Liouville's Theorem (Complex Analysis)/Corollary

Corollary to Liouville's Theorem (Complex Analysis)
Let $f: \C \to \C$ be an entire function such that:


 * $\size {\map f z} \ge M$

for all $z \in \C$ for some real constant $M > 0$.

Then $f$ is constant.

Proof
Note that since:


 * $\size {\map f z} \ge M > 0$

we cannot have:


 * $\map f {z_0} = 0$

for any $z_0 \in \C$.

So, by the Quotient Rule for Complex Derivatives, we have:


 * $\dfrac 1 f$ is entire.

We also have that:


 * $\ds \size {\frac 1 {\map f z} } \le \frac 1 M$

for all $z \in \C$.

So:


 * $\dfrac 1 f$ is a bounded entire function.

So, by Liouville's Theorem (Complex Analysis), we have:


 * $\dfrac 1 f$ is constant.

That is, there exists some $C \in \C \setminus \set 0$ such that:


 * $\dfrac 1 {\map f z} = C$

for all $z \in \C$.

Then, we have:


 * $\map f z = \dfrac 1 C$

for all $z \in \C$.

So $f$ is constant.