[proofplan] We construct the holomorphic extension explicitly. First, we multiply $f$ by $(z - z_0)^2$ to obtain a function $h$ that is holomorphic on a full neighborhood of $z_0$ (the extra factor forces $h$ and $h'$ to vanish at $z_0$, which we verify from boundedness). We then use the Cauchy integral formula on a circle around $z_0$ to represent $f(z)$ for $z \ne z_0$, and observe that the integral formula is holomorphic in $z$ on the entire disk, including $z_0$. This provides the desired extension. Uniqueness follows from the identity theorem. [/proofplan] [step:Define $h(z) = (z - z_0)^2 f(z)$ and show it extends holomorphically across $z_0$ with $h(z_0) = h'(z_0) = 0$] Choose $r > 0$ such that $\overline{B}(z_0, r) \subset \Omega$ and $f$ is bounded on $B(z_0, r) \setminus \{z_0\}$, say $|f(z)| \le M$ for all $z \in B(z_0, r) \setminus \{z_0\}$. Define \begin{align*} h: B(z_0, r) &\to \mathbb{C} \\ z &\mapsto \begin{cases} (z - z_0)^2 f(z) & \text{if } z \ne z_0, \\ 0 & \text{if } z = z_0. \end{cases} \end{align*} We claim $h$ is holomorphic on $B(z_0, r)$. On $B(z_0, r) \setminus \{z_0\}$, $h$ is a product of holomorphic functions, hence holomorphic. At $z_0$, we verify complex differentiability directly: \begin{align*} |h(z) - h(z_0)| = |z - z_0|^2 |f(z)| \le M |z - z_0|^2 \to 0 \quad \text{as } z \to z_0, \end{align*} so $h$ is continuous at $z_0$ with $h(z_0) = 0$. For the derivative: \begin{align*} \left| \frac{h(z) - h(z_0)}{z - z_0} \right| = |z - z_0| \cdot |f(z)| \le M |z - z_0| \to 0 \quad \text{as } z \to z_0, \end{align*} so $h'(z_0) = 0$. Since $h$ is holomorphic on $B(z_0, r) \setminus \{z_0\}$ and complex differentiable at $z_0$, $h \in \mathcal{O}(B(z_0, r))$. [guided] The idea is to "kill" the potential singularity by multiplying by a sufficiently high power of $(z - z_0)$. Why $(z - z_0)^2$ and not just $(z - z_0)$? Because we need $h$ to be not just continuous but complex differentiable at $z_0$. With the single factor $(z - z_0)$, we would get $|h(z)| = |z - z_0| |f(z)| \le M|z - z_0| \to 0$, giving continuity. But the difference quotient $\frac{h(z)}{z - z_0} = f(z)$ need not have a limit as $z \to z_0$ (that is precisely what we are trying to prove). The extra factor of $(z - z_0)$ resolves this: the difference quotient $\frac{(z-z_0)^2 f(z)}{z - z_0} = (z - z_0) f(z)$ is bounded by $M|z - z_0| \to 0$. Formally, with $h(z_0) := 0$: \begin{align*} |h(z)| = |z - z_0|^2 |f(z)| \le M|z - z_0|^2 \to 0 \implies h \text{ continuous at } z_0. \end{align*} \begin{align*} \left|\frac{h(z) - 0}{z - z_0}\right| = |z - z_0| |f(z)| \le M|z - z_0| \to 0 \implies h'(z_0) = 0. \end{align*} A function that is holomorphic on a punctured disk and complex differentiable at the puncture is holomorphic on the entire disk (the Cauchy-Riemann equations and continuity of the partial derivatives can be verified, or one can appeal to Morera's theorem on the full disk). So $h \in \mathcal{O}(B(z_0, r))$ with $h(z_0) = 0$ and $h'(z_0) = 0$. [/guided] [/step] [step:Use the Cauchy integral formula for $f$ on the punctured disk] Fix $0 < \rho < r$. Let $C_\rho$ denote the circle $|w - z_0| = \rho$, traversed counterclockwise. For $z \in B(z_0, \rho) \setminus \{z_0\}$, we apply the [Cauchy Integral Formula](/theorems/???) for $f$ on the simply connected domain $B(z_0, \rho) \setminus \{z_0\}$. Since $h \in \mathcal{O}(B(z_0, r))$ with $h(z_0) = 0$, we may write $h(z) = (z - z_0)^2 f(z)$ for $z \ne z_0$, so $f(z) = h(z)/(z - z_0)^2$. By the Cauchy integral formula applied to $h$: \begin{align*} h(z) = \frac{1}{2\pi i} \oint_{C_\rho} \frac{h(w)}{w - z}\, dw \quad \text{for all } z \in B(z_0, \rho). \end{align*} Since $h(z_0) = 0$ and $h'(z_0) = 0$, by [Holomorphic Functions Are Analytic](/theorems/3354), $h$ has a power series expansion $h(z) = \sum_{n=0}^\infty a_n (z - z_0)^n$ with $a_0 = h(z_0) = 0$ and $a_1 = h'(z_0) = 0$. Therefore $h(z) = (z - z_0)^2 \psi(z)$ where $\psi \in \mathcal{O}(B(z_0, r))$ is defined by \begin{align*} \psi(z) := \sum_{n=0}^{\infty} a_{n+2} (z - z_0)^n = \frac{h(z)}{(z - z_0)^2} \quad \text{for } z \ne z_0. \end{align*} For $z \ne z_0$, $\psi(z) = f(z)$. [/step] [step:Define the holomorphic extension and verify its properties] Define \begin{align*} \tilde{f}: \Omega &\to \mathbb{C} \\ z &\mapsto \begin{cases} \psi(z) & \text{if } z \in B(z_0, r), \\ f(z) & \text{if } z \in \Omega \setminus \{z_0\}. \end{cases} \end{align*} This is well-defined: on $B(z_0, r) \setminus \{z_0\}$, both definitions agree since $\psi(z) = h(z)/(z - z_0)^2 = f(z)$. Moreover: - On $B(z_0, r)$: $\tilde{f} = \psi \in \mathcal{O}(B(z_0, r))$, so $\tilde{f}$ is holomorphic. - On $\Omega \setminus \{z_0\}$: $\tilde{f} = f \in \mathcal{O}(\Omega \setminus \{z_0\})$, so $\tilde{f}$ is holomorphic. Since holomorphicity is a local property and $B(z_0, r) \cup (\Omega \setminus \{z_0\}) = \Omega$, we conclude $\tilde{f} \in \mathcal{O}(\Omega)$. By construction, $\tilde{f}(z) = f(z)$ for all $z \in \Omega \setminus \{z_0\}$. [/step] [step:Prove uniqueness of the extension via the identity theorem] Suppose $\tilde{f}_1, \tilde{f}_2 \in \mathcal{O}(\Omega)$ both extend $f$. Then $\tilde{f}_1 - \tilde{f}_2 \in \mathcal{O}(\Omega)$ vanishes on $\Omega \setminus \{z_0\}$, which is a set with an accumulation point in $\Omega$ (every point of $\Omega \setminus \{z_0\}$ is an accumulation point). By the [Identity Theorem](/theorems/???), a holomorphic function on a connected open set that vanishes on a set with an accumulation point is identically zero. If $\Omega$ is connected, this gives $\tilde{f}_1 = \tilde{f}_2$ on $\Omega$ immediately. If $\Omega$ is not connected, apply the argument to the connected component containing $z_0$ (the extension is uniquely determined on this component by the identity theorem, and on all other components both functions already agree with $f$). Therefore the holomorphic extension is unique. [guided] Why is uniqueness immediate? Any two holomorphic extensions $\tilde{f}_1$ and $\tilde{f}_2$ agree on $\Omega \setminus \{z_0\}$. Their difference $\tilde{f}_1 - \tilde{f}_2$ is holomorphic on $\Omega$ and vanishes everywhere except possibly at $z_0$. But the zero set $\Omega \setminus \{z_0\}$ has accumulation points in $\Omega$ (in fact, every point of $\Omega$ is an accumulation point of $\Omega \setminus \{z_0\}$). By the [Identity Theorem](/theorems/???), a holomorphic function whose zero set has an accumulation point in a connected open set must be identically zero. So $\tilde{f}_1 = \tilde{f}_2$ on each connected component of $\Omega$ that intersects $\Omega \setminus \{z_0\}$ — which is every component (since $z_0$ is a single point and every component of $\Omega$ containing $z_0$ also contains points of $\Omega \setminus \{z_0\}$, and components not containing $z_0$ are entirely contained in $\Omega \setminus \{z_0\}$ where both extensions agree with $f$). [/guided] [/step]