Formulations of Cauchy's theorem that don't seem consistent

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.

• The proof of Theorem 5 made use of Theorem 3, in which the limit condition is required. – trisct May 8 at 15:19
• 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. – Jane Doé May 8 at 15:26
• 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. – trisct May 8 at 15:26
• @trisct Beat you by 4 seconds! – Jane Doé May 8 at 15:27
• 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. – AfterMath May 9 at 8:24