Problem 10 - Complex Analysis for Mathematics and Engineering

Question

Find \(\int_{C} z^{-2}\left(z^{2}-16\right)^{-1} \exp z d z\) along the following contours: (a) The circle \(C_{1}^{+}(0)\). (b) The circle \(C_{1}^{+}(4)\).

Step-by-Step Solution

Answer

Both integrals are zero by Cauchy's Integral Theorem.

Step 1: Identify the Function Singularity

The function we are integrating, \(f(z) = \frac{e^z}{z^2(z^2 - 16)}\), has singularities at \(z = 0\), \(z = 4\), and \(z = -4\). The singularity at \(z = 0\) is a pole of order 2, while \(z = 4\) and \(z = -4\) are simple poles.

Step 2: Apply Cauchy's Integral Theorem for \(C_1^+(0)\)

For the contour \(C_1^+(0)\), a circle centered at the origin with radius 1, the pole at \(z = 0\) lies inside the contour. The poles at \(z = 4\) and \(z = -4\) are outside the contour. According to Cauchy's Integral Theorem, if a function is analytic on and inside a closed contour except at a finite number of singularities all outside the contour, the integral is zero. Therefore, the integral along \(C_1^+(0)\) is zero.

Step 3: Apply Cauchy's Integral Theorem for \(C_1^+(4)\)

For the contour \(C_1^+(4)\), a circle centered at \(z = 4\) with radius 1, the only singularity within the contour is at \(z = 4\). The poles at \(z = 0\) and \(z = -4\) are outside this contour. Offsetting our center by \(z - 4\), so the contour integral simplifies using Cauchy's Integral Formula. Since the contour does not enclose any singularity other than at the center, the integral evaluates to zero, similar to the first case as no non-removable singularity impacts the integral.

Key Concepts

Cauchy's Integral Theorem

Cauchy's Integral Theorem is a fundamental result in complex analysis. It states that if a function is analytic (complex differentiable) on and within a closed contour in a complex plane, then the integral of the function along the contour is zero.
This theorem forms the basis of many other results in complex analysis, including contour integration and residue theory.
In essence, if a contour does not enclose any singularities, or it does, but in such a way that the function remains analytic, the contour integral will result in zero.
**Why is it Important?**

• Provides a tool to evaluate complex integrals.
• Helps in understanding the behavior of analytic functions.
• Forms the basis for more complex theorems and methods, such as Cauchy's Integral Formula and the concept of residues.

In the given exercise, this theorem is used to determine that for both contours, since the functions around those contours are analytic (with exceptions outside the contour), the integral of the function becomes zero.

Singularities

In complex analysis, singularities are points where a complex function is not analytic. These can be places where a function goes to infinity or is not well-defined.
**Types of Singularities**

• **Poles**: Points where a function approaches infinity as the input nears the pole. The order of a pole is the number of times a factor remains in the denominator.
• **Essential Singularities**: Points where the function behaves erratically.
• **Removable Singularities**: Points that can be "removed" by redefining the function to be analytic everywhere.

In the exercise, the function has singularities at \(z = 0\), \(z = 4\), and \(z = -4\):

• \(z = 0\): a pole of order 2.
• \(z = 4\) and \(z = -4\): simple poles (order 1).

By identifying these singularities, it helps to determine how Cauchy's Theorem applies and whether the integral equals zero.

Contour Integration

Contour integration is a technique in complex analysis used to calculate integrals of complex-valued functions along contours in the complex plane.
Essentially, a contour is a path along which we integrate a complex function. The contour can affect the value of the integral, especially if it encloses singularities.
**Key Points to Remember**

• Contour integration is directly influenced by Cauchy's Integral Theorem.
• Only singularities within the contour usually impact the result of the integral.
• The shape or path of the contour can be arbitrarily complex, but its fundamental nature (closing on itself) dictates the result.

In the original exercise, two contours, \(C_1^{+}(0)\) and \(C_1^{+}(4)\), were considered. The contour around \(z = 0\) and \(z = 4\) indicates how the chosen path influences the integral's evaluation with respect to the singularities enclosed, resulting in both integrals becoming zero due to the application of Cauchy's Theorem.