TITLE: How can it be shown that $\mathrm{Aut}(\mathbb{C})=\{f\,|\,f(z)=az+b,a\neq 0\},$ is defined as bijective ...
QUESTION [31 upvotes]: How can it be shown that
$$\mathrm{Aut}(\mathbb{C})=\{f\,|\,f(z)=az+b,a\neq 0\},$$
where an automorphism of $\mathbb{C}$ is defined as a bijective entire function with entire inverse?
If $f$ is of the form $f(z)=az+b$, with $a\neq 0$, then obviously $f$ is in $\mathrm{Aut}(\mathbb{C})$. How can I prove the converse?

REPLY [3 votes]: There is a nice way to do this by using Riemann surfaces, plus the fact that one can view $\Bbb C\mathrm{P}^1$ as "$\Bbb C$ with an added point at infinity" or equivalently, we may view $\Bbb C$ as $\Bbb C\mathrm{P}^1\setminus \{\infty\}$. The key observation is that, if one has an automorphism $f$ of $X\setminus P$, where $P=\{p_1,\dots,p_n\}$ is a finite set and $X$ a Riemann surface, then it must extend to an automorphism of $X$ (not necessarily unique). This is done by taking little disk charts around a puncture $p_j$ and holomorphically continuing $f$ over the point, which is possible by boundedness.
Now, it is a standard result in the theory of Riemann surfaces that the meromorphic function on $\Bbb C\mathrm{P}^1$ are the rational function (i.e. quotients of polynomials), and that meromorphic functions can be viewed as holomorphic maps into $\Bbb C\mathrm{P}^1$. Hence, one just needs to check which rational functions are actually biholomorphisms. Because of the fact that a non-constant holomorphic map (between Riemann surfaces) is locally always of the form $z^n$ for some $n\geq 1$, it suffices to check injectivity. Using this, one establishes easily that the automorphisms of $\Bbb C\mathrm{P}^1$ are precisely the Möbius transformations 
$$ f(z)=\frac{az+b}{cz+d} \qquad \qquad ad-bc\neq 0$$
where $z\in \Bbb C\mathrm{P}^1=\Bbb C\cup \{\infty\}$ (so the domain (and range) include the point "$\infty$"). Now, the above shows that any automorphism of $\Bbb C$ extends to a Möbius transformation which must send $\infty$ to itself. This is equivalent to $c=0$, and you obtain the result you're looking for.