Let $f(z)$ be an entire function with an entire inverse. Prove that as $z$ goes to infinity, $|f(z)|$ goes to infinity.

Question

Prove that $\lim\limits_{z \to \infty} |f(z)| = \infty$ where $f(z)$ is entire and has entire inverse $g(z)$.

I can show that the limit cannot be finite since if it were, then we can use Liouville's theorem to conclude that $f$ is constant. But how do I show that the limit is actually infinity?

Every answer in the other question uses some of Picard's theorem, the Open mapping theorem, Riemann's theorem. The only theorems I've covered which are relevant to the question are Casorati-Weierstrass and Liouville.

Answer 1

Here is an elementary proof which does not even use Casorati-Weierstraß, only the (assumed) existence of a holomorphic inverse.

Assume that $\lim_{z \to \infty} |f(z)| = \infty$ does not hold. Then there is a sequence $(z_n)$ of complex numbers such that $z_n \to \infty$ and $w_n = f(z_n)$ is bounded. A bounded sequence has a convergent subsequence: $w_{n_k} \to w^* \in \Bbb C$.

But the inverse function $g$ is continuous, therefore $$ z_{n_k} = g(w_{n_k}) \to g(w^*) \in \Bbb C $$ in contradiction to the assumption that $z_n \to \infty$.

Answer 2

Hint: If $f(z)$ is entire injective then $f(z)=a\cdot z+b$ (To show this, use the fact that the Taylor expansion must be infinite together with Casoratti-Weierstrass or Picard's theorem).

Answer 3

If $f$ is not a polynomial then it has an essential singularity at infinity. According to Casorati-Weierstraß, $$ G = f(\{z: |z| > 1\}) $$ is dense in $\Bbb C$. Now consider the inverse function $g = f^{-1}$ (which is assumed to exist as an entire function). It follows that $$ g(G) = \{z: |z| > 1\} $$ and since $g$ is continuous and $G$ dense in $\Bbb C$ this implies $$ g(\Bbb C) \subset \{z: |z| \ge 1\} \, . $$ Then $1/g$ is a bounded, entire function and therefore constant, which is not possible (as the inverse of $f$).

So $f$ is necessarily a polynomial, and that implies $\lim_{z \to \infty} |f(z)| = \infty$.