I saw this theorem but the proof says that it follows from real analysis and I can't seem to think of anything that the theorem follows from.
Theorem Suppose that $f : U \to \Bbb{C}$ is complex differentiable and $u,v$ have continuous partial derivatives. Suppose $f'(z_0) \neq 0$ for some $z_0 \in U$.
- Then there exists a disc $U$ about $z_0$ such that $f : U \to f(U)$ is a bijection, $f(U)$ is open and $f^{-1} : f(U) \to U$ is continuous.
Any help would be much appreciated.
It's the Inverse function theorem. Identifying $\mathbb{C}$ with $\mathbb{R}^2$, the real Jacobi matrix at $z_0$ is
$$J_f^{\mathbb{R}}(z_0) = \begin{pmatrix} u_x(z_0) & u_y(z_0) \\ v_x(z_0) & v_y(z_0)\end{pmatrix},$$
and using the Cauchy-Riemann equations, we compute its determinant as
$$u_x(z_0)\underbrace{v_y(z_0)}_{v_y = u_x} - \underbrace{u_y(z_0)}_{u_y = -v_x}v_x(z_0) = u_x(z_0)^2 + v_x(z_0)^2 = \lvert u_x(z_0) + i v_x(z_0)\rvert^2.$$
Now for complex differentiable $f$, we have
$$f'(z_0) = \lim_{h\to 0} \frac{f(z_0+h)-f(z_0)}{h} = \lim_{\substack{h\to 0 \\ h\in \mathbb{R}\setminus \{0\}}} \frac{f(z_0+h) - f(z_0)}{h} = f_x(z_0) = u_x(z_0) + iv_x(z_0),$$
so
$$\det J_f^{\mathbb{R}}(z_0) = \lvert f'(z_0)\rvert^2.$$