It is well known that set of automorphisms (bijective, conformal self maps) of
- $\Bbb{C}_{\infty}$ is $\dfrac{az+b}{cz+d},\,\,\,ad-bc\not=0$ (Möbius transformations),
- $\Bbb{C}$ is $az+b,\,\,\, a\not=0$ (scaling+rotating+translating),
- $\Bbb{H}$ is $\dfrac{az+b}{cz+d},\,\,\,ad-bc\not=0$ with $a,b,c,d\in\Bbb{R}$,
- $\Bbb{D}$ is $e^{i\theta}\dfrac{z-a}{1-\bar{a}z},\,\,\,|a|\lt1,\,\,\,\theta\in\Bbb{R}$,
- $\text{Ann}(0,r_1,r_2)$ is $e^{i\theta}z,\,\,\,\theta\in\Bbb{R}.$
Here I am trying to figure-out the automorphisms of punctured plane $\Bbb{C}\setminus\{0\}.$ My guess is it is the collection of functions of the form $az+\dfrac{b}{z},$ where $a,b$ are any two complex numbers with $|a|^2+|b|^2\not=0.$ Is it correct? If it is so, how can I prove this rigorously?
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 $0$, $b=0$.