Problem 18 - Advanced Engineering Mathematics

Question

(a) By Theorem 18.10 with \(f(z)=\frac{1}{z-4}, f^{\prime}(z)=-\frac{1}{(z-4)^{2}},\) and \(f^{\prime \prime}(z)=\frac{2}{(z-4)^{3}}\) $$\oint_{C} \frac{\frac{1}{z-4}}{z^{3}} d z=\frac{2 \pi i}{2 !}\left(\frac{2}{-64}\right)=-\frac{\pi}{32} i$$ (b) By the Cauchy-Goursat Theorem, Theorem 18.4 \\[ \oint_{C} \frac{1}{z^{3}(z-4)} d z=0 \\]

Step-by-Step Solution

Answer

(a) Integral evaluates to \(-\frac{\pi}{32}i\), (b) Integral evaluates to 0.

Step 1: Understand the Given Problem

We need to evaluate a contour integral involving a complex function. The first integral is \( \oint_{C} \frac{\frac{1}{z-4}}{z^{3}} \, d z \) and the second is \( \oint_{C} \frac{1}{z^{3}(z-4)} \, d z \). The solutions must be found using complex analysis theorems: Theorem 18.10 and the Cauchy-Goursat Theorem (Theorem 18.4).

Step 2: Apply Theorem 18.10 (Taylor and Laurent Series)

Theorem 18.10 likely refers to residue calculus or series expansion which involves determining the coefficients in the Laurent series. For the given function \(f(z) = \frac{1}{z-4}\), its derivative \(f'(z) = -\frac{1}{(z-4)^2}\), and second derivative \(f''(z) = \frac{2}{(z-4)^3}\), the integral is computed as \(\frac{2\pi i}{2!}\left(\frac{2}{-64}\right) = -\frac{\pi}{32}i\). This suggests that we use the coefficient of \(z^{-1}\) from the series expansion in the computation.

Step 3: Understand Theorem 18.4 (Cauchy-Goursat Theorem)

The Cauchy-Goursat Theorem states that if a function is analytic within and on a closed contour, the integral around that contour is zero. For the function \( \frac{1}{z^{3}(z-4)} \), assuming it is analytic on and within \(C\), the theorem directly gives \(\oint_{C} \frac{1}{z^{3}(z-4)} \, d z = 0\).

Step 4: Summarize the Results

For integral (a), using Theorem 18.10 and considering the residues, we find that the integral evaluates to \(-\frac{\pi}{32}i\). For integral (b), since it meets the criteria for Cauchy-Goursat Theorem, it evaluates to 0 as the function is analytic over its contour.

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics focused on studying functions that operate on complex numbers. Complex numbers are numbers that have both real and imaginary parts, expressed in the form \( z = a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit with \( i^2 = -1 \). This field extends real number calculus to the complex plane and explores functions that are differentiable within this multi-dimensional plane.

One key aspect of complex analysis is the concept of contour integration. Contour integration involves integrating complex functions along a curve, known as a contour, in the complex plane. This curve can be any path, but often, problems use closed contours which loop back to the starting point, forming a sort of "circle".

In the exercise, evaluating contour integrals was necessary using complex analysis theorems. The ability to extend these integrals in complex ways opens up a host of applications, ranging from fluid dynamics to engineering problems. This type of analysis provides powerful tools such as analyticity, which refers to functions being differentiable at every point within a domain, and the thoughtful application of theorems like Cauchy-Goursat.

Laurent Series

The Laurent Series is a way of expressing complex functions as an infinite series, similar to a Taylor series but extended to accommodate singularities, or points of discontinuity. Normally, the Taylor series represents a function as a sum of its derivatives around a central point, but it doesn't work well when a function has poles or other singularities, places where the function blows up to infinity.

In such cases, Laurent series comes to the rescue. It includes both positive and negative powers of the variable, allowing it to handle those singular points gracefully.

• Positive powers represent the regular, "nice" part of the function.
• Negative powers capture the behavior near the singularity, allowing for the calculation of what is known as residues.

Residues are vital for evaluating contour integrals using the Residue Theorem, which is part of the core of complex analysis. In the given exercise, the Laurent series expansion was used to find coefficients, specifically the \( z^{-1} \) term, which was critical in the residue calculation process for the integral \( \oint_{C} \frac{\frac{1}{z-4}}{z^{3}} \, d z \).

Cauchy-Goursat Theorem

The Cauchy-Goursat Theorem is a fundamental result in complex analysis that simplifies evaluating contour integrals. It states that if a function is analytic (or holomorphic) within and on a simple closed contour, then the integral of that function over the contour is zero.

Analytic functions are those that are complex differentiable at every point in a domain. They are continuous, and crucially, they adhere to Cauchy's integral theorem, which is another name commonly used for the Cauchy-Goursat Theorem.

• For a function to be analytic over a contour, it means there are no points of singularity within that contour.
• If the analytic condition holds, then the integral will be zero, indicating the absence of "energy" or "information" being transported around the contour, sort of like a perfect balance.

In the exercise's second part, the theorem was applied to the integral \( \oint_{C} \frac{1}{z^{3}(z-4)} \, d z \), which evaluated to zero, reflecting the function's behavior as analytic over the contour \( C \). Understanding and applying the Cauchy-Goursat Theorem can make solving many complex integrals much more straightforward.