Problem: Let $U\subset\mathbb{C}$ be an open set, $f:U\to\mathbb{C}$ an holomorphic function of class $C^1$ and $z_0\in U$. Prove that if $f'(z_0)\neq 0$ then there exists a neighborhood $V$ of $z_0$ such that the restriction of $f$ to $V$ has an holomorphic inverse.
I would like to know if the solution below is correct/acceptable. This exercise was taken of a section that studies the Inverse Function Theorem for mappings from $U\subset\mathbb{R}^m$ to $\mathbb{R}^m$.
Solution: Since
\begin{matrix}
f & : & U& \longrightarrow & \mathbb{C}\\
& & (x,y) & \longmapsto & (u(x,y),v(x,y))
\end{matrix}
is holomorphic, we conclude that $u,v:U\longrightarrow\mathbb{C}$ are differentiable and
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\quad\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$
It follows that
$$\det J_f(z_0)=\left(\frac{\partial u}{\partial x}(z_0)\right)^2+\left(\frac{\partial u}{\partial y}(z_0)\right)^2\neq0$$
because
$$\left(\frac{\partial u}{\partial x}(z_0),-\frac{\partial u}{\partial y}(z_0)\right)=f'(z_0)\neq 0.$$
So, by Inverse Function Theorem, there exists a neighborhood $V$ of $z_0$ and an open set $W\ni f(z_0)$ such that $f|_V:V\to W$ is a diffeomorphism and thus has a differentiable inverse
\begin{matrix}
(f|_V)^{-1} & : & W& \longrightarrow & V\\
& & (x,y) & \longmapsto & (\tilde{u}(x,y),\tilde{v}(x,y))
\end{matrix}
Furthermore, for all $v\in \mathbb{R}^2$,
$$\left((f|_V)^{-1}\right)'(f(z_0))\cdot v=(f'(z_0))^{-1}\cdot v=
\begin{vmatrix}
\frac{\partial u}{\partial x} (z_0)& \frac{\partial u}{\partial y}(z_0)\\
-\frac{\partial u}{\partial y}(z_0)& \frac{\partial u}{\partial x}(z_0)
\end{vmatrix}^{-1}v=
\frac{1}{\det J_f(z_0)}\begin{vmatrix}
\frac{\partial u}{\partial x}(z_0) & -\frac{\partial u}{\partial y}(z_0)\\
\frac{\partial u}{\partial y}(z_0)& \frac{\partial u}{\partial x}(z_0)
\end{vmatrix}v$$
and thus
$$\frac{\partial \tilde{u}}{\partial x}(f(z_0))=\frac{\partial \tilde{v}}{\partial y}(f(z_0)),\quad
\frac{\partial \tilde{u}}{\partial y}(f(z_0))=-\frac{\partial \tilde{v}}{\partial x}(f(z_0)).$$
For any point $p=f(z)\in W$, we have $\det J_f(z)\neq 0$ (because $f|_V$ is a diffeomorphism) so that we can apply a similar argument to conclude that
$$\frac{\partial \tilde{u}}{\partial x}(f(z))=\frac{\partial \tilde{v}}{\partial y}(f(z)),\quad
\frac{\partial \tilde{u}}{\partial y}(f(z))=-\frac{\partial \tilde{v}}{\partial x}(f(z)).$$
Hence, $(f|_V)^{-1}$ is holomorphic. $\blacksquare$
Thanks.
Best Answer
Your proof is correct. Alternatively, you could prove that $(f|_V)^{-1}$ is holomorphic by proving that it is differentiable (as a complex function):
Let $w=f(z)\in f(V)$. Notice that $f'(z)\neq 0$ (because $|f'(z)|^2=\det J_f(z)\neq 0$, since $f|_V$ is a diffeomorphism in the real sense) (if we wanted, we could just restrict $V$ a little more and assume this, but this isn't necessary). Then $$\dfrac{1}{f'(f^{-1}(w))}=\dfrac{1}{f'(z)}=\lim_{h\to 0}\dfrac{h}{f(z+h)-f(z)}.$$
We already know that $f|_V:V\to f(V)$ is a diffeomorphism, so in particular it is a homeomorphism. Thus, we may make the substitution $k=f(z+h)-f(z)=f(z+h)-w$, or equivalently $h=f^{-1}(w+k)-z=f^{-1}(w+k)-f^{-1}(w)$, in the limit above, and obtain $$\dfrac{1}{f'(f^{-1}(w))}=\lim_{k\to 0}\dfrac{f^{-1}(w+k)-f^{-1}(w)}{k},$$ which means that $(f|_V)^{-1}$ is differentiable (at every $w$) in the complex sense, that is, $(f|_V)^{-1}$ is holomorphic.
This argument (using Invariance of Domain) actually shows that if $f$ is continuous, injective, and differentiable (in the complex sense) at a point $z$ with $f'(z)\neq 0$, then $f^{-1}$ is differentiable in the complex sense at $f(z)$.