Automorphisms of punctured plane

The automorphism group of $\mathbb{C}\setminus \{0\}$ is generated by linear maps $z\mapsto az$ (with $a\ne 0$) and inversion $z\mapsto 1/z$. So, every element of the group can be written as $z\mapsto az^{\pm 1}$.

Indeed, suppose $f$ is an automorphism of $\mathbb{C}\setminus \{0\}$. Since $f$ is injective, $0$ cannot be a point of essential singularity (recall Picard's theorem), so it's either a pole or removable singularity. By replacing $f$ with $1/f$ if necessary, we can assume $0$ is removable. Then $f$ extends to an automorphism of $\mathbb{C}$, so $f(z)=az+b$. And since it fixes $0$, $b=0$.