Why a Certain Entire Function Must Be Constant When Its Imaginary Part is Bounded

In this blog entry, we’ll explore a fascinating result from complex analysis, similar to the one we discussed previously about the real part of an entire function. This time, we’ll focus on the imaginary part and show that if it is bounded, the function must be constant.

Here’s the statement we aim to prove:

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

As before, we’ll break the proof into manageable steps and reference key theorems and ideas to make it easy to follow, even for those with just a basic background in complex analysis.

---

Step 1: Understanding the Problem

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

$$\operatorname{Im}(f(z)) \leq M \quad \text{for all } z \in \mathbb{C}.$$

Our task is to show that this condition implies $f(z)$ is constant. The result follows from Liouville's Theorem and the harmonic properties of $\operatorname{Im}(f(z))$.

---

Step 2: Define a New Function

To take advantage of boundedness, define the following function:

$$G(z) = e^{i f(z)}.$$

Why this choice? The exponential function ensures that both the real and imaginary parts of $f(z)$ contribute in a manageable way, while its modulus can be analyzed to reveal important properties.

---

Step 3: Show That $G(z)$ is Entire

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

---

Step 4: Analyze $|G(z)|$

The modulus of $G(z)$ is given by:

$$|G(z)| = \left| e^{i f(z)} \right| = e^{-\operatorname{Im}(f(z))}.$$

Since $\operatorname{Im}(f(z)) \leq M$ for all $z$, it follows that:

$$|G(z)| = e^{-\operatorname{Im}(f(z))} \geq e^{-M}.$$

Thus, $|G(z)|$ is bounded below by $e^{-M} > 0$. Furthermore, because $\operatorname{Im}(f(z))$ is bounded above, $|G(z)|$ is also bounded above. Therefore, $|G(z)|$ is bounded on $\mathbb{C}$.

---

Step 5: Apply Liouville's Theorem

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

$$G(z) = c \quad \text{for all } z \in \mathbb{C}.$$

---

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

Recall that $G(z) = e^{i f(z)}$. If $G(z) = c$, then:

$$e^{i f(z)} = c.$$

Taking the logarithm (using the complex logarithm), we have:

$$i f(z) = \ln(c),$$

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

Dividing through by $i$, we find:

$$f(z) = -i \ln(c).$$

Thus, $f(z)$ is constant.

---

Step 7: Final Thoughts

This result complements our earlier proof about the real part of an entire function. It demonstrates how boundedness of either the real or imaginary part of an entire function forces the function to be constant. This is a testament to the rigidity of analytic functions over the complex plane.

Thanks for reading! If you have questions or comments, feel free to leave them below. Let’s keep exploring the beauty of complex analysis together!