Problem 29 - Advanced Engineering Mathematics

Question

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. \(\oint_{C} \cot \pi z d z, C\) is the rectangle defined by \(x=\frac{1}{2}, x=\pi, y=-1\), \(y=1\)

Step-by-Step Solution

Answer

The integral is \( 2i \).

Step 1: Understanding the Problem Statement

We are tasked with finding the integral \( \oint_{C} \cot \pi z \, dz \) where \( C \) is a rectangular contour defined by the vertices \( x = \frac{1}{2}, \ x = \pi, \ y = -1, \ y = 1 \). The contour is traversing counterclockwise.

Step 2: Identifying Singularities

The function \( \cot \pi z \) has singularities where \( \sin \pi z = 0 \). This happens at \( z = 0, \pm 1, \pm 2, \ldots \) or generally \( z = n \) for any integer \( n \). Check which integers fall within the contour's range defined by \( \frac{1}{2} \leq x \leq \pi \) and \( -1 \leq y \leq 1 \). Only \( z = 1 \) is within the contour \( C \).

Step 3: Applying Cauchy's Residue Theorem

According to Cauchy's residue theorem, the value of the integral is \( 2 \pi i \) times the sum of the residues of the singularities inside the contour. Since the singularity inside the contour is \( z = 1 \), we need to find the residue of \( \cot \pi z \) at \( z = 1 \).

Step 4: Calculating the Residue

The residue of \( \cot \pi z \) at any integer \( n \) can be found using the limit \( \lim_{z \to n} (z - n) \cot \pi z = \lim_{z \to n} \frac{z - n}{\tan \pi z} = \frac{1}{\pi} \). Thus, the residue of \( \cot \pi z \) at \( z = 1 \) is \( \frac{1}{\pi} \).

Step 5: Computing the Integral

Plugging the residue back into Cauchy's residue theorem gives: \( \oint_{C} \cot \pi z \, dz = 2 \pi i \times \frac{1}{\pi} = 2i \).

Key Concepts

Complex Integration

Complex integration is similar in concept to real integration but occurs over the complex plane. It involves integrating complex-valued functions over contours in the complex plane, often represented as paths or curves. In our case, we are integrating the function \( \cot \pi z \) over a specified contour. Integrating complex functions follows the same principles as real integration, such as respecting the direction of the contour path. The contour or path, denoted by \( C \), is traversed in a particular direction - counterclockwise, for this exercise.

• The direction is important because it can affect the sign of the integral's result.
• When working with complex integration, we also encounter unique considerations, like singularities.

This is where Cauchy's Residue Theorem plays a crucial role by simplifying the integration process over contours.

Singularities

In complex analysis, singularities are points where a function exhibits a form of behavior that's not typical of most numbers, often leading to undefined or infinitely large values. For the function \( \cot \pi z \), singularities occur at points where \( \sin \pi z = 0 \), which are integers \( z = n \). Essentially, these are points that make the denominator zero in the function \( \cot \pi z = \frac{\cos \pi z}{\sin \pi z} \).

• Identifying singularities is the first step in solving integrals involving complex functions.
• We need to determine if any singularities are inside our chosen contour.
• Only singularities inside the contour affect the integral's result.

In this specific exercise, the contour is described by the rectangle with boundaries given by \( x = \frac{1}{2}, \ x = \pi, \ y = -1, \ y = 1 \), which include \( z = 1 \) within it.

Residue Calculation

Cauchy's Residue Theorem provides a powerful tool in evaluating complex integrals, especially when dealing with functions that have singularities. Instead of performing a direct integration, the theorem allows us to focus on calculating the residues of the singularities within the contour.

• Residue is essentially a coefficient of \( \frac{1}{z - z_0} \) in the function's Laurent series expansion around the singularity \( z_0 \).
• It can also be calculated using specific formulas or limits, as applied in this scenario.
• For \( \cot \pi z \), the residue at an integer \( n \) is \( \frac{1}{\pi} \).

By knowing the residue at \( z = 1 \), we can directly apply Cauchy's Residue Theorem to find the integral's value by simply multiplying by \( 2 \pi i \). Thus, the result of the integral \( \oint_{C} \cot \pi z \, dz \) is \( 2i \), showcasing how residues simplify complex integrations significantly.