Problem 8 - 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 \(|z|=1\). \(f(z)=\frac{z^{2}-9}{\cosh z}\)

Step-by-Step Solution

Answer

The integral is zero because the function is analytic on and inside the unit circle.

Step 1: Identify the Singularities

The function given is \( f(z) = \frac{z^2 - 9}{\cosh z} \). To find where this function might have singularities, check \( \cosh z = 0 \). Recall that \( \cosh z = \frac{e^z + e^{-z}}{2} = 0 \). Setting this equation to zero, we get \( e^z + e^{-z} = 0 \). Solving this gives \( z = i\left(n + \frac{1}{2}\right)\pi \) for \( n \in \mathbb{Z} \). In particular, for the unit circle \(|z| = 1\), we check these singularities.

Step 2: Check if Singularities are Inside the Unit Circle

The unit circle is defined by \(|z| = 1\). The points \( z = i(n + \frac{1}{2})\pi \) are purely imaginary, and their magnitude depends on \( n \). Specifically, we consider \( |i(n + \frac{1}{2})\pi| = |(n + \frac{1}{2})\pi| \). For \(|z|=1\), we look at \( |(n + \frac{1}{2})\pi| \leq 1 \). Solving this inequality shows there are no integer values for \( n \) satisfying the condition. Thus, all such singularities lie outside the unit circle.

Step 3: Apply Cauchy's Theorem

Since all the singularities of \( f(z) \) lie outside the unit circle \( |z| = 1 \), the function \( f(z) \) is analytic inside and on the contour \( C \). Cauchy's theorem can now be applied, which states that if a function is analytic within and on a closed contour, then the integral of the function around that contour is zero. Therefore, \( \oint_{C} f(z) \, dz = 0 \).

Key Concepts

Cauchy's Theorem

Cauchy's Theorem is a fundamental result in complex analysis, particularly when dealing with analytic functions. This theorem states that if a function is analytic and defined on a simple, closed contour and throughout its interior, the contour integral around this closed path is zero. In simple terms, for a function without any singularities within the path of integration, the total "accumulated" value around the composed path cancels out, leading to a net result of zero. Key points to remember include:

• A function must be analytic in the region enclosed by the contour, and the contour itself.
• No singularities should lie within this closed path.

Understanding Cauchy's Theorem not only simplifies many problems in complex analysis but also forms the foundation for several other important theorems, such as Cauchy's Integral Formula and the Residue Theorem.

Singularities

Singularities refer to points where a function ceases to be analytic. These are crucial as they determine the conditions and applicability of various theorems in complex analysis. For example, if a function has singularities within or on the boundary of a contour, Cauchy's Theorem cannot be directly applied. Singularities can be classified into a few types:

• Removable Singularities: Ones where the function can be redefined to make the function analytic.
• Pole: A point where the function approaches infinity.
• Essential Singularities: Points where the function exhibits chaotic behavior, without a limit.

Analyzing the location and type of singularities a function possesses provides deep insights into the behavior of the function within a given region. Taking the example of the function in our exercise, its singularities were determined from where the denominator of the function equals zero.

Unit Circle

The unit circle is a circle in the complex plane centered at the origin with a radius of one. Mathematically, it is described by the set of all complex numbers satisfying \(|z| = 1\). In complex analysis, the unit circle serves as a commonly used contour for evaluating complex integrals, especially when utilizing Cauchy's Theorem or the residues at singularities. The exercise focuses on determining whether any of the singularities are enclosed by the unit circle. Since in the solution we found all singularities lie outside the unit circle, this indicates that the region inside is free of singularities, making Cauchy's Theorem applicable. It is this establishment of a singularity-free region within the contour that crucially allows one to conclude that the integral around the unit circle evaluates to zero.

Analytic Functions

An analytic function in complex analysis is one that is differentiable at every point in its domain. Such functions exhibit smoothness, allowing for differentiation just like real-valued calculus functions. Furthermore, analytic functions can be represented by power series, which exemplifies their importance and utility. Analytic functions are crucial because they:

• Satisfy the conditions needed to apply many core theorems in complex analysis, such as Cauchy's Theorem.
• Extensively describe complex variable behaviors, facilitating solutions to integrals along paths or contours.
• Provide insights into the nature and classification of singular points.

In our exercise, determining whether the given function is analytic on and within the contour provides a direct application of Cauchy's Theorem. Without any singularities inside the contour, the function is analytic, leading to the integral of the function around the contour equating to zero.