How can it be shown that $\mathrm{Aut}(\mathbb{C})=\{f\,|\,f(z)=az+b,a\neq 0\},$ is defined as bijective ... 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?

You can start by noting that Big Picard's Theorem will tell you that you can not have an essential singularity at infinity because otherwise you will find multiple points mapping to the same value in different neighborhoods of infinity. This tells you that $f$ is necessarily a polynomial
Then note that if a function is degree greater greater than 1 it will have multiple roots and therefore have more than 1 value mapping to 0. Then you just eliminate constant functions to keep injectivity.
