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.  
 A: The condition that $\lim_{z \to \zeta_i} (z - \zeta_i) f(z) = 0$ is used in showing that the function $F(z)$ is an indefinite integral of $f(z)$. Specifically, Ahlfors invokes Theorem 3, which says the following:

Theorem 3. Let $f(z)$ be analytic on the set $R'$ obtained from a rectangle $R$ by omitting a finite number of interior points $\zeta_j$. If it is true that
$$
\lim_{z \to \zeta_j} (z - \zeta_j) f(z) = 0
$$
for all $j$, then
$$
\int_{\partial R} f(z)\, dz = 0.
$$

Indeed, if this condition does not hold, then it need not be true that $\int_\gamma f(z)\, dz = 0$ for every closed curve $\gamma$ in a region $\Delta'$ obtained from an open disk $\Delta$ by omitting a finite number of points, even when $f(z)$ is analytic in $\Delta'$.
For example, take $\Delta$ to be the open unit disk centered at $0$, and $\Delta'$ to be the punctured unit disk $\Delta \setminus \{ 0 \}$. Let $f(z) = 1/z$ on $\Delta'$. Then, $\lim_{z \to 0} (z - 0) f(z) = 1 \neq 0$. Moreover,
$$
\int_C \frac{1}{z}\, dz = 2\pi i \neq 0
$$
for any circle $C$ centered at $0$ and lying in $\Delta'$. (This very example is discussed at the end of $\S$4.1.3 by Ahlfors.)
The given condition on the $\zeta_j$'s is intended to eliminate such behavior. Specifically, see Theorem 7 in $\S$4.3.1:

Theorem 7. Suppose that $f(z)$ is analytic in the region $\Omega'$ obtained by omitting a point $a$ from a region $\Omega$. A necessary and sufficient condition that there exist an analytic function in $\Omega$ which coincides with $f(z)$ in $\Omega'$ is that $\displaystyle\lim_{z \to a} (z - a) f(z) = 0$. The extended function is uniquely determined.

Such points are also called removable singularities, and Theorem 7 above is also known as Riemann's Removable Singularity Theorem.
