Conformal equivalence

Two plane domains are conformally equivalent when there is a bijective holomorphic map between them whose inverse is holomorphic.

A conformal map from ${\mathbb{C}}^{\ast }$ into the punctured unit disc would be bounded near its isolated singularity at zero, hence extend to a bounded entire function. The Liouville theorem would make it constant.

Composing a hypothetical conformal equivalence $\mathbb{C}\to \mathbb{H}$ with a Cayley transform $\mathbb{H}\to \mathbb{D}$ would give a nonconstant bounded entire function, contradicting the Liouville theorem.

The radial map $z\mapsto z/(1+|z|)$ is a homeomorphism from $\mathbb{C}$ to the unit disc, with inverse $w\mapsto w/(1-|w|)$. A conformal equivalence in that direction would be a nonconstant bounded entire function, contradicting the Liouville theorem.

The mapmaps $\{z:|z|<1,\mathrm{Im}z>0\}$ conformally onto the upper half-plane. The Möbius transformation maps the half-disc onto the first quadrant, and squaring maps that quadrant onto the upper half-plane.