Entire Function with Bounded Real Part is Constant

Theorem
Let $f : \C \to \C$ be an entire function.

Let the real part of $f$ be bounded.

That is, there exists a positive real number $M$ such that:


 * $\cmod {\map \Re {\map f z} } < M$

for all $z \in \C$, where $\map \Re {\map f z}$ denotes the real part of $\map f z$.

Then $f$ is constant.

Proof
Let $g : \C \to \C$ be a complex function with:


 * $\displaystyle \map g z = e^{\map f z}$

By Derivative of Complex Composite Function, $g$ is entire with derivative:


 * $\displaystyle \map {g'} z = \map {f'} z e^{\map f z}$

We have:

So $g$ is a bounded entire function, and is therefore constant by Liouville's Theorem.

We therefore have:


 * $\map {f'} z e^{\map f z} = 0$

for all $z \in \C$.

Since the exponential function is non-zero, we must have:


 * $\map {f'} z = 0$

for all $z \in \C$.

From Zero Derivative implies Constant Complex Function, we then have that $f$ is constant on $\C$.