entire bijection of $\mathbb{C}$ with 2 fixed points

Solution 1:

No. Let $f$ be an entire function, and consider the singularity of $f$ at infinity. If it's removable, then by Liouville's theorem $f$ is constant and so not a bijection. If it's essential, then by Picard's great theorem $f$ is not injective. If it's a pole, then $f$ is a polynomial. If the degree of $f$ is greater than one, then $f$ is not injective. So $f$ is a linear polynomial, and the only linear polynomial that fixes two points is the identity.

Solution 2:

Since invoking Picard seems like a bit of an overkill, here's a simple argument (essentially the same as Chris's answer, but replacing Picard by Casorati-Weierstrass and the open mapping theorem):

Claim. If $f: \mathbb{C} \to \mathbb{C}$ is entire and injective then $f$ is linear and non-constant, that is $f(z) = az + b$ for some $a \neq 0$.

It is easy to check that a linear function with two fixed points must be the identity, so we are left with proving the claim.

Proof of the claim. By hypothesis $f: \mathbb{C}^{\times} \to \mathbb{C}$ is injective where $\mathbb{C}^{\times} = \{z \neq 0\}$ is the set of non-zero complex numbers. Now $g(z) = f(1/z)$ is also injective $\mathbb{C}^{\times} \to \mathbb{C}$ and has a singularity at $0$. By injectivity, $g(\{|z| \gt 1\}) \cap g(\{0 \lt |z| \lt 1\}) = \emptyset$. On the other hand, both sets are open by the open mapping theorem, and by the Casorati-Weierstrass theorem $g(\{0 \lt |z| \lt 1\})$ would be dense if $g$ had an essential singularity at $0$. Therefore $g$ has a pole at $0$. But this implies that $f$ is a polynomial and hence it is linear and non-constant by injectivity.