Why a Certain Entire Function Must Be Constant

If you’ve studied some complex analysis, you might have encountered Liouville’s Theorem or the Maximum Modulus Principle. These theorems help us understand the behavior of analytic functions, and today we’ll explore a fascinating result about entire functions.

We’re going to prove the following statement:

If f(z)f(z) is an entire function on the complex plane, and Re⁡(f(z))\operatorname{Re}(f(z)) (the real part of f(z)f(z)) is bounded above, then f(z)f(z) must be constant.

Let’s break this proof into digestible steps so that you can follow the logic easily. We’ll also highlight which theorems or concepts are being used and why they are crucial.

Step 1: Understanding the Problem

An entire function is a complex function that is holomorphic (analytic) everywhere on the complex plane. For f(z)f(z), we’re given that Re⁡(f(z))\operatorname{Re}(f(z)) is bounded above, meaning there exists a real number MM such that:

Re⁡(f(z))≤Mfor all z∈C.\operatorname{Re}(f(z)) \leq M \quad \text{for all } z \in \mathbb{C}.

The goal is to show that f(z)f(z) must be constant. This conclusion aligns with Liouville's Theorem, which states that a bounded entire function is constant, and with the Maximum Modulus Principle, which we’ll use indirectly.

Step 2: Define a New Function

To leverage existing results, we’ll consider the exponential function. Define:

F(z)=e−f(z).F(z) = e^{-f(z)}.

Why this choice? The exponential function is never zero, and its growth behavior simplifies analysis when dealing with bounded real parts.

Step 3: Show That F(z)F(z) is Entire

Since f(z)f(z) is entire, the composition e−f(z)e^{-f(z)} is also entire. This follows from the fact that the exponential function is entire and the composition of two entire functions is entire. Therefore, F(z)F(z) is holomorphic on C\mathbb{C}.

Step 4: Analyze ∣F(z)∣|F(z)|

Next, let’s compute the modulus of F(z)F(z):

∣F(z)∣=∣e−f(z)∣=e−Re⁡(f(z)).|F(z)| = \left| e^{-f(z)} \right| = e^{-\operatorname{Re}(f(z))}.

Since Re⁡(f(z))≤M\operatorname{Re}(f(z)) \leq M for all zz, it follows that:

∣F(z)∣=e−Re⁡(f(z))≥e−M.|F(z)| = e^{-\operatorname{Re}(f(z))} \geq e^{-M}.

Thus, ∣F(z)∣|F(z)| is bounded below by e−M>0e^{-M} > 0.

On the other hand, because Re⁡(f(z))\operatorname{Re}(f(z)) is bounded above, ∣F(z)∣|F(z)| is also bounded above. Therefore, ∣F(z)∣|F(z)| is bounded on C\mathbb{C}.

Step 5: Apply Liouville's Theorem

By Liouville’s Theorem, any bounded entire function must be constant. Since F(z)F(z) is bounded and entire, F(z)F(z) is constant. Let’s denote this constant by cc:

F(z)=cfor all z∈C.F(z) = c \quad \text{for all } z \in \mathbb{C}.

Step 6: Conclude That f(z)f(z) is Constant

Recall that F(z)=e−f(z)F(z) = e^{-f(z)}. If F(z)=cF(z) = c, then:

e−f(z)=c⇒f(z)=−ln⁡(c),e^{-f(z)} = c \quad \Rightarrow \quad f(z) = -\ln(c),

where ln⁡(c)\ln(c) is a constant (since cc is constant and nonzero).

Thus, f(z)f(z) is constant.

Step 7: Final Thoughts

This result is a specific application of Liouville’s Theorem and highlights the power of boundedness in analyzing entire functions. If you’re exploring more about entire functions, take time to appreciate how results like the Maximum Modulus Principle and properties of exponential functions interplay in these proofs.

Thanks for following along! If you have questions or want to dive deeper, feel free to ask in the comments below.

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