[proofplan] We expand the Cauchy kernel $\frac{1}{w - z}$ as a geometric series in $\frac{z - z_0}{w - z_0}$, which converges when $|z - z_0| < |w - z_0| = \rho$. Substituting this expansion into the Cauchy integral formula and interchanging summation with integration (justified by uniform convergence on the contour) yields the power series representation. The coefficient formula then follows from the Cauchy integral formula for derivatives. Absolute and uniform convergence on compact subsets of $B(z_0, r)$ follows from the geometric series estimate. [/proofplan] [step:Choose the radius and set up the Cauchy integral formula] Fix $z_0 \in \Omega$. Since $\Omega$ is open, choose $r > 0$ such that $\overline{B}(z_0, r) \subset \Omega$. For any $0 < \rho < r$, let $C_\rho$ denote the circle $|w - z_0| = \rho$, traversed counterclockwise. By the [Cauchy Integral Formula](/theorems/???), for every $z \in B(z_0, \rho)$, \begin{align*} f(z) = \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(w)}{w - z}\, dw. \end{align*} [/step] [step:Expand the Cauchy kernel as a geometric series] For $w \in C_\rho$ and $z \in B(z_0, \rho)$, we have $|z - z_0| < \rho = |w - z_0|$. Factor the Cauchy kernel: \begin{align*} \frac{1}{w - z} = \frac{1}{(w - z_0) - (z - z_0)} = \frac{1}{w - z_0} \cdot \frac{1}{1 - \frac{z - z_0}{w - z_0}}. \end{align*} Define $\zeta := \frac{z - z_0}{w - z_0}$. Since $|\zeta| = \frac{|z - z_0|}{|w - z_0|} = \frac{|z - z_0|}{\rho} < 1$, the geometric series $\frac{1}{1 - \zeta} = \sum_{n=0}^{\infty} \zeta^n$ converges absolutely. Substituting: \begin{align*} \frac{1}{w - z} = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(w - z_0)^{n+1}}. \end{align*} [guided] The key idea is to exploit the fact that $z$ is closer to $z_0$ than $w$ is (since $z$ is inside the disk and $w$ is on its boundary). This means the ratio $\zeta = \frac{z - z_0}{w - z_0}$ has modulus strictly less than $1$, so the geometric series $\sum \zeta^n$ converges. For $w \in C_\rho$, we have $|w - z_0| = \rho$, and for $z \in B(z_0, \rho)$, we have $|z - z_0| < \rho$. Writing \begin{align*} \frac{1}{w - z} = \frac{1}{(w - z_0)(1 - \zeta)}, \quad \zeta = \frac{z - z_0}{w - z_0}, \end{align*} we have $|\zeta| < 1$, so the geometric series $\frac{1}{1 - \zeta} = \sum_{n=0}^\infty \zeta^n$ converges absolutely. Therefore \begin{align*} \frac{1}{w - z} = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(w - z_0)^{n+1}}, \end{align*} with absolute convergence for each fixed $z$ and $w$. [/guided] [/step] [step:Interchange summation and integration to obtain the power series] We substitute the geometric series expansion into the Cauchy integral formula: \begin{align*} f(z) = \frac{1}{2\pi i} \oint_{C_\rho} f(w) \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(w - z_0)^{n+1}}\, dw. \end{align*} To interchange the sum and integral, we verify uniform convergence in $w \in C_\rho$. Fix $z \in B(z_0, \rho)$ and let $q := |z - z_0|/\rho < 1$. Since $f$ is continuous on the compact set $C_\rho$, there exists $M > 0$ with $|f(w)| \le M$ for all $w \in C_\rho$. The $n$-th term of the series satisfies \begin{align*} \left| f(w) \cdot \frac{(z - z_0)^n}{(w - z_0)^{n+1}} \right| \le \frac{M \cdot |z - z_0|^n}{\rho^{n+1}} = \frac{M}{\rho} \cdot q^n \end{align*} for all $w \in C_\rho$. Since $\sum_{n=0}^\infty q^n < \infty$, the Weierstrass $M$-test guarantees uniform convergence of the series in $w$ on $C_\rho$. We therefore interchange sum and integral: \begin{align*} f(z) = \sum_{n=0}^{\infty} \left( \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(w)}{(w - z_0)^{n+1}}\, dw \right) (z - z_0)^n. \end{align*} Define the coefficients \begin{align*} a_n := \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(w)}{(w - z_0)^{n+1}}\, dw. \end{align*} Then $f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n$ for all $z \in B(z_0, \rho)$. [guided] The interchange of summation and integration is the central step. It requires the series $\sum_n f(w) \frac{(z - z_0)^n}{(w - z_0)^{n+1}}$ to converge uniformly in the integration variable $w$ on $C_\rho$. Why does uniform convergence suffice? Because a uniformly convergent series of continuous functions can be integrated term-by-term along a contour of finite length. We verify the hypothesis of the Weierstrass $M$-test. Let $M := \max_{w \in C_\rho} |f(w)|$ (this maximum exists because $f$ is continuous and $C_\rho$ is compact). For $w \in C_\rho$: \begin{align*} \left| f(w) \cdot \frac{(z - z_0)^n}{(w - z_0)^{n+1}} \right| \le M \cdot \frac{|z - z_0|^n}{\rho^{n+1}} = \frac{M}{\rho} \cdot q^n, \end{align*} where $q = |z - z_0|/\rho < 1$. The bound $\frac{M}{\rho} q^n$ is independent of $w$ and summable in $n$, so the Weierstrass $M$-test applies. We conclude \begin{align*} f(z) &= \frac{1}{2\pi i} \sum_{n=0}^\infty \oint_{C_\rho} \frac{f(w)}{(w - z_0)^{n+1}}\, dw \cdot (z - z_0)^n = \sum_{n=0}^\infty a_n (z - z_0)^n, \end{align*} where $a_n = \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(w)}{(w - z_0)^{n+1}}\, dw$. [/guided] [/step] [step:Identify the coefficients with $f^{(n)}(z_0)/n!$ via the derivative formula] By [Holomorphic Implies Infinitely Differentiable](/theorems/3353), $f$ is infinitely differentiable on $\Omega$ and the $n$-th derivative at $z_0$ is given by \begin{align*} f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_{C_\rho} \frac{f(w)}{(w - z_0)^{n+1}}\, dw. \end{align*} Comparing with the definition of $a_n$: \begin{align*} a_n = \frac{1}{2\pi i} \oint_{C_\rho} \frac{f(w)}{(w - z_0)^{n+1}}\, dw = \frac{f^{(n)}(z_0)}{n!}. \end{align*} Note that the integral $\oint_{C_\rho} \frac{f(w)}{(w - z_0)^{n+1}}\, dw$ is independent of the choice of $\rho \in (0, r)$, since the integrand $\frac{f(w)}{(w - z_0)^{n+1}}$ is holomorphic on the annulus $0 < |w - z_0| < r$ minus the point $z_0$, and any two circles $C_{\rho_1}$, $C_{\rho_2}$ are homotopic in this annulus. (Alternatively, this follows directly from the [Cauchy Integral Theorem](/theorems/3352) applied to the holomorphic function $w \mapsto \frac{f(w)}{(w-z_0)^{n+1}}$ on the simply connected annular region between the two circles.) [/step] [step:Establish absolute and uniform convergence on compact subsets] Let $K \subset B(z_0, r)$ be compact. Choose $\rho$ with $\sup_{z \in K} |z - z_0| < \rho < r$. Such a $\rho$ exists because the continuous function $z \mapsto |z - z_0|$ attains its maximum on the compact set $K$, and this maximum is strictly less than $r$. Define $q := \sup_{z \in K} |z - z_0|/\rho < 1$. For all $z \in K$ and all $n \ge 0$: \begin{align*} |a_n (z - z_0)^n| &= |a_n| \cdot |z - z_0|^n. \end{align*} By the contour integral formula for $a_n$ with the circle $C_\rho$: \begin{align*} |a_n| \le \frac{1}{2\pi} \cdot \frac{M_\rho}{\rho^{n+1}} \cdot 2\pi\rho = \frac{M_\rho}{\rho^n}, \end{align*} where $M_\rho := \max_{|w - z_0| = \rho} |f(w)|$. Therefore, for $z \in K$: \begin{align*} |a_n (z - z_0)^n| \le M_\rho \cdot \left(\frac{|z - z_0|}{\rho}\right)^n \le M_\rho \cdot q^n. \end{align*} Since $q < 1$, $\sum_{n=0}^\infty M_\rho q^n = \frac{M_\rho}{1 - q} < \infty$. By the Weierstrass $M$-test, $\sum_{n=0}^\infty a_n (z - z_0)^n$ converges absolutely and uniformly on $K$. Since the power series converges for all $z \in B(z_0, \rho)$ and $\rho < r$ was arbitrary (subject to $\overline{B}(z_0, \rho) \subset \Omega$), the series converges on all of $B(z_0, r)$. [/step]