Suppose $f(z)$ is an entire function in $C$, and $|f(z)| ≥ M$, $∀z ∈ C$ for a positive constant $M$.
Prove that $f(z)$ must be a constant function.
Hint: Liouville’s theorem may be applied in a novel way.
I am assuming that this question is basically asking to prove Liouville's theorem because that is exactly what the theorem says.
I copied this from Wikipedia to make sure I am not forgetting anything or looking at a different theorem.
In complex analysis, Liouville's theorem, named after Joseph Liouville, states that every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number $M$ such that $|f(z)| <= M$ for all $z \in {C}$ is constant. Equivalently, non-constant holomorphic functions on $C$ have dense images.
Best Answer
If $|f(z)|\geq M>0$, then $g(z):=\frac{1}{f(z)}$ is entire (in particular, because $f$ is entire and $f(z)\neq0$) and $$|g(z)|=\frac{1}{|f(z)|}\leq\frac{1}{M}:=C.$$
Now apply the conclusion of Liouville's theorem to $g$ in order to conclude the same for $f$.
P.S. - You can see how to interpret the statement non-constant entire functions have dense images in this context (and use it for an alternative proof): if $f$ is continuous and strictly greater than $0$, then it avoids a neighborhood $V$ of $0$ and so the image of $f$ is a subset of $\mathbb{C}-V$, which is of course not dense in $\mathbb{C}$.