Problem 27 - Advanced Engineering Mathematics

Question

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. \(\oint_{C} \frac{\tan z}{z} d z, C:|z-1|=2\)

Step-by-Step Solution

Answer

The integral evaluates to \( 2\pi i \).

Step 1: Identify the singularities

First, identify the singularities of the function \( f(z) = \frac{\tan z}{z} \) inside the contour \( C: |z-1|=2 \). The tangent function, \( \tan z = \frac{\sin z}{\cos z} \), has poles where \( \cos z = 0 \), which occur at \( z = \frac{(2n+1)\pi}{2} \), where \( n \) is an integer. Additionally, there is a pole at \( z = 0 \) because of the \( \frac{1}{z} \) term. Check which of these are inside the contour \( |z-1| = 2 \).

Step 2: Analyze location of singularities relative to contour

The contour \( |z-1| = 2 \) is centered at \( z = 1 \) with radius 2. For the poles of \( \tan z \), convert \( z = \frac{(2n+1)\pi}{2} \) into an approximate numerical value to determine whether they lie within this contour. We need singularities such as \( z = \frac{\pi}{2} \approx 1.57 \) and \( z = \rac{3\pi}{2} \approx 4.71 \). However, only \( z = \frac{\pi}{2} \) and \( z = 0 \) (since \( |0-1| < 2 \) ) lie inside this contour.

Step 3: Calculate the residues

Calculate the residues at the singularities inside the contour: \( z = 0 \) and \( z = \frac{\pi}{2} \). For \( z = 0 \), the residue can be found by multiplying through by \( z \) and taking the limit as \( z \to 0 \), resulting in \( \lim_{z \to 0} \tan z = 0 \). For \( z = \frac{\pi}{2} \), since it is a simple pole of \( \tan z \), compute the residue using the formula for poles \( \text{Residue of } \tan z = \lim_{z \to \frac{\pi}{2}} (z - \frac{\pi}{2})\tan z = 1 \). In fact, due to symmetry and periodicity, the residue of \( \tan z \) at any \( \frac{(2n+1)\pi}{2} \) is 1.

Step 4: Apply Cauchy's residue theorem

Apply Cauchy's residue theorem: \( \oint_{C} f(z) \, dz = 2\pi i \times \sum \text{Residues of } f(z) \text{ inside } C \). The sum of residues inside the contour \( C \) is \( 0 + 1 = 1 \). Thus, the integral evaluates to \( 2\pi i \times 1 = 2\pi i \).

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics that deals with functions of a complex variable. It combines real and imaginary numbers to analyze the behavior and properties of complex functions. A complex number is expressed in the form \( z = x + yi \), where \( x \) and \( y \) are real numbers and \( i \) is the imaginary unit, defined by \( i^2 = -1 \).

• Complex functions are functions that take complex numbers as inputs and output complex numbers.
• The study of these functions includes concepts like limits, continuity, and differentiation.
• Complex analysis also explores the unique properties of analytic functions, which are functions that are differentiable at every point in their domain.

Understanding complex analysis is crucial for topics such as contour integration and the evaluation of integrals involving complex functions.

Contour Integration

Contour integration is a method used in complex analysis to evaluate integrals of complex functions along a specific path—or contour—in the complex plane. This technique is particularly useful for functions with singularities, as it allows the integral to be calculated without actually integrating over the singularities themselves.

• A contour is any continuous curve in the complex plane.
• Choosing the appropriate contour is key to simplifying the calculation of complex integrals.
• Contour integrals can be closed, meaning they form a loop that returns to the starting point.

By applying principles like Cauchy's residue theorem, complex integrals can often be solved in a straightforward manner, even when dealing with difficult to integrate functions.

Singularities

Singularities are points at which a complex function is not defined or not analytic. In the context of functions of a complex variable, singularities take on special importance, as they often determine the behavior of a function around those points.

• Poles are a common type of singularity, where a function approaches infinity.
• Essential singularities are points where the behavior of the function is more erratic, with no limit as you approach the point.
• Branch points are where a function cannot return to a single value, leading to multi-valued results.

Identifying the type and location of singularities helps in planning the contour for integration and applying theorems like the residue theorem.

Residue Calculation

Residue calculation involves determining the residues of a function at its singularities. This is an essential step in applying Cauchy's residue theorem for evaluating complex integrals.

• The residue of a function at a pole is the coefficient of \( \frac{1}{z-a} \) when the function is expressed in its Laurent series around the pole \( a \).
• For simple poles, the residue can be found directly by taking limits or using residue formulas.
• Residue calculation simplifies the evaluation of contour integrals by reducing the problem to finding sums of residues inside the contour.

In the exercise example, we calculated the residues at the singularities \( z = 0 \) and \( z = \frac{\pi}{2} \), which contributed directly to the evaluation of the integral.