Problem 27 - Advanced Engineering Mathematics

Question

In Problems 21-32, use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C}^{\tan z} d z, C:|z-1|=2 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Singularity

The function we are trying to integrate is \(f(z) = \tan(z)\). The function \(\tan(z)\) is equal to \(\frac{\sin(z)}{\cos(z)}\), which has singularities where \(\cos(z) = 0\), i.e., at points \(z = \frac{\pi}{2} + n\pi\), where \(n\) is an integer.

Step 2: Check Singularity Inside Contour

The contour \(|z-1| = 2\) is a circle centered at 1 with radius 2. We need to check which singularities of \(f(z)\) lie inside this contour. To do so, check if \(|z-1| < 2\) is satisfied. It includes singularities at \(z = \frac{\pi}{2}\) and \(z = -\frac{\pi}{2} + \pi \approx 2.14\). Thus, \(z = \frac{\pi}{2}\) is the only singularity inside the contour.

Step 3: Find the Residue

Since \(\tan(z)\) has a simple pole at \(z = \frac{\pi}{2}\), we compute the residue at this point. The residue of \(\tan(z)\) at a simple pole \(z = a\) is given by \(\text{Res}(\tan(z), z=a) = \lim_{z \to a} (z-a)\tan(z)\). For \(a = \frac{\pi}{2}\), compute \(\lim_{z \to \frac{\pi}{2}} (z-\frac{\pi}{2})\frac{\sin(z)}{\cos(z)}\). The residue turns out to be 1.

Step 4: Apply the Residue Theorem

By Cauchy’s residue theorem, the integral of a function around a closed contour is \(2\pi i\) times the sum of the residues of the function's singularities inside the contour. The integral is given as \(\oint_{C} \tan(z) \, dz = 2\pi i \cdot \text{Res}(\tan(z), z=\frac{\pi}{2}) = 2\pi i\).

Key Concepts

Complex Integration

Complex integration is a fundamental concept in complex analysis, allowing us to evaluate integrals over complex-valued functions. In contrast to real integration, complex integration takes place on paths in the complex plane, often known as contours. These paths can be straight lines, circles, or more complex shapes depending on the scenario.

In the context of Cauchy's residue theorem, integration is performed around closed contours, which can be considered as loops or circuits in the plane. One reason why contour integration is interesting is that it connects the integral of a function in the complex plane to properties of the function such as singularities and residues. This forms a core part of complex integration, enabling powerful applications in both mathematics and physics.

When integrating over, say, the function \(\tan(z)\), the path is defined by the contour \(C:|z-1|=2\). This denotes a circle with a center at \(z=1\) and radius 2. Understanding contour paths like this is crucial in computing path integrals and applying theorems such as Cauchy's residue theorem.

Singularities

Singularities in complex analysis are points at which a given function ceases to be well-behaved in some manner—typically by not being differentiable or by taking an infinite value. For the function \(f(z) = \tan(z)\), singularities occur where the function is undefined, which in this case is where \(\cos(z) = 0\).

These points are specifically defined by the equation \(z = \frac{\pi}{2} + n\pi\), where \(n\) is an integer. Understanding the locations of these singularities is essential because only the singularities within the specified contour contribute residues that impact the integral's value when applying Cauchy's residue theorem.

In the given problem, the contour \( |z-1| = 2 \) encloses certain singularities. We find \(z = \frac{\pi}{2}\) is inside the contour, as it gives us a valid point for calculating residues, while others, such as \(z = \frac{5\pi}{2}\), are outside.

Residue Calculation

Residue calculation is a method used in complex analysis to simplify the integration of complex functions. It involves identifying and calculating the residues of the function's singularities within a given contour. A residue at a point is essentially the coefficient of \(\frac{1}{z - a}\) in the Laurent series expansion of the function around the singularity.

For a simple pole, as seen in the function \(\tan(z)\) at \(z = \frac{\pi}{2}\), the residue can be found using the formula:

• \[ \text{Res}(\tan(z), z=a) = \lim_{z \to a} (z-a)\tan(z) \]

Applying this for our specific point \(a = \frac{\pi}{2}\), we calculate the residue to be 1. This means that when combined with Cauchy's residue theorem, the integral becomes a straightforward calculation involving multiplying this residue by \(2\pi i\).

• \( \oint_{C} \tan(z) \, dz = 2\pi i \times 1 = 2\pi i \)

This simplified process demonstrates the power of residue calculation in evaluating complex integrals efficiently.