[proofplan] We prove holomorphicity pointwise. At an arbitrary point $z_0 \in \Omega$, holomorphicity of $f$ and $g$ gives the existence of their complex derivatives, and therefore also their continuity at $z_0$. The sum, scalar multiple, and product rules follow by rewriting the corresponding difference quotients and taking limits. For the quotient, we first prove that the reciprocal $1/g$ is holomorphic when $g$ has no zeros, then multiply by $f$ using the product rule already established. [/proofplan] [step:Fix a point and record the derivative limits available there] Let $z_0 \in \Omega$ be arbitrary. Since $\Omega$ is open, there exists $\rho > 0$ such that \begin{align*} \{z \in \mathbb{C} : |z - z_0| < \rho\} \subset \Omega. \end{align*} Define the punctured increment set $P_{z_0,\rho} \subset \mathbb{C}$ by \begin{align*} P_{z_0,\rho} := \{\xi \in \mathbb{C} : 0 < |\xi| < \rho\}. \end{align*} For every $\xi \in P_{z_0,\rho}$, the point $z_0+\xi$ belongs to $\Omega$. By the [definition of holomorphicity](/page/Holomorphic%20Function), the complex derivatives $f'(z_0) \in \mathbb{C}$ and $g'(z_0) \in \mathbb{C}$ exist and satisfy \begin{align*} \lim_{\xi \to 0} \frac{f(z_0+\xi)-f(z_0)}{\xi} &= f'(z_0),\\ \lim_{\xi \to 0} \frac{g(z_0+\xi)-g(z_0)}{\xi} &= g'(z_0), \end{align*} where the limits are taken over $\xi \in P_{z_0,\rho}$. Moreover $f$ and $g$ are continuous at $z_0$. Indeed, \begin{align*} f(z_0+\xi)-f(z_0) = \xi \frac{f(z_0+\xi)-f(z_0)}{\xi} \to 0 \end{align*} as $\xi \to 0$, because $\xi \to 0$ and the difference quotient tends to the finite number $f'(z_0)$. The same argument gives \begin{align*} g(z_0+\xi)-g(z_0) = \xi \frac{g(z_0+\xi)-g(z_0)}{\xi} \to 0. \end{align*} [guided] We work at a single point because holomorphicity is a local pointwise condition: a function is holomorphic on $\Omega$ precisely when it is complex differentiable at every point of $\Omega$. Let $z_0 \in \Omega$ be fixed. Since $\Omega$ is open, the definition of openness gives a radius $\rho > 0$ such that \begin{align*} \{z \in \mathbb{C} : |z - z_0| < \rho\} \subset \Omega. \end{align*} This radius ensures that every sufficiently small increment stays inside the domain. Define \begin{align*} P_{z_0,\rho} := \{\xi \in \mathbb{C} : 0 < |\xi| < \rho\}. \end{align*} Thus, for each $\xi \in P_{z_0,\rho}$, the expression $z_0+\xi$ lies in $\Omega$, so all difference quotients below are well-defined. Because $f: \Omega \to \mathbb{C}$ and $g: \Omega \to \mathbb{C}$ are holomorphic, the [definition of holomorphicity](/page/Holomorphic%20Function) says that the complex derivative exists at $z_0$. Hence there are complex numbers $f'(z_0)$ and $g'(z_0)$ such that \begin{align*} \lim_{\xi \to 0} \frac{f(z_0+\xi)-f(z_0)}{\xi} &= f'(z_0),\\ \lim_{\xi \to 0} \frac{g(z_0+\xi)-g(z_0)}{\xi} &= g'(z_0). \end{align*} We will also need continuity of $f$ and $g$ at $z_0$, especially in the product and quotient arguments. This continuity is not an extra theorem: it follows directly from differentiability. For $f$, rewrite the increment as \begin{align*} f(z_0+\xi)-f(z_0) = \xi \frac{f(z_0+\xi)-f(z_0)}{\xi}. \end{align*} As $\xi \to 0$, the first factor tends to $0$, and the second factor tends to the finite number $f'(z_0)$. By the product law for limits in $\mathbb{C}$, the product tends to $0$. Therefore $f(z_0+\xi) \to f(z_0)$, which is continuity at $z_0$. The same computation with $g$ gives \begin{align*} g(z_0+\xi)-g(z_0) = \xi \frac{g(z_0+\xi)-g(z_0)}{\xi} \to 0, \end{align*} so $g$ is continuous at $z_0$. [/guided] [/step] [step:Pass derivatives through sums and scalar multiples] Define the sum and scalar multiple maps \begin{align*} s: \Omega &\to \mathbb{C}, & z &\mapsto f(z)+g(z),\\ m_a: \Omega &\to \mathbb{C}, & z &\mapsto a f(z). \end{align*} For $\xi \in P_{z_0,\rho}$, \begin{align*} \frac{s(z_0+\xi)-s(z_0)}{\xi} &= \frac{f(z_0+\xi)-f(z_0)}{\xi} + \frac{g(z_0+\xi)-g(z_0)}{\xi}. \end{align*} Taking the limit and using the sum law for limits in $\mathbb{C}$ gives \begin{align*} \lim_{\xi \to 0}\frac{s(z_0+\xi)-s(z_0)}{\xi} = f'(z_0)+g'(z_0). \end{align*} Thus $s=f+g$ is complex differentiable at $z_0$. Similarly, \begin{align*} \frac{m_a(z_0+\xi)-m_a(z_0)}{\xi} = a\frac{f(z_0+\xi)-f(z_0)}{\xi}. \end{align*} Taking the limit and using the scalar multiple law for limits gives \begin{align*} \lim_{\xi \to 0}\frac{m_a(z_0+\xi)-m_a(z_0)}{\xi} = a f'(z_0). \end{align*} Thus $m_a=af$ is complex differentiable at $z_0$. [/step] [step:Rewrite the product difference quotient so each factor has a limit] Define the product map \begin{align*} p: \Omega &\to \mathbb{C}, & z &\mapsto f(z)g(z). \end{align*} For $\xi \in P_{z_0,\rho}$, add and subtract $f(z_0+\xi)g(z_0)$ to obtain \begin{align*} \frac{p(z_0+\xi)-p(z_0)}{\xi} &= \frac{f(z_0+\xi)g(z_0+\xi)-f(z_0)g(z_0)}{\xi}\\ &= f(z_0+\xi)\frac{g(z_0+\xi)-g(z_0)}{\xi} + g(z_0)\frac{f(z_0+\xi)-f(z_0)}{\xi}. \end{align*} As $\xi \to 0$, continuity of $f$ at $z_0$ gives $f(z_0+\xi) \to f(z_0)$, and the derivative limits give the remaining two limits. Therefore the product law for limits in $\mathbb{C}$ yields \begin{align*} \lim_{\xi \to 0}\frac{p(z_0+\xi)-p(z_0)}{\xi} = f(z_0)g'(z_0)+g(z_0)f'(z_0). \end{align*} Thus $p=fg$ is complex differentiable at $z_0$. [guided] The product rule requires one algebraic rearrangement before the limits can be taken. Define \begin{align*} p: \Omega &\to \mathbb{C}, & z &\mapsto f(z)g(z). \end{align*} For $\xi \in P_{z_0,\rho}$, the product difference quotient is \begin{align*} \frac{p(z_0+\xi)-p(z_0)}{\xi} = \frac{f(z_0+\xi)g(z_0+\xi)-f(z_0)g(z_0)}{\xi}. \end{align*} The expression has two functions changing at once, so we separate the change in $g$ from the change in $f$. Add and subtract the intermediate term $f(z_0+\xi)g(z_0)$: \begin{align*} f(z_0+\xi)g(z_0+\xi)-f(z_0)g(z_0) &= f(z_0+\xi)\bigl(g(z_0+\xi)-g(z_0)\bigr)\\ &\quad+ g(z_0)\bigl(f(z_0+\xi)-f(z_0)\bigr). \end{align*} Dividing by $\xi$ gives \begin{align*} \frac{p(z_0+\xi)-p(z_0)}{\xi} &= f(z_0+\xi)\frac{g(z_0+\xi)-g(z_0)}{\xi} + g(z_0)\frac{f(z_0+\xi)-f(z_0)}{\xi}. \end{align*} Each factor now has a known limit. From the previous step, $f$ is continuous at $z_0$, so \begin{align*} f(z_0+\xi) \to f(z_0). \end{align*} By holomorphicity, \begin{align*} \frac{g(z_0+\xi)-g(z_0)}{\xi} \to g'(z_0), \qquad \frac{f(z_0+\xi)-f(z_0)}{\xi} \to f'(z_0). \end{align*} Using the product and sum laws for limits in $\mathbb{C}$, we obtain \begin{align*} \lim_{\xi \to 0}\frac{p(z_0+\xi)-p(z_0)}{\xi} = f(z_0)g'(z_0)+g(z_0)f'(z_0). \end{align*} The limit exists, so $p=fg$ is complex differentiable at $z_0$. [/guided] [/step] [step:Differentiate the reciprocal when the denominator has no zeros] Assume now that $g(z) \ne 0$ for every $z \in \Omega$. Define the reciprocal map \begin{align*} r: \Omega &\to \mathbb{C}, & z &\mapsto \frac{1}{g(z)}. \end{align*} Since $g(z_0) \ne 0$ and $g(z_0+\xi) \to g(z_0)$, the quotient law for limits gives \begin{align*} \frac{1}{g(z_0+\xi)} \to \frac{1}{g(z_0)}. \end{align*} For $\xi \in P_{z_0,\rho}$, \begin{align*} \frac{r(z_0+\xi)-r(z_0)}{\xi} &= \frac{\frac{1}{g(z_0+\xi)}-\frac{1}{g(z_0)}}{\xi}\\ &= -\frac{1}{g(z_0+\xi)g(z_0)} \frac{g(z_0+\xi)-g(z_0)}{\xi}. \end{align*} Taking limits, using $g(z_0) \ne 0$, continuity of multiplication, and the derivative limit for $g$, gives \begin{align*} \lim_{\xi \to 0}\frac{r(z_0+\xi)-r(z_0)}{\xi} = -\frac{g'(z_0)}{g(z_0)^2}. \end{align*} Thus $r=1/g$ is complex differentiable at $z_0$. [guided] Now assume that $g$ has no zeros on $\Omega$. This hypothesis is needed so the reciprocal is defined everywhere on $\Omega$. Define \begin{align*} r: \Omega &\to \mathbb{C}, & z &\mapsto \frac{1}{g(z)}. \end{align*} We first check the denominator behavior near $z_0$. Since $g(z_0) \ne 0$ and $g$ is continuous at $z_0$, we have \begin{align*} g(z_0+\xi) \to g(z_0). \end{align*} Because inversion is continuous on $\mathbb{C} \setminus \{0\}$, equivalently by the quotient law for limits with nonzero limiting denominator, \begin{align*} \frac{1}{g(z_0+\xi)} \to \frac{1}{g(z_0)}. \end{align*} We compute the reciprocal difference quotient. For $\xi \in P_{z_0,\rho}$, \begin{align*} \frac{r(z_0+\xi)-r(z_0)}{\xi} &= \frac{\frac{1}{g(z_0+\xi)}-\frac{1}{g(z_0)}}{\xi}\\ &= \frac{g(z_0)-g(z_0+\xi)}{\xi\, g(z_0+\xi)g(z_0)}\\ &= -\frac{1}{g(z_0+\xi)g(z_0)} \frac{g(z_0+\xi)-g(z_0)}{\xi}. \end{align*} This factorization isolates the known derivative quotient for $g$. As $\xi \to 0$, \begin{align*} \frac{g(z_0+\xi)-g(z_0)}{\xi} \to g'(z_0), \end{align*} and \begin{align*} \frac{1}{g(z_0+\xi)g(z_0)} \to \frac{1}{g(z_0)^2}. \end{align*} Using the product law for limits in $\mathbb{C}$, we get \begin{align*} \lim_{\xi \to 0}\frac{r(z_0+\xi)-r(z_0)}{\xi} = -\frac{g'(z_0)}{g(z_0)^2}. \end{align*} Therefore the reciprocal map $r=1/g$ is complex differentiable at $z_0$. [/guided] [/step] [step:Combine the reciprocal with the product rule to obtain the quotient] Assume again that $g(z) \ne 0$ for every $z \in \Omega$. Define the quotient map \begin{align*} q: \Omega &\to \mathbb{C}, & z &\mapsto \frac{f(z)}{g(z)}. \end{align*} With $r: \Omega \to \mathbb{C}$ defined by $r(z)=1/g(z)$, we have $q=fr$. The previous step shows that $r$ is complex differentiable at $z_0$, and the product rule proved above shows that $fr$ is complex differentiable at $z_0$. Hence $q=f/g$ is complex differentiable at $z_0$. Since $z_0 \in \Omega$ was arbitrary, $f+g$, $fg$, and $af$ are holomorphic on $\Omega$ for every $a \in \mathbb{C}$, and, under the nonvanishing hypothesis on $g$, $f/g$ is holomorphic on $\Omega$. [/step]