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:

$\ds \map g z = e^{\map f z}$

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

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

We have:

\(\ds \cmod {\map g z}\)

\(=\)

\(\ds e^{\map \Re {\map f z} }\)

Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part

\(\ds \)

\(\le\)

\(\ds e^{\cmod {\map \Re {\map f z} } }\)

Exponential is Strictly Increasing

\(\ds \)

\(<\)

\(\ds e^M\)

Exponential is Strictly Increasing

So $g$ is a bounded entire function.

By Liouville's Theorem, $g$ is therefore a constant function.

We therefore have, by Derivative of Constant: Complex:

$\map {g'} z = 0$

for all $z \in \C$.

That is:

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

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 $\C$.

$\blacksquare$