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

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 $$\gamma$$.

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 $$a$$.

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: $$\int_\gamma f(z) \, d z = 0. \tag{18}$$
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 $$\Omega$$.