Problem 3 - Advanced Engineering Mathematics

Question

In Problems 1-6, use a Laurent series to find the indicated residue. $$ f(z)=\frac{4 z-6}{z(2-z)} ; \operatorname{Res}(f(z), 0) $$

Step-by-Step Solution

Answer

The residue of \( f(z) \) at \( z = 0 \) is \( -3 \).

Step 1: Identify the poles

The function given is \( f(z) = \frac{4z - 6}{z(2-z)} \). We need to find the poles by setting the denominator to zero: \( z(2-z) = 0 \). This gives us the poles at \( z = 0 \) and \( z = 2 \).

Step 2: Find the residue at the pole

Since we are interested in the residue at \( z = 0 \), we rewrite the function to focus on this pole. This function can be expressed as \( \frac{4z - 6}{2} \cdot \frac{1}{z} + \frac{4z - 6}{2} \cdot \frac{1}{2-z} \). The first term can be simplified to \( \frac{4z - 6}{2z} = 2 - \frac{3}{z} \).

Step 3: Extract the Laurent series term

For the first term \( 2 - \frac{3}{z} \), the constant and higher terms do not affect the residue, only the \( \frac{-3}{z} \) term does. For the second term, \( \frac{4z - 6}{2(2-z)} \), we check the series centered at \( z = 0 \). Since it doesn't contribute any residue at this pole, we focus only on \( \frac{-3}{z} \).

Step 4: Determine the residue

The residue of \( f(z) \) at \( z = 0 \) is the coefficient of \( \frac{1}{z} \) in the Laurent series expansion around \( z = 0 \), which is \( -3 \). Thus, \( \operatorname{Res}(f(z), 0) = -3 \).

Key Concepts

Residue theorem

In complex analysis, the residue theorem is a powerful tool used to evaluate contour integrals. The main attraction of the residue theorem is its ability to simplify complex integration tasks involving meromorphic functions. A meromorphic function is one that is holomorphic (complex differentiable) on an open set except for a set of isolated points known as poles.

The residue theorem states that if a function is holomorphic in a given region except for a finite number of isolated singularities inside a closed contour, then the integral of the function around that contour is equal to \( 2\pi i \) times the sum of residues of the function inside the contour. This means:

• If \( f(z) \) has poles inside the contour \( C \), then \( \int_C f(z) \; dz = 2\pi i \times \sum \text{Residues of } f(z) \text{ inside } C \).

Understanding residues at these points becomes essential for calculating these integrals, especially along paths that would be difficult to integrate directly. Therefore, mastering the calculation of residues is crucial for efficiently using the residue theorem.

Complex analysis

Complex analysis is the study of complex-valued functions of a complex variable. This field is rich with tools that provide deeper insights into properties such as differentiability and integrability when extended into the complex plane.

Key concepts in complex analysis include:

• Complex Differentiability: Functions that are differentiable in the complex sense are called holomorphic. These functions have derivatives that are independent of direction in the complex plane.
• Analytic Functions: These are functions that are locally given by a convergent power series.
• Contour Integrals: These are integrals taken over a path in the complex plane, providing a way to accumulate values around paths of interest.

Complex analysis extends real analysis into the complex plane, offering powerful techniques like the Cauchy-Riemann equations and Cauchy's integral theorem, which simplify solving complex problems. This framework facilitates understanding the behavior of functions near singularities through tools like Laurent series and residues.

Poles in complex functions

Poles in complex functions are specific types of singularities where a function's value approaches infinity as the variable approaches a certain point.

A function \( f(z) \) has a pole at point \( z = a \) if its behavior near \( a \) can be described as:

• \( f(z) = \frac{g(z)}{(z-a)^m} \) where \( g(z) \) is analytic and non-zero at \( z = a \), and \( m \) is the order of the pole.

The order of the pole indicates how the function diverges as it approaches the pole. Pole of order 1 is called a simple pole, the most common occurrence in problems.

Poles play a significant role in the residue theorem as they are the points at which residues are calculated. Analyzing poles involves finding where the denominator of a function equals zero, as seen in the solution process with the function \( \frac{4z - 6}{z(2-z)} \). Identifying and classifying these poles help determine the function’s behavior and are essential in evaluating boundary integrals within the complex plane.