[Math] Inverse function of a conformal mapping

Question

I'm trying to prove that, if $f$ is a conformal mapping at $z_0$, then it has an inverse $g$ that is conformal at $w_0=f(z_0)$.

I proved the existence of $g$ using the Inverse Function Theorem. Since $f$ is conformal at $z_0$, then $f$ is analytic in $z_0$ and so $u,v\in C^1(D)$.

Tha Jacobian is

$\begin{equation*}
J =\begin{vmatrix}
u_x & u_y\\
v_x & v_y\\
\end{vmatrix}
=u_xv_y-u_yv_x
\end{equation*}$

and using the Cauchy-Riemann equations, we have that

\begin{equation*}
J=(u_x)^2+(v_x)^2=|f'(z)|^2
\end{equation*}

Since $f'(z_0)\neq 0$, $J\neq 0$. So, by the Inverse Function Theorem, $f$ has an inverse $g$.

How can I prove that $g$ is analytic in $w_0$?

That's the only thing that is missing, since I also proved that
\begin{equation*}
g'(w_0)=\frac{1}{f'(z_0)}
\end{equation*}

and since $f'(z_0)\neq 0$, then $g'(w_0)\neq 0$. This, coupled with the fact that $g$ is analytic in $w_0$, implies that $g$ is conformal at $w_0$.

Answer 1

This would be basically an application of the Lagrange inversion theorem, which states the following:

Suppose $w$ is defined as a function of $z$ by an equation of the form $$f(z)=w,$$ where $f$ is analytic at a point $z_0$ and $f'(z_0)\neq 0$. Then it is possible to invert (or solve) the equation for $z$: $$z=g(w),$$ on a neighborhood of $f(z_0)=w_0$, where $g$ is analytic at the point $w_0$.

Since you have (or have proved) all the hypothesis, you can just apply the theorem.