Inverse function theorem in complex analysis

For functions of one real variable, the proof is simpler because nonvanishing derivative implies strict monotonicity, and we get inverse function at once. But this can be adapted to complex variables by interpreting strict monotonicity as $$\operatorname{Re}\frac{f(a)-f(b)}{a-b}>0,\quad a\ne b \tag1$$ (Of course this really mimics increasing rather than monotone functions). Based on this, one can give a proof that does not rely on the inverse function theorem for several real variables.

Indeed, if $f'(z_0)\ne 0$, then the function $g(z)=f(z)/f'(z_0)$ satisfies $g'(z_0)=1$. Since $g$ is $C^1$, it follows that $\operatorname{Re}g'\ge \frac12$ in some neighborhood $U=\{z:|z-z_0|<r\}$. Therefore, $$\operatorname{Re}\frac{g(a)-g(b)}{a-b} = \int_0^1 \operatorname{Re} g'(ta+(1-t)b)\,dt \ge \frac12,\quad a,b\in U, \ a\ne b \tag2$$ Hence, $|g(a)-g(b)|\ge \frac12 |a-b|$ for $a,b\in U$. We conclude that $g^{-1}$ is defined and Lipschitz continuous in $g(U)$ (which is an open set by the open mapping theorem for analytic functions). It remains to compute the derivative of $g^{-1}$ in the usual way, by flipping the difference quotient.

Let $f(z):D \to \mathbb{C}$ be holomorphic in a domain $D$ of the complex plane. and suppose that $f'(z_0) \neq 0$, then it is locally invertible.

Because of the following, let $f(z)=u(x,y)+iv(x,y)$, then we can think of this function as a function from $f(x,y) : \mathbb{R}^2 \to \mathbb{R}^2$ where $f(x,y)=\left(u(x,y),v(x,y)\right)$. now using the inverse function theorem for functions from $\mathbb{R}^n$ to $\mathbb{R}^2$, we form the Jacobian of this function: $$ J_f(x,y) = \det \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix} = u_xv_y-u_yv_x$$ and note that the condition of nonzero Jacobian at the point $(x_0,y_0) \equiv z_0$ is the same as $f'(z_0) \neq 0$. Because by Cauchy-Riemann equations we have $$ 0 \neq u_x(x_0,y_0)v_y(x_0,y_0)-u_y(x_0,y_0)v_x(x_0,y_0) = \left( u_x(x_0,y_0) \right)^2+\left( v_x(x_0,y_0) \right)^2= |f'(z_0)|^2 \ \ \blacksquare $$