[proofplan]
We convert the implicit equation $F(z,w)=0$ into an inverse function problem by introducing the holomorphic map $\Phi(z,w)=(z,F(z,w))$. The hypothesis says precisely that the derivative of $\Phi$ at $(a,b)$ is invertible, because its block form is triangular with identity in the $z$-block and the given invertible matrix in the $w$-block. The [Holomorphic Inverse Function Theorem](/theorems/4950) then gives a local holomorphic inverse near $(a,0)$, and the desired function $g$ is the second component of this inverse along the slice $\mathbb{C}^m \times \{0\}$. The graph identity follows because $F(z,w)=0$ is exactly the statement that $\Phi(z,w)=(z,0)$.
[/proofplan]
[step:Build a holomorphic map whose inverse encodes the zero set]
Define the map
\begin{align*}
\Phi: U \to \mathbb{C}^m \times \mathbb{C}^n, \quad (z,w) \mapsto (z,F(z,w)).
\end{align*}
Since the coordinate projection $(z,w)\mapsto z$ is holomorphic and $F$ is holomorphic by hypothesis, $\Phi$ is holomorphic. Also
\begin{align*}
\Phi(a,b)=(a,F(a,b))=(a,0).
\end{align*}
[/step]
[step:Verify that the complex derivative of $\Phi$ is invertible at $(a,b)$]
Let
\begin{align*}
D_zF_{(a,b)}: \mathbb{C}^m \to \mathbb{C}^n
\end{align*}
denote the complex derivative of $F$ at $(a,b)$ in the $z$-variables, and let
\begin{align*}
D_wF_{(a,b)}: \mathbb{C}^n \to \mathbb{C}^n
\end{align*}
denote the complex derivative of $F$ at $(a,b)$ in the $w$-variables. The matrix of $D_wF_{(a,b)}$ in the standard bases is precisely
\begin{align*}
\left(\frac{\partial F_i}{\partial w_j}(a,b)\right)_{1 \leq i,j \leq n},
\end{align*}
so $D_wF_{(a,b)}$ is invertible by hypothesis.
The complex derivative
\begin{align*}
d\Phi_{(a,b)}: \mathbb{C}^m \times \mathbb{C}^n \to \mathbb{C}^m \times \mathbb{C}^n
\end{align*}
is given by
\begin{align*}
d\Phi_{(a,b)}(u,v)=(u,D_zF_{(a,b)}u+D_wF_{(a,b)}v)
\end{align*}
for every $(u,v) \in \mathbb{C}^m \times \mathbb{C}^n$. To prove invertibility, let $(p,q) \in \mathbb{C}^m \times \mathbb{C}^n$. The equation $d\Phi_{(a,b)}(u,v)=(p,q)$ forces $u=p$, and then forces
\begin{align*}
D_wF_{(a,b)}v=q-D_zF_{(a,b)}p.
\end{align*}
Since $D_wF_{(a,b)}$ is invertible, this has the unique solution
\begin{align*}
v=(D_wF_{(a,b)})^{-1}(q-D_zF_{(a,b)}p).
\end{align*}
Thus $d\Phi_{(a,b)}$ is a bijective complex-[linear map](/page/Linear%20Map), hence is invertible.
[guided]
The reason for defining $\Phi(z,w)=(z,F(z,w))$ is that it keeps the parameter $z$ unchanged and replaces the unknown $w$ by the value of the equation $F(z,w)$. If $\Phi$ can be inverted near $(a,b)$, then solving $F(z,w)=0$ becomes the same as asking for the preimage of $(z,0)$.
We now check the derivative carefully. Define
\begin{align*}
D_zF_{(a,b)}: \mathbb{C}^m \to \mathbb{C}^n
\end{align*}
to be the complex derivative of $F$ at $(a,b)$ in the $z$-variables, and define
\begin{align*}
D_wF_{(a,b)}: \mathbb{C}^n \to \mathbb{C}^n
\end{align*}
to be the complex derivative of $F$ at $(a,b)$ in the $w$-variables. The given matrix
\begin{align*}
\left(\frac{\partial F_i}{\partial w_j}(a,b)\right)_{1 \leq i,j \leq n}
\end{align*}
is the standard-basis matrix of $D_wF_{(a,b)}$. Hence the hypothesis that this matrix is invertible is exactly the statement that the complex-linear map $D_wF_{(a,b)}$ is invertible.
For a tangent vector $(u,v) \in \mathbb{C}^m \times \mathbb{C}^n$, differentiating the first component of $\Phi$ gives $u$, because the first component is the identity map in the $z$-variable. Differentiating the second component gives the first-order variation of $F$ in both variables. Therefore
\begin{align*}
d\Phi_{(a,b)}(u,v)=(u,D_zF_{(a,b)}u+D_wF_{(a,b)}v).
\end{align*}
To see why this map is invertible, solve explicitly for the input from the output. Given $(p,q) \in \mathbb{C}^m \times \mathbb{C}^n$, the equation
\begin{align*}
d\Phi_{(a,b)}(u,v)=(p,q)
\end{align*}
first gives $u=p$. Substituting this into the second component gives
\begin{align*}
D_wF_{(a,b)}v=q-D_zF_{(a,b)}p.
\end{align*}
Since $D_wF_{(a,b)}$ is invertible, there is exactly one solution:
\begin{align*}
v=(D_wF_{(a,b)})^{-1}(q-D_zF_{(a,b)}p).
\end{align*}
Thus every target $(p,q)$ has exactly one preimage under $d\Phi_{(a,b)}$, so $d\Phi_{(a,b)}$ is invertible. This is the precise point where the invertibility of the $w$-Jacobian is used.
[/guided]
[/step]
[step:Apply the holomorphic inverse function theorem and shrink to product neighbourhoods]
We invoke the Holomorphic [Inverse Function Theorem](/theorems/51) (citing a result not yet in the wiki: Holomorphic Inverse Function Theorem) for the holomorphic map $\Phi$ at $(a,b)$. The hypotheses are satisfied because $U$ is open, $\Phi$ is holomorphic on $U$, and $d\Phi_{(a,b)}$ is invertible. Hence there exist open neighbourhoods $W \subseteq U$ of $(a,b)$ and $V \subseteq \mathbb{C}^m \times \mathbb{C}^n$ of $(a,0)$ such that
\begin{align*}
\Phi|_W: W \to V
\end{align*}
is biholomorphic. Let
\begin{align*}
\Psi: V \to W
\end{align*}
denote its holomorphic inverse.
Since $W$ is open and contains $(a,b)$, choose open neighbourhoods $A_1 \subseteq \mathbb{C}^m$ of $a$ and $B \subseteq \mathbb{C}^n$ of $b$ such that $A_1 \times B \subseteq W$. Since $V$ is open and contains $(a,0)$, choose an open neighbourhood $A_2 \subseteq \mathbb{C}^m$ of $a$ such that $A_2 \times \{0\} \subseteq V$. Write $\Psi_2: V \to \mathbb{C}^n$ for the second coordinate function of $\Psi$. Since $\Psi_2$ is continuous and $\Psi_2(a,0)=b$, shrink $A_2$ if necessary so that $\Psi_2(z,0) \in B$ for every $z \in A_2$. Define
\begin{align*}
A:=A_1 \cap A_2.
\end{align*}
Then $A$ is an open neighbourhood of $a$, $A \times B \subseteq U$, $A \times \{0\} \subseteq V$, and $\Psi_2(z,0) \in B$ for every $z \in A$.
[/step]
[step:Define the implicit function as the second component of the inverse]
Define
\begin{align*}
g: A \to B, \quad z \mapsto \Psi_2(z,0).
\end{align*}
The map $z \mapsto (z,0)$ from $A$ to $V$ is holomorphic, $\Psi_2$ is holomorphic as a coordinate function of the holomorphic map $\Psi$, and therefore $g$ is holomorphic as a composition of holomorphic maps.
For each $z \in A$, write
\begin{align*}
\Psi(z,0)=(\Psi_1(z,0),\Psi_2(z,0)),
\end{align*}
where $\Psi_1:V\to\mathbb{C}^m$ is the first coordinate function of $\Psi$. Since $\Phi(\Psi(z,0))=(z,0)$ and $\Phi(\zeta,\eta)=(\zeta,F(\zeta,\eta))$, we get
\begin{align*}
\Psi_1(z,0)=z.
\end{align*}
Thus
\begin{align*}
\Psi(z,0)=(z,g(z))
\end{align*}
for every $z \in A$.
[/step]
[step:Identify the zero set with the graph of $g$]
Let $(z,w) \in A \times B$. Since $A \times B \subseteq W$, the restriction $\Phi|_W$ is injective at $(z,w)$.
First suppose $F(z,w)=0$. Then
\begin{align*}
\Phi(z,w)=(z,F(z,w))=(z,0).
\end{align*}
Because $(z,0) \in A \times \{0\} \subseteq V$, applying the inverse map $\Psi$ gives
\begin{align*}
(z,w)=\Psi(z,0)=(z,g(z)).
\end{align*}
Conversely, suppose $(z,w)=(z,g(z))$ for some $z \in A$. By the definition of $g$ and the previous step,
\begin{align*}
(z,g(z))=\Psi(z,0).
\end{align*}
Applying $\Phi$ gives
\begin{align*}
\Phi(z,g(z))=(z,0).
\end{align*}
Since $\Phi(z,g(z))=(z,F(z,g(z)))$, the second coordinate equality gives
\begin{align*}
F(z,g(z))=0.
\end{align*}
Therefore
\begin{align*}
\{(z,w) \in A \times B : F(z,w)=0\}=\{(z,g(z)) : z \in A\}.
\end{align*}
This is the asserted local holomorphic graph representation of the zero set.
[/step]
