Theorem
Let $\Omega\subseteq \mathbb{C}$ be open and $f\in H(\Omega)$ ($f$ analytic on $\Omega$). If $ C(z_{0},R)\subseteq\Omega$ (where $C(z_0,R)$ is the circle with origin $z_0$ and radius $R$), then we can represent $f$ on $C(z_{0},R)$ as a power series with convergence radius $\geq R$.
Proof
Let $0
Because $C(z_{0},R)$ is convex and $\gamma$ is in this circle, we have the following from the Cauchy integral formula: $ f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in C(z_{0},\ r)) $
$\color{red}{\text{(1) Why not directly on } C(z_{0},\ R)? }$
Because $C(z_{0},\ r)\subseteq \mathbb{C}\backslash \gamma^{*}$ it follows that $f(z)=\displaystyle \sum_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ for a power series with convergence radius $\geq r$. Because $ r\in (0,\ R)$ is chosen arbitrarily and because the the coefficients $c_{n}$ are determined by the differentials $f^{(n)}(z_{0})$, the power series $\displaystyle \Sigma_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ has the convergence radius $\geq R$.
$\color{red}{\text{(2) Why are the coefficients calculated via }f^{(n)}(z_{0})? }$
$\color{red}{\text{(3) Why does it follow that the convergence radius is } \geq R? }$