Problem 9 - Complex Analysis

Question

Let \(f\) be analytic in \(\dot{U}_{r}(0):=U_{r}(0) \backslash\\{0\\}, r>0\). Show: \(\operatorname{Res}\left(f^{\prime} ; 0\right)=0\).

Step-by-Step Solution

Answer

The residue of \( f' \) at zero is zero: \( \operatorname{Res}(f'; 0) = 0 \).

Step 1: Understand the Definitions

The residue of a function's derivative, denoted as \( \operatorname{Res}(f'; 0) \), is what we're trying to prove is zero. The function \( f \) is given as analytic in \( \dot{U}_{r}(0)\), meaning it's analytic in the punctured disk centered at the origin with radius \( r \) but excluding the origin.

Step 2: Use the Definition of Residue

The residue at a point \( z_0 \) of a function \( g(z) \) is the coefficient of \( (z-z_0)^{-1} \) in its Laurent series expansion about that point. For \( g(z) = f'(z) \), we are considering the series around the origin \( 0 \).

Step 3: Analyze the Analytic Nature of \( f \)

Since \( f \) is analytic in \( \dot{U}_r(0) \), it can be represented by a Laurent series which, in this region, has no negative power terms except potentially at \( z=0 \). However, since \( f \) is not specified to have a singularity at \( z = 0 \), assume it's removable there or that \( f \) is an entire function.

Step 4: Differentiate the Laurent Series of \( f \)

Assume \( f(z) = a_0 + a_1z + a_2z^2 + \ldots \) (a Taylor series since analytical at all non-zero points around origin). Differentiating term-by-term gives: \( f'(z) = a_1 + 2a_2z + 3a_3z^2 + \ldots \).

Step 5: Evaluate the Coefficient of \( z^{-1} \) in \( f'(z) \)

The residue of \( f'(z) \) at \( z=0 \) is the coefficient of \( z^{-1} \) in its Laurent series. From the differentiated series, there is no \( z^{-1} \) term. Thus, \( \operatorname{Res}(f'; 0) = 0 \).

Step 6: Conclusion

Since \( f'(z) \) has no negative power terms, especially the \( z^{-1} \) term, in the derived series representation, its residue at the origin is indeed zero.

Key Concepts

Residue Theorem

The Residue Theorem plays a significant role in complex analysis. It allows us to evaluate certain types of complex integrals by examining residues, which are specific coefficients within a function’s Laurent series. Here's how it works:

• For a given function, residues at poles correspond to the coefficients of the \( (z-z_0)^{-1} \) terms in its Laurent series expansion.
• The theorem states that the integral of a function around a closed contour can be determined by summing these residues within the enclosed region.

In our exercise, we focus on proving that the residue of the derivative of an analytic function is zero when defined in a punctured disk around the origin. This is consistent with the residue definition since the term \( z^{-1} \) does not appear in the Taylor series of the derivative. Therefore, the integral of \( f'(z) \) around a closed curve that encircles the origin results in zero due to the absence of a \( z^{-1} \) term.

Laurent Series

A Laurent series is a tool used in complex analysis to represent functions. It extends the concept of a Taylor series to include terms with negative powers, which are crucial for handling points where the function is not analytic.

• Laurent series can be expressed as \(f(z) = \sum_{n=- ext{infty}}^{ ext{infty}} a_n (z-z_0)^n\).
• For regions excluding the singularity point (a non-analytic point), the negative power terms capture the behavior of the function.

In our problem, the function \( f \) is analytic and represented by a series without negative powers, except at \( z = 0 \). Differentiating this series removes any negative power terms. Since the derivative does not produce a \( z^{-1} \) term, the coefficient corresponding to this is zero, confirming that \( \operatorname{Res}(f'; 0) = 0 \). This outcome illustrates how the Laurent series directly influences calculations of residues.

Analytic Functions

Analytic functions are at the heart of complex analysis. These functions are differentiable at each point in their domain, allowing them to be expanded into a power series.Key characteristics include:

• Their derivative is also analytic wherever the original function is analytic.
• Around any point within their domain, they can be expressed as a Taylor series, which includes only non-negative powers of \( z \).

For the provided exercise, the function \( f \) is analytic on a punctured disk around the origin, meaning it can be expressed without negative powers outside \( z = 0 \). Therefore, its derivative \( f'(z) \) retains the analytic nature and ensures no \( z^{-1} \) term appears. This confirms that \( \operatorname{Res}(f'; 0) \) is zero, emphasizing the seamless nature of derivatives within analytic domains.