any closed curve $\gamma$ in $\Delta'$.

Now, I have a question about the consistency between these two theorems. In theorem $5$, the set $\Delta'$ is an open disc with a finite number of punctures. Thus, $\Delta'$ is a non-empty, open connected subset of $\mathbb{C}$, a region. Thus, according to the second formulation of theorem $1$, if we could show that $f$ is the derivative of an analytic function $F$ on $\Delta'$, we could conclude that $\int_{\gamma}fdz$ depends only on the endpoints of $\gamma \iff \int_{\gamma}f\ dz=0$ for any closed curve $\gamma$ in $\Delta'$.

Now, in the book, they define the function $F$ on $\Delta'$ as $$F(z)=\int_{S}^{z}f(\sigma)d\sigma,$$ where the integral is taken from the center $S$ of the disc (or from a different fixed point $S$ if the center is a $\zeta$-point), and along a sequence of straight lines parallel to the coordinate axes, not passing through any of the $\zeta$'s. They then conclude that $F(z)$ is indefinite integral of $f(z)$.

My question is: Where exactly is the condition $\lim\limits_{z \to \zeta_i}(z-\zeta_i)f(z)=0$ required for this / used in the proof of theorem $5$? As I see it, both theorem $1$ and $5$ cannot both be correct at the same time as $F(z)$ is actually an antiderivative for $f$, because then the limit condition would be unnessecary according to theorem $1$, right?

And so this also raises the question if theorem $5$ applies if the path is enclosing one of the points $\zeta$...

Can anyone see the cause of my confusion?

Thanx, - R.

  • 2
    $\begingroup$ The proof of Theorem 5 made use of Theorem 3, in which the limit condition is required. $\endgroup$ – trisct May 8 at 15:19
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    $\begingroup$ To see that the condition is necessary, you can look at the function $1/z$ over a disc with the origin removed. Basically the condition disallows poles (or worse singularities) at the removed point. Later it will be phrased as "$f(z)$ has removable singularities" at the omitted points. $\endgroup$ – Jane Doé May 8 at 15:26
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    $\begingroup$ If the condition is not true, for example $f(z)=1/z$ in some disk near $0$, then its integral along any circuit around $0$ is $2\pi i$. In further study you will learn about such points $\zeta$ as removable singularities, meaning you can re-designate the value of $f$ at that point to make $f$ analytic. $\endgroup$ – trisct May 8 at 15:26
  • $\begingroup$ @trisct Beat you by 4 seconds! $\endgroup$ – Jane Doé May 8 at 15:27
  • $\begingroup$ Okay, so after reading your comments and thinking some more about it i think i get it now. The limit condition is used in the proof of Theorem 5 in the sense that, when defining the supposed function F to use as an indefinite integral for f, the assumption that the definition of F is the same wheter or not the final segment from S to z is horisontal / vertical, relies (in the book) on Cauchy's theorem for rectangels with punctures, which further relies on the limit condition. $\endgroup$ – AfterMath May 9 at 8:24

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