Automorphisms of the complex plane

It depends on how much complex analysis you know. Let me give you a short proof (using heavy machinery).

If $f$ has an essential singularity at $\infty$, it can't be injective on $\mathbb{C}$ due to big Picard. If the singularity at $\infty$ is removable, then $f$ is constant by Liouville. Hence $f$ has a pole at $\infty$ which means that $f$ is a polynomial. If $\deg f \ge 2$, it won't be injective (fundamental theorem of algebra). That only leaves you with polynomials of degree exactly 1.

If $f : \mathbb{C} \rightarrow \mathbb{C}$ is an automorphism, it is entire and hence has a globally convergence power series $f(z) = \sum_{n=0}^\infty a_n z^n$. We can extend this function to a map $\tilde{f} : \mathbb{C} \rightarrow \bar{\mathbb{C}}$ by considering it as a function into the extended complex plane $\bar{\mathbb{C}} = \mathbb{C} \cup \lbrace \infty \rbrace$ which has a singularity at $\infty$.

If we take a coordinate chart about $\infty$ as $U_\infty = \lbrace \infty \rbrace \cup \lbrace z : |z| > 1 \rbrace$ with coordinate map $\phi : U \rightarrow \mathbb{D}$ given by $\phi(z) = 1/z$ for $z \neq \infty$ and $\phi(\infty) = 0$ then we can explore the nature of $\tilde{f}$'s singularity at infinity by composing it with $\phi$:

$ \tilde{f}(\phi(z)) = \sum_{n=0}^\infty \frac{a_n}{z^n} $

Its almost by definition that the above formula has an essential singularity at $z=0$ unless the $a_n$ eventually are all zero. If there were an essential singularity then by Cassorati-Weirestrass $f$ maps $U_\infty \setminus \lbrace \infty \rbrace$ to a dense subset of $\mathbb{C} $. But then $ f(\mathbb{D}) $ would necessarily intersect that dense set non-trivially, so $f$ would not be injective and hence not an automorphism. Therefore the above series $\tilde{f}(\phi(z))$ is a rational function, which implies that $\tilde{f}$ is a polynomial, so $f$ is a polynomial.

Finally, by the fundamental theorem of algebra, a polynomial is injective if and only if its degree is one, so $f(z) = az + b$.

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.