Show that $\mathbb{C}$ is homogeneous with respect $Aut(\mathbb{C})$ and $\mathbb{C}\setminus{\{0}\}$ is homogeneous with respect $Aut(\mathbb{C}\setminus{\{0}\})$
Idea: A biholomorphic mapping of a domain D onto itself is called an automorphism of D. $Aut (D)$ is a group with respect to the composition of mappings and the identity mapping is its neutral element.
A domain $D$ in $\mathbb{C}$ is called homogeneous with respect to a subgroup $L$ of $Aut D$, if for every two points $z_1, z_2\in D$ there is an automorphism $h\in L$ with $h(z_1) =z_2 $.
For Show that $\mathbb{C}$ is homogeneous with respect $Aut(\mathbb{C})$ just keep in mind $Aut(\mathbb{C})={\{az+b:a\in\mathbb{C}\setminus{\{0}\}, b \in\mathbb{C}}\}$ I don't know how to prove that $\mathbb{C}\setminus{\{0}\}$ is homogeneous with respect $Aut(\mathbb{C}\setminus{\{0}\})$. Can someone help me?
Suppose $w\in\mathbb{C}$ then $h(z)=z+w$ is automorphism of $\mathbb{C}$ Then as $h(0)=w$, the orbit ${\{h(0):h(z)=z+w}\}$ fills $\mathbb{C}$ hence $\mathbb{C}$ is homegeneous with respect $Aut(\mathbb{C})$.
In the other hand Suppose $w\in\mathbb{C}\setminus{\{0}\}$ then $h(z)=wz$ is automorphism of $\mathbb{C}$ Then as $h(1)=w$, the orbit ${\{h(1):h(z)=z+w}\}$ fills $\mathbb{C}\setminus{\{0}\}$ hence $\mathbb{C}\setminus{\{0}\}$ is homegeneous with respect $Aut(\mathbb{C}\setminus{\{0}\})$