Using Liouville's theorem to conclude that if $|f(z)| \le |\sin^3 z|$ then $f$ is a multiple of $\sin^3 z$

Question

Let $f(z)$ be an entire function and suppose that $ |f(z)| \leq |\sin^3(z)|$ for all $z \in \mathbf{C}$. Show that $f(z) = \lambda \sin^3(z)$ for some $\lambda \in \mathbf{C}$.

Answer 1

It is a theorem that if a function $g$ is holomorphic in a (deleted) neighborhood of a point and bounded in that neighborhood, then $g$ has only a removable singularity. (And analogously, if $|g(z)| \to \infty$ as $z \to z_0$, then $g$ has a pole there; the remaining case is that $g$ has an essential singularity.) In this case $g$ has a holomorphic extension across that point.

Following your comments, this deals with the issue of concluding that $f$ has zeros of multiplicity at least $3$ at every zero of $\sin^3 z$. So setting $g(z) = f(z) / \sin^3 z$ leads to a holomorphic function with a discrete set of removable singularities. Replacing $g$ with an entire extension and invoking Liouville completes the proof.