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

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

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