[proofplan] We prove analyticity locally at an arbitrary point $z_0 \in U$. For sums and products, we write the local [power series](/page/Power%20Series) expansions of $f$ and $g$ about $z_0$ and combine them by termwise addition and the Cauchy product. For the composition, we first shrink the neighbourhood so that $g(z)$ stays inside a disk on which $h$ has a power series expansion, then substitute the power series for $g(z)-g(z_0)$ into the power series for $h$. Since the point $z_0$ is arbitrary, the resulting local power series expansions prove analyticity on all of $U$. [/proofplan] [step:Choose local power series expansions at an arbitrary point] Fix $z_0 \in U$. Since $U$ is open, choose $r_U > 0$ such that $B(z_0,r_U) \subset U$. Since $f:U \to \mathbb{C}$ and $g:U \to \mathbb{C}$ are analytic at $z_0$, there exist radii $r_f,r_g \in (0,r_U]$ and coefficient sequences $(a_n)_{n=0}^{\infty}$ and $(c_n)_{n=0}^{\infty}$ in $\mathbb{C}$ such that, for every $z \in B(z_0,r_f)$, \begin{align*} f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n, \end{align*} and, for every $z \in B(z_0,r_g)$, \begin{align*} g(z)=\sum_{n=0}^{\infty} c_n (z-z_0)^n. \end{align*} Set $r_0:=\min\{r_f,r_g\}$. Then both series converge on $B(z_0,r_0)$ and represent $f$ and $g$ there. [/step] [step:Add the two local power series] Define a coefficient sequence $(\alpha_n)_{n=0}^{\infty}$ in $\mathbb{C}$ by \begin{align*} \alpha_n:=a_n+c_n. \end{align*} For every $z \in B(z_0,r_0)$, absolute convergence of the two representing series gives \begin{align*} (f+g)(z)=\sum_{n=0}^{\infty} \alpha_n (z-z_0)^n. \end{align*} Thus $f+g$ is represented by a convergent power series in a neighbourhood of $z_0$. [/step] [step:Multiply the two local power series by the Cauchy product] Define a coefficient sequence $(\beta_n)_{n=0}^{\infty}$ in $\mathbb{C}$ by \begin{align*} \beta_n:=\sum_{k=0}^{n} a_k c_{n-k}. \end{align*} For every $z \in B(z_0,r_0)$, the two series \begin{align*} \sum_{n=0}^{\infty} a_n(z-z_0)^n \end{align*} and \begin{align*} \sum_{n=0}^{\infty} c_n(z-z_0)^n \end{align*} are absolutely convergent. Therefore their product is given by the Cauchy product, so \begin{align*} (fg)(z)=\sum_{n=0}^{\infty} \beta_n (z-z_0)^n. \end{align*} Thus $fg$ is represented by a convergent power series in a neighbourhood of $z_0$. [guided] We want a power series for the pointwise product $fg$ near $z_0$. The local expansions give two absolutely convergent series on $B(z_0,r_0)$: \begin{align*} f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n \end{align*} and \begin{align*} g(z)=\sum_{n=0}^{\infty} c_n (z-z_0)^n. \end{align*} The correct coefficient of $(z-z_0)^n$ in the product is obtained by collecting all pairs of powers whose exponents add to $n$. Thus we define \begin{align*} \beta_n:=\sum_{k=0}^{n} a_k c_{n-k}. \end{align*} Because both original series are absolutely convergent at the fixed point $z \in B(z_0,r_0)$, the Cauchy product theorem for absolutely convergent numerical series applies to the two numerical series \begin{align*} \sum_{n=0}^{\infty} a_n(z-z_0)^n \end{align*} and \begin{align*} \sum_{n=0}^{\infty} c_n(z-z_0)^n. \end{align*} It gives \begin{align*} f(z)g(z)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_k c_{n-k}\right)(z-z_0)^n. \end{align*} By the definition of $\beta_n$, this is \begin{align*} (fg)(z)=\sum_{n=0}^{\infty}\beta_n(z-z_0)^n. \end{align*} Hence $fg$ has a convergent power series expansion in a neighbourhood of $z_0$, which is exactly analyticity at $z_0$. [/guided] [/step] [step:Prepare the local substitution for the composition] Assume now that $V \subset \mathbb{C}$ is open, $g(U)\subset V$, and $h:V \to \mathbb{C}$ is analytic. Let $w_0:=g(z_0)$. Since $w_0 \in V$ and $V$ is open, choose $\rho_V>0$ such that $B(w_0,\rho_V)\subset V$. Since $h$ is analytic at $w_0$, there are $\rho \in (0,\rho_V]$ and a coefficient sequence $(b_m)_{m=0}^{\infty}$ in $\mathbb{C}$ such that, for every $w \in B(w_0,\rho)$, \begin{align*} h(w)=\sum_{m=0}^{\infty} b_m(w-w_0)^m. \end{align*} The local expansion of $g$ at $z_0$ gives $c_0=g(z_0)=w_0$. Define the map $G: B(0,r_g) \to \mathbb{C}$ by \begin{align*} G(\zeta):=\sum_{n=1}^{\infty} c_n\zeta^n \end{align*} for every $\zeta \in B(0,r_g)$. Then for every $z \in B(z_0,r_g)$, \begin{align*} g(z)-w_0=G(z-z_0). \end{align*} The power series defining $G$ is continuous at $0$ and satisfies $G(0)=0$. Hence choose $s \in (0,r_g)$ such that \begin{align*} |G(\zeta)|<\rho \end{align*} for every $\zeta \in B(0,s)$. [/step] [step:Expand the composition as a power series] Choose $s_1 \in (0,s)$ and $\eta \in (0,\rho)$ such that $|G(\zeta)|\leq \eta$ for every $\zeta \in B(0,s_1)$; this is possible by the continuity of $G$ at $0$. Define the map $H:B(0,s_1)\to\mathbb{C}$ by \begin{align*} H(\zeta):=\sum_{m=0}^{\infty} b_m G(\zeta)^m. \end{align*} For $\zeta \in B(0,s_1)$, the bound $|G(\zeta)|\leq \eta<\rho$ implies that this series converges absolutely and equals $h(w_0+G(\zeta))$. For each $m\geq 0$, the $m$-fold product $G(\zeta)^m$ is represented on $B(0,s_1)$ by an absolutely convergent power series in $\zeta$, using repeated applications of the Cauchy product theorem for absolutely convergent power series. Since the power series for $h$ has radius at least $\rho$, the numerical series \begin{align*} \sum_{m=0}^{\infty} |b_m|\eta^m \end{align*} converges. The [Weierstrass M-test](/theorems/261) applied to the functions $\zeta\mapsto b_mG(\zeta)^m$ on $B(0,s_1)$ gives [uniform convergence](/page/Uniform%20Convergence) of \begin{align*} \sum_{m=0}^{\infty} b_mG(\zeta)^m. \end{align*} Therefore the locally uniformly convergent sum of these power-series functions is represented by a power series on a possibly smaller disk $B(0,s_2)$ with $0<s_2<s_1$. Hence there is a coefficient sequence $(d_n)_{n=0}^{\infty}$ in $\mathbb{C}$ such that, for every $\zeta \in B(0,s_2)$, \begin{align*} H(\zeta)=\sum_{n=0}^{\infty}d_n\zeta^n. \end{align*} Substituting $\zeta=z-z_0$, we obtain, for every $z \in B(z_0,s_2)$, \begin{align*} (h\circ g)(z)=h(g(z))=H(z-z_0)=\sum_{n=0}^{\infty}d_n(z-z_0)^n. \end{align*} Thus $h\circ g$ is represented by a convergent power series in a neighbourhood of $z_0$. [guided] The delicate point is not pointwise convergence of the substituted series, but the legitimacy of collecting the infinitely many power series that arise from the powers of $G$. We first make the substitution uniformly small. Since $G$ is continuous at $0$ and $G(0)=0$, choose $s_1\in(0,s)$ and a number $\eta\in(0,\rho)$ such that \begin{align*} |G(\zeta)|\leq \eta \end{align*} for every $\zeta\in B(0,s_1)$. Define the map $H:B(0,s_1)\to\mathbb{C}$ by \begin{align*} H(\zeta):=\sum_{m=0}^{\infty}b_mG(\zeta)^m. \end{align*} For each $\zeta\in B(0,s_1)$, the inequality $|G(\zeta)|\leq\eta<\rho$ places $w_0+G(\zeta)$ inside the disk on which the power series for $h$ converges. Therefore \begin{align*} H(\zeta)=h(w_0+G(\zeta)). \end{align*} Now we justify why $H$ has a power series in $\zeta$. For each fixed $m\geq0$, the function $\zeta\mapsto G(\zeta)^m$ is a finite product of power-series functions, so repeated use of the Cauchy product theorem for absolutely convergent power series gives a convergent power series representation on $B(0,s_1)$. To sum over all $m$, we need uniform control. Since the power series for $h$ converges for every complex number of modulus less than $\rho$ and $\eta<\rho$, the majorant series \begin{align*} \sum_{m=0}^{\infty}|b_m|\eta^m \end{align*} converges. Also, for every $\zeta\in B(0,s_1)$, \begin{align*} |b_mG(\zeta)^m|\leq |b_m|\eta^m. \end{align*} The [Weierstrass M-test](/theorems/264) therefore gives uniform convergence of \begin{align*} \sum_{m=0}^{\infty}b_mG(\zeta)^m \end{align*} on $B(0,s_1)$. A locally uniformly convergent sum of power-series functions is represented by a power series on a sufficiently small disk about $0$; hence there are $s_2\in(0,s_1)$ and a coefficient sequence $(d_n)_{n=0}^{\infty}$ in $\mathbb{C}$ such that \begin{align*} H(\zeta)=\sum_{n=0}^{\infty}d_n\zeta^n \end{align*} for every $\zeta\in B(0,s_2)$. Finally, putting $\zeta=z-z_0$ gives, for every $z\in B(z_0,s_2)$, \begin{align*} (h\circ g)(z)=h(g(z))=H(z-z_0)=\sum_{n=0}^{\infty}d_n(z-z_0)^n. \end{align*} This is a convergent power series expansion of $h\circ g$ near $z_0$. [/guided] [/step] [step:Conclude analyticity on the whole open set] We have shown that, for the arbitrary point $z_0 \in U$, each of the pointwise maps $f+g$ and $fg$, and, under the additional hypotheses, the composition map $h\circ g$, is represented by a convergent power series on some neighbourhood of $z_0$. Hence each function is analytic at every point of $U$. Therefore $f+g$ and $fg$ are analytic on $U$, and if $V$ is open, $g(U)\subset V$, and $h:V\to\mathbb{C}$ is analytic, then $h\circ g$ is analytic on $U$. [/step]