Spring 2021 - Problem 11

Solution.

1. Suppose otherwise, and let $n$ be minimal, i.e., $f^{\p{n}}$ is polynomial but $f^{\p{n-1}}$ is not. Then $f^{\p{n-1}}$ has an essential singularity at $\infty$, so by Casorati-Weierstrass, there exists a sequence $\set{z_k}_k$ such that $z_k \to \infty$ and $f^{\p{n-1}}\p{z_k} \to z_0 \in \C$. But by continuity,

   $$f^{\p{n}}\p{z_k} = f\p{f^{\p{n-1}}\p{z_k}} \xrightarrow{k\to\infty} f\p{z_0} \neq \infty,$$

   but this contradicts the fact that polynomials have a pole at $\infty$. Thus, no $n$ could have existed to begin with.

2. Suppose otherwise, so that $f^{\p{n}}\p{z} = e^z$. Then

   $$f\p{e^z} = f\p{f^{\p{n}}\p{z}} = f^{\p{n}}\p{f\p{z}} = e^{f\p{z}}.$$

   Thus, if $g\p{z} = f\p{e^z} = e^{f\p{z}}$, then we see that $g$ must omit $0$, since $e^{f\p{z}} \neq 0$ for any $z$. Similarly, since $e^z \neq 0$, we also see that $g$ omits $f\p{0}$. Since $f$ is non-constant, little Picard implies that $f\p{0} = 0$. But this means

   $$1 = e^0 = f^{\p{n}}\p{0} = 0,$$

   a contradiction. Thus, no $n$ could have existed to begin with.