Problem 29 - A First Course in Complex Analysis with Applications

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \(6i\).

Step 1: Analyze the Function

The integrand is given as \( \cot(\pi z) \), which is \( \frac{\cos(\pi z)}{\sin(\pi z)} \). This function has poles where the sine function is zero, i.e., at \( z = n \) for \( n \in \mathbb{Z} \).

Step 2: Identify the Poles within the Contour

The contour \( C \) is a rectangle defined by \( x=\frac{1}{2}, x=\pi, y=-1, y=1 \). Therefore, it encompasses points in the complex plane with \( \frac{1}{2} \leq \text{Re}(z) \leq \pi \) and \( -1 \leq \text{Im}(z) \leq 1 \). The integer values of \( z \) within this region are \( z = 1, 2, 3 \).

Step 3: Calculate the Residues at the Poles

The residues of \( \cot(\pi z) \) at its poles, \( z = n \), are given by \( \text{Res}(\cot(\pi z), n) = \lim_{z \to n} (z-n) \cot(\pi z) = \frac{1}{\pi} \) because the residue of \( \cot \) at any integer \( n \) is \( \frac{1}{\pi} \). Thus, for \( z = 1, 2, 3 \), the residues are \( \frac{1}{\pi} \) each.

Step 4: Apply Cauchy's Residue Theorem

Cauchy's residue theorem states that \( \oint_{C} f(z) \, dz = 2\pi i \sum \text{Res}(f, a_i) \), where \( a_i \) are the poles inside \( C \). In this case, there are three poles, each with a residue of \( \frac{1}{\pi} \). Therefore, the integral becomes \[ \oint_{C} \cot(\pi z) \, dz = 2\pi i \left( \frac{1}{\pi} + \frac{1}{\pi} + \frac{1}{\pi} \right) = 2\pi i \cdot \frac{3}{\pi} = 6i. \]

Step 5: Conclusion

The value of the integral, based on the application of the Residue theorem, is \( 6i \).

Key Concepts

Cauchy's Residue Theorem

Cauchy's Residue Theorem is a powerful tool in complex analysis used to evaluate integrals of certain functions over closed contours in the complex plane. At its core, the theorem states that if a function \( f(z) \) is analytic inside and on some simple closed contour \( C \), except for a finite number of singularities inside \( C \), the integral of \( f(z) \) around \( C \) can be computed using the residues of these singularities.

Let's summarize the steps involved when using Cauchy's Residue Theorem to evaluate an integral:

• Identify the poles of the function within the contour. Poles are points where the function becomes unbounded.
• Compute the residue of the function at each pole. A residue is essentially the coefficient that gives the behavior of the function near the pole.
• Apply the theorem: \( \oint_{C} f(z) \, dz = 2\pi i \sum_{a_i} \text{Res}(f, a_i) \), where \( a_i \) are the poles, and \( \text{Res} \) indicates the residue at each pole.

The beauty of this theorem lies in its ability to reduce complex contour integrals into simple calculations involving the locations and nature of singularities. This transformation is extremely valuable, especially for functions that are difficult to integrate directly.

Contour Integration

Contour integration is a technique of evaluating complex integrals along specified paths, or contours, in the complex plane. This method is particularly useful in complex analysis because it considers both the magnitude and direction of the path traced out in the plane.

When setting up a contour integration, consider these key points:

• Choose a Contour: The choice of contour \( C \) depends on the region in which you are interested. Common choices include circles, squares, and other closed shapes. These paths allow the use of the powerful Cauchy's Residue Theorem.
  
• Ensure Analyticity: A function must be analytic (differentiable and continuous) on and inside the contour, except at some isolated singularities (poles), to apply the residue theorem effectively.
• Integrate Along the Path: Use the chosen contour to evaluate the integral by summing contributions from different parts of the contour, often breaking it into segments if it helps in manual evaluation.

Contour integration not only aids in evaluating complex integrals but also in solving real integrals through the application of certain theorems that connect real and complex analysis, like the residue theorem. This methodology is essential for understanding more intricate aspects of complex analysis.

Residues in Complex Analysis

Residues, in the context of complex analysis, provide crucial information about the behavior of functions near their singularities. They are instrumental in the application of Cauchy's Residue Theorem to compute complex integrals.

Here's what you need to know about residues:

• Definition: A residue is essentially the coefficient of the \( \frac{1}{(z-a)} \) term in the Laurent series expansion of a function around a singularity \( a \).
  
• Calculating Residues: For simple poles, the residue can often be found directly by evaluating \( \lim_{z \to a} (z-a)f(z) \). For higher-order poles, differentiation might be required to find the residue.
  
• Use in Integrals: Residues summarize the "effect" of singularities when integrating a function along a closed contour. They show how much each singularity contributes to the total value of the contour integral.
  

Understanding residues is a fundamental step in mastering complex analysis, as these values significantly simplify the evaluation of many complex integrals, converting potentially difficult problems into straightforward computations.