Without using Reimann Mapping theorem , How to argue conformal maps does not exist?

Question

Without using Reimann Mapping theorem , How to argue conformal maps does not exist ?

1) For $\mathbb C/${0}$\to \mathbb D$

I know first set is not simply connected .But need to argue without using RMT.
I tried to Liovellies theorem but as function is not entire, I could not applied. I also tried reverse direction I know that any non vanishing function is of form $e^{f(z)}$ for some holomorphic function f(z). Please suggest some natural strategy to tackle such problem?
Any Help will be appreciated

Answer 1

Let $f : \mathbb C \setminus \{0\} \to \mathbb D$ holomorphic. Then $0$ is an isolated singularity of $f$ and $f$ is bounded. Hence, $0$ is a removable singularity of $f$. Therefore there is a entire function $g$ such that $f(z)=g(z)$ for all $z \ne 0$. Then $g$ is bounded and by Liouville $g$ is constant. Thus $f$ is constant.

Answer 2

You can extend such a map to the whole $\mathbb{C}$ by the classification of singularities. So you get a contradiction from Louville's Theorem.

Answer 3

If a conformal map existed then it would be invertible, under which function the image of $\mathbb D$ would have to be simply connected.