Problem 5 - Advanced Engineering Mathematics

Question

Prove that \(\oint_{c} f(z) d z=0\), where \(f\) is the given function and \(C\) is the unit circle \(|z|=1\). \(f(z)=\frac{\sin z}{\left(z^{2}-25\right)\left(z^{2}+9\right)}\)

Step-by-Step Solution

Answer

Since all singularities are outside the unit circle, \( \oint_{C} f(z) \, dz = 0 \) by Cauchy's Integral Theorem.

Step 1: Review Cauchy's Integral Theorem

Cauchy's Integral Theorem states that if a function \( f(z) \) is analytic (i.e., complex differentiable) everywhere inside and on some closed contour \( C \), then \( \oint_C f(z) \, dz = 0 \). We need to determine if \( f(z) \) is analytic on \( C \), which is \(|z| = 1\).

Step 2: Identify Singularities of f(z)

The function \( f(z) = \frac{\sin z}{(z^2 - 25)(z^2 + 9)} \) has singularities where the denominator is zero: \( z^2 - 25 = 0 \) yields \( z = \pm 5 \); \( z^2 + 9 = 0 \) yields \( z = \pm 3i \). These points \( z = \pm 5, \pm 3i \) are the poles of the function.

Step 3: Analyze Location of Singularities in Relation to the Unit Circle

The unit circle \(|z| = 1\) is around the origin with radius 1. The singularities \( z = \pm 5 \) and \( z = \pm 3i \) do not lie inside or on the unit circle since their magnitudes are greater than 1 (specifically, 5 and 3, respectively).

Step 4: Apply Cauchy's Integral Theorem

Since none of the singularities are on or inside the unit circle, \( f(z) \) is analytic inside and on \( C \). According to Cauchy's Integral Theorem, \( \oint_{C} f(z) \, dz = 0 \). Therefore, the integral over the unit circle of \( f(z) \) is zero.

Key Concepts

Analytic Functions

Analytic functions are a fundamental concept in complex analysis. These functions are complex differentiable in a particular domain, meaning they have a well-defined derivative at each point within that area. This unique property implies that they are infinitely differentiable and can be represented by a power series around any point in their domain. Some important aspects of analytic functions include:

• They are smooth and have no abrupt changes in their behavior.
• They provide a way to extend real analysis techniques to the complex plane.
• They are crucial in applications ranging from physics to engineering, where modeling of smooth and continuous phenomena is desired.

In exercises involving the contour integral of a function like the one seen here, checking if the function is analytic helps verify the conditions of Cauchy's Integral Theorem. For the given function, we examined this criterion by analyzing the presence and locations of any singularities.

Complex Integration

Complex integration involves the process of integrating a complex function along a given path or contour. This powerful tool allows us to compute integrals of functions over complex plane curves.Key elements of complex integration include:

• Path Selection: The contour or path of integration is crucial as it directly influences the integral's value.
• Contour: A closed contour forms a loop where the starting point equals the endpoint.
• Use of Theorems: Theorems like Cauchy's Integrals and Residue Theorem simplify complex integration.

For the complex integral over the unit circle, understanding the contour's impact and utilizing the theorems available in complex analysis are vital. Here, the unit circle \( |z| = 1 \) as the contourprovides a specific boundary to check for any singularities of the function.

Poles and Singularities

Poles and singularities are points in the complex plane where a function ceases to be analytic. Identifying these points is essential for solving integration problems in complex analysis.Considerations when dealing with poles and singularities:

• Types: Simple poles, higher order poles, and essential singularities.
• Locations: Determine where these singularities fall concerning the contour
• Residues: At poles, residues help calculate integrals using the residue theorem.

In the given exercise, the function had singularities resulting from zeros of the denominator: \( z = \pm 5 \) and \( z = \pm 3i \). However, these poles were well outside the unit circle,ensuring that the function remains analytic within and on the closed contour. This allows us to apply Cauchy's Integral Theorem effectively, concluding that the integral over this contour is zero.