Let $\mathbb{B}_r$ denote the closed disk $\{z \in \mathbb{C} : |z| \le r \}$. State whether $\infty$ is a removable singularity , pole , or essential singularity in the following statement
$f$ is an entire function for which $f^{-1}(B_r)$ is bounded for all $r >0$
My attempt :Here $\infty$ are both removable singularity and pole
Inverse image of closed unit disk of $f$ is bounded so $f$ can not have essential singularity at $\infty$
Removable singularity :-
Suppose that $f$ is an entire function that has a removable singularity at infinity.Then there exist an entire function $g$ such that $g(z)= f(1/z)$ for all $z \in \mathbb{B_r}-\{0\}$.This implies that $\lim_{z\to \infty} f(1/z)=f(0) $ which in turn implies that $g$ is bounded.Since g is a bounded entire function, by liouville's theorem , $g$ is constant.Hence $ f$ is constant $\implies f^{-1}(B_r)$ is bounded for all $r >0$
For pole
take $f(z)= z$ , $g(z)=f(1/z)=1/z$ where $|z|\le r >0$. $g(z)$ has pole at $z=0 \implies f$ has a pole at $\infty$
$\implies f^{-1}(B_r)$ is bounded for all $r >0$
Best Answer
Your second example is fine. For removable singularity the argument is simpler. Any entire function with removable singularty at $\infty$ is a constant by Liouville's Theorem because it is bounded.
If $f$ has an essential singularity at $\infty$ then it would take values in $B_1$ in every neighborhood of $\infty$ so $f^{-1} (B_1)$ is not bounded. Hence, $f$ cannot have an essential singularity at $\infty$.