Proof of implicit function theorem for a complex function not necessarily holomorphic

Question

Let $U,V\subset\mathbb{C}$ be domains, and $F(z,w):U\times V\to\mathbb{C}$ be continuous, and holomorphic in $z$ for every fixed $w\in V$. Let $(z_0,w_0)\in U\times V$ be s.t $F(z_0,w_0)=0$ and $\frac{\partial F}{\partial z}(z_0,w_0)=F_1(z_0,w_0)\neq0$. Let $r>0$ be s.t for every $z_0\neq z\in\overline{\mathbb{D}_r(z_0)}, F(z,w_0)\neq0$. Prove that there exists $\delta>0$ s.t for every $w\in\mathbb{D}_\delta(w_0)$ there exists a unique $z=g(w)\in\overline{\mathbb{D}_r(z_0)}$ s.t $F(z,w)=0$.

Edit: I now see that I cannot use the implicit function theorem for real functions as $F$ is not necessarily continuiously differentiable as a real function. I tried using the argument theorem but this didnt get me too far. Any help would be appreciated.

Answer 1

For any $z \in \mathbb C$ and $b \gt 0$, let $C_b(z)$ be the circle centered on $z$ of radius $b$ and $D_b(z)$ the associated open disk.

Let's first prove that it exists $r_1,r_2 \gt 0$ such that $F$ doesn't vanish on $C_{r_1}(z_0) \times D_{r_2}(w_0)$

Let $a = \frac{\partial F}{\partial z}(z_0,w_0)= F_1(z_0,w_0) \neq 0$. As $F(\cdot, w_0)$ is supposed to be holomorphic, from $F(z_0,w_0) = 0$ and $a \neq 0$, it follows that it exists $r_1 \gt 0$ such that $F(z,w_0) \neq 0$ in $\overline{D_{r_1}(z_0)} \setminus \{z_0\}$. As $C_{r_1}(z_0)$ is compact, $\lvert F \rvert$ attains a positive minimum $b$ on $C_{r_1}(z_0)$. From the continuity of $F$ on $U \times V$ and again the compactness of $C_{r_1}(z_0)$, we can find $r_2 \gt 0$ such that $\lvert F(z,w) \rvert \gt b/2 \gt 0$ for $(z,w) \in C_{r_1}(z_0) \times D_{r_2}(w_0)$.

About the desired result

From the formula

$$\frac{\partial F}{\partial z}(z,w) = \frac{1}{2 i \pi} \int_{C_{\delta}(z_0)} \frac{F(\zeta,w)}{(\zeta -z)^2} \ d\zeta$$ which holds for $\delta \gt 0$ small enough, $z \in D_\delta(z_0)$ and the continuity of $F$, it follows that $\frac{\partial F}{\partial z}(z,w)$ is also continuous. Therefore the map $$\eta(w)=\frac{1}{2\pi i}\int_{C_{r_1}(z_0)}\frac{F_1(z,w)\,dz}{F(z,w)} $$ is well defined and continuous in the open disk $D_{r_2}(w_0)$. According to the argument principle, $\eta(w)$ is the number of zeros of the equation $F(z,w)=0$ in the open disk $D_{r_1}(z_0)$. Also $\eta(w_0) = 1$ as $z_0$ is the only root of $F(z,w_0) = 0$ in $D_{r_1}(z_0)$. We finally get the desired conclusion that for any $w \in D_{r_2}(w_0)$, the equation $F(z,w) = 0$ has a unique solution in the open disk $D_{r_1}(z_0)$.