[proofplan] We prove [complex differentiability](/page/Complex%20Differentiability) pointwise. Fix an arbitrary point $a \in U$ and use analyticity to represent $f$ on a disk about $a$ by a convergent [power series](/page/Power%20Series) centered at $a$. The termwise differentiation theorem for power series then gives a complex derivative for that local power-series representative. Since the representative agrees with $f$ on a neighbourhood of $a$, the same derivative is the complex derivative of $f$ at $a$; arbitrariness of $a$ proves holomorphicity on all of $U$. [/proofplan] [step:Choose a local power series representation around an arbitrary point] Fix $a \in U$. Since $f$ is analytic on $U$, there exist a radius $r>0$ and coefficients $(c_n)_{n=0}^{\infty} \subset \mathbb{C}$ such that $B(a,r) \subset U$ and \begin{align*} f(z)=\sum_{n=0}^{\infty} c_n(z-a)^n \end{align*} for every $z \in B(a,r)$. Define the local power-series representative $F_a: B(a,r) \to \mathbb{C}$ by \begin{align*} F_a(z)=\sum_{n=0}^{\infty} c_n(z-a)^n. \end{align*} Then $F_a(z)=f(z)$ for every $z \in B(a,r)$. [/step] [step:Differentiate the local power series at the chosen point] The power series defining $F_a$ converges on $B(a,r)$, so its [radius of convergence](/theorems/262) $R_a$ satisfies $R_a \ge r>0$. By [citetheorem:8249], the function represented by this power series is complex differentiable on $B(a,R_a)$, hence in particular at $a$, and \begin{align*} F_a'(a)=\sum_{n=1}^{\infty} n c_n(a-a)^{n-1}. \end{align*} Thus $F_a'(a)$ exists as a complex number. [guided] We need to convert the local power-series expression into an actual complex derivative. The relevant input is [citetheorem:8249], which applies to a power series \begin{align*} \sum_{n=0}^{\infty} c_n(z-a)^n \end{align*} with positive [radius of convergence](/theorems/265). Here the series converges for every $z \in B(a,r)$, where $r>0$, so its [radius of convergence](/theorems/273) $R_a$ is at least $r$. In particular $R_a>0$, and the point $a$ lies in the disk $B(a,R_a)$. Therefore [citetheorem:8249] applies to the power series centered at $a$. It says that the represented function is complex differentiable throughout $B(a,R_a)$ and that its derivative is obtained by differentiating term by term: \begin{align*} F_a'(z)=\sum_{n=1}^{\infty} n c_n(z-a)^{n-1} \end{align*} for every $z \in B(a,R_a)$. Evaluating this formula at $z=a$ gives \begin{align*} F_a'(a)=\sum_{n=1}^{\infty} n c_n(a-a)^{n-1}. \end{align*} The important conclusion is not the simplified value of this sum, but the existence of the complex derivative $F_a'(a)$. [/guided] [/step] [step:Transfer the derivative from the local representative to $f$] Since $B(a,r) \subset U$ and $F_a=f$ on $B(a,r)$, for every $h \in \mathbb{C}$ with $0<|h|<r$ we have $a+h \in B(a,r)$ and \begin{align*} \frac{f(a+h)-f(a)}{h}=\frac{F_a(a+h)-F_a(a)}{h}. \end{align*} Taking the limit as $h \to 0$ in $\mathbb{C}$ gives \begin{align*} \lim_{h \to 0}\frac{f(a+h)-f(a)}{h}=F_a'(a). \end{align*} Therefore $f$ is complex differentiable at $a$. [/step] [step:Conclude holomorphicity on the whole open set] The point $a \in U$ was arbitrary. Hence $f$ is complex differentiable at every point of $U$. By the definition of holomorphicity on an open subset of $\mathbb{C}$, $f$ is holomorphic on $U$. [/step]