If $f'(z_0)\neq 0$ then $f$ has an holomorphic inverse. 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.
 A: 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 $f(z)$.
A: (Note: I am a little inconsistent with notation below. For parameters to
a function, I write $(a,b)$ rather than $(a,b)^T$. No confusion should arise.)
Let $C = \left\{ \begin{bmatrix} a & b \\ -b & a\end{bmatrix}\right\}_{a,b \in \mathbb{R}} $. If $A \in C$ is invertible, then it is straightforward to show that $A^{-1} \in C$ (in fact we can write down the
inverse explicitly, but that is not the focus here).
Consider $f$ as a mapping $\phi:\mathbb{R}^2 \to \mathbb{R}^2$. 
That is, let $\phi((x_1,x_2)) = (\operatorname{re} f(x_1+i x_2), \operatorname{im} (f x_1+i x_2))^T$.
It is clear that any function $\gamma: \mathbb{R}^2 \to \mathbb{R}^2$ has an associated function (not necessarily analytic) $g:\mathbb{C} \to \mathbb{C}$ given by $g(z) = \gamma_1((\operatorname{re} z, \operatorname{im} z)) + i \gamma_2( ( \operatorname{re} z, \operatorname{im} z ) )$.
It is straightforward to show that $f'(z) \neq 0$ iff $D\phi((\operatorname{re} z, \operatorname{im} z))$ is invertible and $D\phi((\operatorname{re} z, \operatorname{im} z)) \in C$.
Hence if $f'(z) \ne 0$, the inverse function theorem gives the existence of some $\gamma$ such that 
$\gamma(\phi(x)) = x$ in some neighbourhood of $(\operatorname{re} z, \operatorname{im} z)$. All that remains is to show that the corresponding $g$ satisfies the Cauchy Riemann equations.
We see that $f$ satisfies the
Cauchy Riemann equations at $z$ iff $D\phi(\operatorname{re} z, \operatorname{im} z) \in C$. 
Since $\gamma$ satisfies $\gamma(\phi(x)) = x$ in some neighbourhood, we have $D \gamma(\phi(x)) D \phi(x) = I$, in particular,
$D \gamma(\phi(x)) =(D \phi(x) )^{-1}$ (hence $D \gamma(\phi(x)) \in C$). Hence $\gamma$ (or rather $g$) satisfies the Cauchy Riemann equations.
