Problem 7
Question
Let \(f, g: \mathbb{C} \rightarrow \mathbb{C}\) be two analytic functions. We assume $$ f(g(z))=0 \text { for all } z \in \mathbb{C} $$ Show: If \(g\) is non-constant, then \(f \equiv 0\)
Step-by-Step Solution
Verified Answer
If \( g \) is non-constant, then \( f \equiv 0 \) by analytic properties and identity theorem.
1Step 1: Understanding the Given
The problem tells us that the composition of two complex analytic functions, \( f \) and \( g \), results in zero for all inputs \( z \) in the complex plane. We need to show that the only way this can be true, without \( g \) being constant, is if \( f \) is identically zero.
2Step 2: Analyzing the Functions
Since \( f(g(z)) = 0 \) for all \( z \in \mathbb{C} \), \( g(z) \) must only take on values \( w \) such that \( f(w) = 0 \). If \( g \) is non-constant, it does not map the complex plane \( \mathbb{C} \) to a single point.
3Step 3: Using Liouville's Theorem
Since \( g \) is not constant and analytic, its image is an unbounded subset of \( \mathbb{C} \). If \( f \) is not zero, it would only vanish at isolated points unless identically zero. However, \( f \) must vanish for an entire set in \( \mathbb{C} \) due to \( g \), which means by analytic property, \( f \) must be identically zero.
4Step 4: Conclusion by Identity Theorem
The identity theorem for analytic functions states that if an analytic function vanishes on a set that has a limit point, then it must be identically zero. So, the only way \( f(g(z)) = 0 \) for all \( z \) when \( g \) is not constant is if \( f \equiv 0 \).
Key Concepts
Analytic FunctionsLiouville's TheoremIdentity Theorem for Analytic Functions
Analytic Functions
An analytic function is a special type of function that is differentiable at every point in its domain. In complex analysis, these are not just differentiable in the sense of calculus, but infinitely differentiable, and they can be represented as power series. This gives them very nice and predictable behavior.
When we say a function \( f : \mathbb{C} \rightarrow \mathbb{C} \) is analytic, it means that for every point \( z_0 \) in its domain, there exists a power series in the complex plane that converges to \( f \) near \( z_0 \).
When we say a function \( f : \mathbb{C} \rightarrow \mathbb{C} \) is analytic, it means that for every point \( z_0 \) in its domain, there exists a power series in the complex plane that converges to \( f \) near \( z_0 \).
- Analytic functions are always smooth and continuous.
- They can be added, multiplied, and composed while remaining analytic.
- Their power series expansion makes them extremely powerful for computations and proofs.
Liouville's Theorem
Liouville's Theorem is a foundational result in complex analysis that states if a function is both entire (analytic across the whole complex plane) and bounded, then it must be constant. This means that any non-constant entire function must be unbounded.
This theorem is essential in proving other results, such as the fact that there are no non-constant analytic functions that are bounded, reshaping our understanding of how functions behave across the entire complex plane. For example:
This theorem is essential in proving other results, such as the fact that there are no non-constant analytic functions that are bounded, reshaping our understanding of how functions behave across the entire complex plane. For example:
- If a complex function like \( g(z) \) is non-constant and entire, it cannot be bounded across the whole complex plane.
- The theorem helps us assert that non-constant analytic functions must take a wide range of values, displaying a crucial property of function behavior.
Identity Theorem for Analytic Functions
The Identity Theorem is a powerful tool that tells us about when two analytic functions must be identical. It asserts that if two analytic functions agree on a set of points with a limit point in the complex plane, then they must be identical across their whole domain.
This theorem is especially important in complex analysis because it conveys how tightly constrained analytic functions are:
This theorem is especially important in complex analysis because it conveys how tightly constrained analytic functions are:
- If \( f(g(z)) = 0 \) for a non-constant analytic function \( g \), then \( f \) must be zero everywhere if it vanishes on a set with a limit point.
- This theorem underpins arguments like the one in the exercise, ensuring that if one function replicates conditions over a significant set of points, it dictates global behavior.
Other exercises in this chapter
Problem 6
Compute the following integrals: (a) \(I:=\oint_{|\zeta|=2} \frac{1}{(\zeta-3)\left(\zeta^{13}-1\right)} d \zeta\) (b) \(I:=\oint_{|\zeta|=10} \frac{\zeta^{3}}{
View solution Problem 7
In which domain \(D \subset \mathbb{C}\) defines the following series $$ \sum_{n=1}^{\infty} \frac{\sin (n z)}{2^{n}} $$ an analytic function? \((\) Answer: \(D
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For \(\nu \in \mathbb{Z}\), and \(w \in \mathbb{C}\) let \(\mathcal{J}_{\nu}(w)\) be the coefficient of \(z^{\nu}\) in the LAURENT power series expansion of $$
View solution Problem 7
If \(f\) has at \(a \in \mathbb{C}\) a pole of order 1 , and if \(g\) is analytic in an open neighborhood of \(a\), then $$ \operatorname{Res}(f g ; a)=g(a) \op
View solution