# If $f$ is entire and $g$ has an essential singularity, must $f\circ g$ have an essential singularity?

I was trying to prove:

Let $f:\mathbb{C}\to\mathbb{C}$ be entire. If $f(f(z))$ is a polynomial, prove that $f$ is polynomial.

Assuming the contrary, we would get that $f(1/z)$ has an essential singularity at $z=0$ while $f(f(1/z))$ has a pole at $z=0$. So I conjectured something stronger:

If $f(z)$ is entire, non-constant and $g(z)$ has an essential singularity at $z=a$, then $f(g(z))$ has an essential singularity at $z=a$.

I think this should be true since the Laurent expansion will not terminate. Casorati-Weierstrass Theorem seems to be useful. If $f(g(z))$ does not have an essential singularity at $a$, we can find $n$ such that $(z-a)^nf(g(z))$ is bounded near $a$. On the other hand $g(z)$ should be "wild". The difficulty is to make precise this wildness to ensure that the factor $(z-a)^n$ is not enough to tame it.

The quick way to see that this is true is to use the Picard theorems. $g$ takes every value except possibly one in any punctured neighbourhood of $a$; and $f$ is onto, again possibly excepting one point; so $f\circ g$ maps any punctured neighbourhood of $a$ onto $\mathbb{C}$ minus a point or two.
You don't really need Picard even; you just need the Casorati–Weierstrass theorem, which states that $g$ maps any punctured neighbourhood of $a$ onto a dense subset of $\mathbb{C}$. You also need to know that an entire nonconstant function has a dense range. For if $b$ is not in the closure of the range of $f$, then $1/(f(z)-b)$ is a bounded, non-constant entire function, which contradicts the Liouville theorem. (The proof of the Casorati–Weierstrass theorem uses the same idea, but instead of the Liouville theorem, employ the fact that an isolated singularity near which the function is bounded, is a removable singularity.)
• ... namely the Casorati-Weierstrass theorem. Also you need that $f$ has dense range, which is a corollary of Liouville's theorem. – Robert Israel May 3 '13 at 7:00