How is Ahlfors applying Cauchy's Integral Formula when deriving Taylor's Theorem?

Question

From Ahlfors' Complex Analysis (by 'analytic' he means 'complex-differentiable' or 'holomorphic'; he does not mean that the function can be expanded as a power series, as that is what he sets out to prove):

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 $\lim_{z\to a}(z - a)f(z) = 0$. The extended function is uniquely determined.

At the beginning of the proof, he writes:

To prove the sufficiency we draw a circle $C$ about $a$ so that $C$ and its inside are contained in $\Omega$. Cauchy's formula is valid, and we can write $$f(z) = \frac{1}{2\pi i}\int_C\frac{f(\xi)}{\xi - z}\text{d}\xi$$ for all $z \ne a$ inside of $C$.

How and why does Cauchy formula apply here? From my understanding, the function $f$ needs to be holomorphic (and thus defined) in the entirety of an open disk. As Ahlfors wrote a few pages back:

We have thus proved: Theorem 6: Suppose that $f(z)$ is analytic in an open disk $D$, and let $\gamma$ be a closed curve in $D$. For any point $a$ not on $\gamma$ $$n(\gamma,a)\cdot f(a) = \frac{1}{2\pi i}\int_\gamma\frac{f(z)}{z-a}\text{d}z$$ where $n(\gamma,a)$ is the index [winding number] of $a$ with respect to $\gamma$.

Answer 1

Immediately proceeding Ahlfors's Theorem 6:

It is clear that Theorem 6 remains valid for any region $\Omega$ to which Theorem 5 can be applied. The presence of exceptional points $\zeta_j$ is permitted, provided none of them coincides with $a$.

Here Theorem 5 reads as:

Theorem 5. Let $f(z)$ be analytic in the region $\Delta'$ obtained by omitting a finite number of points $\zeta_j$ from an open disk $\Delta$. If $f(z)$ satisfies the condition $\lim_{z \to \zeta_j} (z - \zeta_j) f(z) = 0$ for all $j$, then (18) holds for any closed curve $\gamma$ in $\Delta'$.

Here (18) is the following: $$ \int_\gamma f(z) \, d z = 0. \tag{18} $$

Answer 2

Since we are here to prove the sufficiency part of the statement, we assume that $f$ satisfies the condition $\lim_{z \to a} (z-a) f(z) = 0$.

Given $z \neq a$ inside the circle $C$, draw another small circle $C' = C'(r)$ of radius $r$ centred at $a$, lying inside $C$ and such that $z$ lies outside of $C'$. Connecting $C$ to $C'$ via a radial line $\gamma$ avoiding $z$, we obtain a contour $\Gamma = C + \gamma - C' - \gamma$ which lies (as well as its inside) within $\Omega'$ and winding once around $z$. Hence Cauchy's formula is applicable for $\Gamma$, so that $$ f(z) = \frac{1}{2\pi i} \oint_{\Gamma} \frac{f(\xi)}{\xi - z} d\xi = \frac{1}{2\pi i} \oint_{C} \frac{f(\xi)}{\xi - z} d\xi - \frac{1}{2\pi i} \oint_{C'} \frac{f(\xi)}{\xi - z} d\xi \, . $$ We note that the last two integrals are meaningful, since $f(\xi)$ and $1/(\xi - z)$ are continuous, hence bounded, over the compact sets $C$ and $C'$.

Hence, to prove Ahlfors' claim, we only need to prove that $\lim_{r \to 0^+} \oint_{C'(r)} \frac{f(\xi)}{\xi - z} d\xi = 0$. Parametrizing $C'(r)$ as $\xi = a + re^{i\theta}$ for $\theta \in [0, 2\pi]$, and since $r < |z-a|$, we compute $$ \left| \oint_{C'(r)} \frac{f(\xi)}{\xi - z} d\xi \right| = \left| \int_{0}^{2\pi} \frac{f(a + re^{i\theta})}{a + re^{i\theta} - z} ire^{i \theta} d\theta \right| \le \int_{0}^{2\pi} \frac{| r f(a + re^{i\theta})|}{| z- a| - r} d\theta $$ whence $$ 0 \le \lim_{r \to 0^+} \left| \oint_{C'(r)} \frac{f(\xi)}{\xi - z} d\xi \right| \le \int_{0}^{2\pi} \frac{| \lim_{r \to 0^+} r f(a + re^{i\theta})|}{\lim_{r \to 0^+} (| z- a| - r)} d\theta = \int_{0}^{2\pi} \frac{0}{| z- a|} d\theta = 0. $$

The idea exposed here extends to the case when a finite number of points are removed from $\Omega$.