Holomorphic functions are open mapping via inverse function theorem

Question

I have been reading about the open mapping theorem for non constant holomorphic functions, all the proofs involve theorems of complex analysis (like Rouche, argument principle). What if we just applied the inverse function theorem to obtain that holomorphic functions are local diffeomorphism and therefore open, is this valid?

Answer 1

Yes, you can prove it using the Inverse Function Theorem:

Consider a non-constant holomorphic function $f:\Omega\longrightarrow \mathbb{C}$, where $\Omega$ is a connected open set. Let $z_0\in \Omega$.

Since $f$ is not constant, the zero $z_0$ from the function $f-f(z_0)$ can be factored out. That is, there exists a function $g$ holomorphic and non-zero in some disk $D(z_0,r)$ such that $$f(z)-f(z_0)=(z-z_0)^ng(z),$$ where $n$ is the multiplicity of $z_0$.

Now, since $g$ is non-zero, it has an holomorphic logarithm: there exists a function $h\in \mathcal{H}(D(z_0,r))$ such that $$g=e^h.$$

Therefore, $$f(z)-f(z_0)=(z-z_0)^ne^{h(z)}=\left((z-z_0)e^{\frac{h(z)}{n}}\right)^n.$$

So $f$ can be expressed locally as $$f=f(z_0)+F^n,$$ where $F:=(z-z_0)e^{h/n}$. You can check that $F'$ doesn't annihilate in $z_0$. Thus, $F$ is locally open by the Inverse Function Theorem. Furthermore, $F^n$ is also open since it is the composition of two open functions, $F$ and $z\mapsto z^n$. Finally, adding the constant $f(z_0)$ doesn't change anything, so $f$ is open.