Give an example of a function f(z) that is analytic in a region, bounded throughout the complex plane
but is not a constant function. Does this contradict the Liouville’s theorem? Explain.

To provide an example of a function \( f(z) \) that is analytic in a region, bounded throughout the complex plane, but is not a constant function, we can consider the following:

Let \( f(z) = e^{-z^2} \).

1. **Analyticity**:
   - The function \( e^{-z^2} \) is entire, meaning it is analytic everywhere in the complex plane. This can be verified by noting that \( e^{-z^2} \) can be represented as a power series that converges everywhere in the complex plane.

2. **Boundedness**:
   - To determine if \( f(z) = e^{-z^2} \) is bounded, we observe the behavior of \( e^{-z^2} \) as \( |z| \to \infty \). For large values of \( |z| \), the term \( z^2 \) becomes very large, and hence \( e^{-z^2} \) tends to zero. Therefore, \( |e^{-z^2}| \leq 1 \) for all \( z \) in the complex plane, implying that the function is bounded.

3. **Not Constant**:
   - The function \( f(z) = e^{-z^2} \) is clearly not a constant function, as its value changes with \( z \).

Now, to address the question of whether this contradicts Liouville’s theorem:

**Liouville's Theorem** states:
- If a function is entire (analytic everywhere in the complex plane) and bounded, then it must be constant.

In this case, \( f(z) = e^{-z^2} \) is indeed entire and bounded, but it is not a constant function. At first glance, this might seem to contradict Liouville's theorem. However, upon closer inspection, we realize that \( f(z) = e^{-z^2} \) is bounded but not bounded by a finite constant as \( z \) approaches infinity; it tends to zero. 

**Explanation**:
Liouville's theorem requires that the boundedness condition holds uniformly, meaning the function must be bounded by some finite constant \( M \) such that \( |f(z)| \leq M \) for all \( z \) in the complex plane. The function \( e^{-z^2} \) is not uniformly bounded by a single finite constant \( M \); instead, it approaches zero as \( |z| \) increases. Therefore, it does not satisfy the conditions of Liouville's theorem in the strict sense, and there is no contradiction.

In conclusion, \( f(z) = e^{-z^2} \) is an entire and bounded function but not constant, and this does not contradict Liouville's theorem because the function is not bounded by a finite constant over the entire complex plane.

Would you like further details or have any questions about this explanation?

Here are some related questions to explore further:

1. What is the formal statement of Liouville's theorem?
2. Can you prove Liouville's theorem using the Cauchy integral formula?
3. What are the implications of Liouville's theorem in complex analysis?
4. What are some other examples of entire functions that are not bounded?
5. How can one show that a function is entire?
6. What role does the maximum modulus principle play in the proof of Liouville's theorem?
7. Can a function be analytic in a region but not in the entire complex plane?
8. What are the differences between entire and meromorphic functions?

**Tip**: When dealing with complex functions, always verify the region of analyticity and boundedness to ensure accurate application of theorems like Liouville's.

Explore an example of a function f(z) that is analytic in a region, bounded throughout the complex plane, but is not constant. Learn about Liouville's theorem and how this example fits within its framework. Understand the concepts of analyticity, boundedness, and the implications for complex functions.

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