Problem 7 - Complex Analysis

Question

If \(f\) has in \(a \in \mathbb{C}\) a pole of order 1 , and if \(g\) is analytic in an open neighborhood of \(a\), then $$ \operatorname{Res}(f g ; a)=g(a) \operatorname{Res}(f ; a) $$

Step-by-Step Solution

Answer

The residue of \( fg \) at \( a \) is \( g(a) \operatorname{Res}(f ; a) \).

Step 1: Understand the Residue Definition

The residue of a function at a point is a complex number that describes the behavior of the function near that point. For a pole of order 1 at \( a \), the residue of \( f \) at \( a \), denoted \( \operatorname{Res}(f;a) \), is the coefficient \( b_{-1} \) in the Laurent series expansion of \( f \) around \( a \).

Step 2: Express \( f \) as a Laurent Series

Since \( f \) has a pole of order 1 at \( a \), it can be expressed as a Laurent series around \( a \): \[ f(z) = \frac{b_{-1}}{z-a} + b_0 + b_1 (z-a) + \cdots \] Here, \( b_{-1} \) is the residue of \( f \) at \( a \).

Step 3: Analyze \( g \) Near \( a \)

The function \( g \) is analytic in a neighborhood of \( a \), so it can be expanded into a Taylor series around \( a \): \[ g(z) = g(a) + g'(a)(z-a) + \frac{g''(a)}{2}(z-a)^2 + \cdots \]

Step 4: Multiply \( f \) and \( g \)

Multiply the Laurent series of \( f(z) \) with the Taylor series of \( g(z) \). The product is: \[ (f \cdot g)(z) = \left(\frac{b_{-1}}{z-a} + b_0 + b_1(z-a) + \cdots \right) \left(g(a) + g'(a)(z-a) + \cdots\right) \] Focus on the term involving \( \frac{1}{z-a} \) since the residue is the coefficient of this term.

Step 5: Identify the Residue of \( fg \) at \( a \)

The term \( \frac{b_{-1}}{z-a} \) from \( f(z) \) combined with the constant term \( g(a) \) from \( g(z) \) results in the residue at \( a \) of the product \( fg \): \[ \frac{b_{-1}g(a)}{z-a} \] Thus, the residue is \( b_{-1}g(a) \).

Step 6: Conclude the Expression for the Residue

From the above results and calculations, the residue of the product \( fg \) at \( a \) is given by: \[ \operatorname{Res}(f g ; a) = g(a) \operatorname{Res}(f ; a) \] with \( \operatorname{Res}(f ; a) = b_{-1} \) and \( g(a) \) being the value of \( g \) at \( a \).

Key Concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis that connects contour integrals with the sum of residues of a function's singular points inside the contour. Essentially, it allows us to calculate complex integrals by simplifying them into algebraic sums.

Consider a complex function with several isolated singular points inside a closed curve. The theorem states that the integral of this function around the curve is equal to (up to a factor of \( 2\pi i\)) the sum of the residues at these singular points.

• This theorem is particularly useful in scenarios where direct integration is tedious or complex.
• Residue computation is integral to solving several real-world problems in physics and engineering.

By utilizing the Residue Theorem, complex integrals involving poles can be efficiently evaluated, making it a bedrock concept in the field of complex analysis.

Laurent Series

The Laurent Series extends the idea of a Taylor Series to include terms with negative powers. This series expansion is especially useful in describing complex functions near singularities, such as poles.

For a function with a pole at a point \( a \), the Laurent series is represented as:

• \[ f(z) = \frac{b_{-1}}{z-a} + b_0 + b_1 (z-a) + \cdots \]
• The term \( \frac{b_{-1}}{z-a} \) defines the nature of the singularity and is often referred to as the principal part of the series.

Unlike Taylor Series that only accommodate analytic functions, Laurent Series accommodate nonsingular functions by accounting for negative powers. This is why it can consider functions with poles, making it essential for residue calculations.

Analytic Functions

Analytic functions in complex analysis are functions characterized by their smoothness and infinite differentiability within their domain.

These functions can be expressed through power series—functions like the Taylor series—that converge within a neighborhood. Analytic functions have several unique properties:

• Vanishing derivative: If all derivatives at a point are zero, the function itself is zero within that domain.
• Uniqueness: If two analytic functions are identical over a set of points, they are identical everywhere in that interval.

The function \( g \) in the problem is analytic at \( a \), indicating a smooth behavior with no abrupt changes or singularities in its neighborhood, making it crucial in multiplying with \( f \) in the exercise context.

Poles in Complex Functions

Poles are a specific type of singularity in the realm of complex functions. They are points where a function's value grows unbounded or undefined.

Poles can be of different orders, from simple (order 1) to higher ones. A pole of order 1, as mentioned in the exercise, implies that the function behaves like \( \frac{1}{z-a} \) near the pole. Key insights about poles include:

• They are isolated, meaning there are no other singularities infinitely close to a given pole.
• Analysis and residue calculation at poles help integrate complex functions with ease.

Understanding the role and behavior of poles is central to applying the Residue Theorem and dealing with Laurent series, offering profound insights into how complex functions evolve around singularities.