[proofplan] Part (1) uses Morera's theorem: since each $f_n$ is holomorphic, contour integrals over triangles vanish, and uniform convergence passes the limit through the integral, so $f$ satisfies Morera's criterion. Part (2) uses the [Cauchy integral formula](/theorems/345) for derivatives to express $f_n' - f'$ as a contour integral of $f_n - f$, then bounds it uniformly using the ML inequality. [/proofplan] [step:Verify Morera's criterion to show $f$ is holomorphic] Fix a triangle $T \subset U$. Since $T$ is compact and contained in $U$, the convergence $f_n \to f$ is uniform on $T$ (by the hypothesis of locally uniform convergence). Each $f_n$ is holomorphic, so by [Cauchy's Theorem](/theorems/349), $\int_{\partial T} f_n\,dz = 0$. Uniform convergence on the compact set $\partial T$ permits passing the limit through the integral: \begin{align*} \int_{\partial T} f\,dz = \lim_{n \to \infty}\int_{\partial T} f_n\,dz = 0. \end{align*} Since $f$ is continuous (as a uniform limit of continuous functions) and $\int_{\partial T} f\,dz = 0$ for every triangle $T \subset U$, [Morera's Theorem](/theorems/???) gives that $f$ is holomorphic on $U$. [guided] The strategy is to avoid differentiating $f$ directly and instead use Morera's theorem, which characterises holomorphicity by the vanishing of contour integrals. Why does this work? Morera's theorem states: if $f$ is continuous on an open set $U$ and $\int_{\partial T} f\,dz = 0$ for every triangle $T \subset U$, then $f$ is holomorphic. We verify the two hypotheses. First, continuity: $f_n \to f$ locally uniformly implies $f_n \to f$ uniformly on each compact subset. A uniform limit of continuous functions is continuous, so $f$ is continuous on $U$. Second, vanishing contour integrals: fix a triangle $T \subset U$. Since $T$ is compact, $f_n \to f$ uniformly on $T$. Each $f_n$ is holomorphic, so [Cauchy's Theorem](/theorems/349) gives $\int_{\partial T} f_n\,dz = 0$. Uniform convergence on $\partial T$ justifies: \begin{align*} \left|\int_{\partial T} f\,dz - \int_{\partial T} f_n\,dz\right| = \left|\int_{\partial T}(f - f_n)\,dz\right| \leq \ell(\partial T)\cdot \sup_{z \in \partial T}|f(z) - f_n(z)| \to 0, \end{align*} where $\ell(\partial T)$ denotes the length of $\partial T$. Hence $\int_{\partial T} f\,dz = \lim_{n \to \infty} \int_{\partial T} f_n\,dz = 0$, and Morera's theorem applies. [/guided] [/step] [step:Bound $f_n' - f'$ uniformly on compact subsets via the Cauchy integral formula] Fix $a \in U$ and choose $r > 0$ with $\overline{B(a, 2r)} \subset U$. For any $z \in B(a, r)$, the circle $\{w : |w - z| = r\}$ lies inside $\overline{B(a, 2r)} \subset U$. By the [Cauchy Integral Formula](/theorems/345) for derivatives applied to both $f_n$ and $f$: \begin{align*} f_n'(z) - f'(z) = \frac{1}{2\pi i}\int_{|w - z| = r} \frac{f_n(w) - f(w)}{(w - z)^2}\,dw. \end{align*} Define $\varepsilon_n = \sup_{w \in \overline{B(a, 2r)}}|f_n(w) - f(w)|$, which tends to $0$ by locally uniform convergence. For $z \in B(a, r)$ and $|w - z| = r$, the point $w$ lies in $\overline{B(a, 2r)}$, so $|f_n(w) - f(w)| \leq \varepsilon_n$. The ML inequality gives: \begin{align*} |f_n'(z) - f'(z)| \leq \frac{1}{2\pi}\cdot 2\pi r \cdot \frac{\varepsilon_n}{r^2} = \frac{\varepsilon_n}{r}. \end{align*} This bound is uniform in $z \in B(a, r)$, so $f_n' \to f'$ uniformly on $B(a, r)$. [guided] The key idea is that Cauchy's integral formula lets us control derivatives of holomorphic functions by the values of the functions themselves on a nearby contour. For $z \in B(a, r)$, the circle $|w - z| = r$ is contained in $B(a, 2r)$ (since $|w - a| \leq |w - z| + |z - a| < r + r = 2r$). We chose $r$ so that $\overline{B(a, 2r)} \subset U$, ensuring $f_n$ and $f$ are defined on the contour. By the [Cauchy Integral Formula](/theorems/345), $f_n'(z) = \frac{1}{2\pi i}\int_{|w-z|=r}\frac{f_n(w)}{(w-z)^2}\,dw$ and similarly for $f'$. Subtracting: \begin{align*} f_n'(z) - f'(z) = \frac{1}{2\pi i}\int_{|w-z|=r}\frac{f_n(w) - f(w)}{(w-z)^2}\,dw. \end{align*} On the contour, $|w - z| = r$ (exactly), so $|1/(w-z)^2| = 1/r^2$. The supremum of $|f_n - f|$ on $\overline{B(a,2r)}$ is $\varepsilon_n \to 0$. The contour has length $2\pi r$. The ML inequality: \begin{align*} |f_n'(z) - f'(z)| \leq \frac{1}{2\pi}\cdot 2\pi r \cdot \frac{\varepsilon_n}{r^2} = \frac{\varepsilon_n}{r}. \end{align*} Since this bound does not depend on $z \in B(a, r)$, the convergence $f_n' \to f'$ is uniform on $B(a, r)$. Since $a \in U$ was arbitrary, $f_n' \to f'$ locally uniformly on $U$. The same argument with $(w-z)^{k+1}$ in the denominator and the $k$-th derivative Cauchy formula gives $f_n^{(k)} \to f^{(k)}$ locally uniformly for all $k \geq 0$. [/guided] [/step]