Problem 15 - Advanced Engineering Mathematics

Question

(a) By Theorem 18.9 with \(f(z)=\frac{2 z+5}{z-2}\) $$\oint_{C} \frac{\frac{2 z+5}{z-2}}{z} d z=2 \pi i\left(-\frac{5}{2}\right)=-5 \pi i$$ (b) since the circle \(|z-(-1)|=2\) encloses only \(z=0\), the value of the integral is the same as in part (a). (c) From Theorem 18.9 with \(f(z)=\frac{2 z+5}{z}\) $$\oint_{C} \frac{\frac{2 z+5}{z}}{\frac{z}{z-2}} d z=2 \pi i\left(\frac{9}{2}\right)=9 \pi i$$ (d) since the circle \(|z-(-2 i)|=1\) encloses neither \(z=0\) nor \(z=2\) it follows from the Cauchy-Goursat Theorem, Theorem \(18.4,\) that $$\oint_{C} \frac{2 z+5}{z(z-2)} d z=0$$

Step-by-Step Solution

Answer

(a) -5πi; (b) -5πi (same as a); (c) 9πi; (d) 0.

Step 1: Analyze Part (a)

In part (a), we are using Theorem 18.9, which is likely referring to the Residue Theorem. The function given is \( f(z) = \frac{2z + 5}{z-2} \). The integral around the contour \( C \) is given by \( \oint_{C} \frac{\frac{2z+5}{z-2}}{z} \, dz \). The poles are at \( z=2 \) (from \( f(z) \)) and \( z=0 \) from \( \frac{1}{z} \). The integral calculates the residue at \( z=0 \) because the contour \( C \) likely encloses \( z=0 \) only, leading to the expression \( 2\pi i \times \text{Res}(f(z),0) \), where \( \text{Res}(f(z),0) = -\frac{5}{2} \). Thus, the answer is \(-5\pi i\).

Step 2: Evaluate Part (b)

In part (b), the circle \(|z+1|=2\) is considered. It encloses only the pole at \(z=0\), not \(z=2\). Therefore, the evaluation of the integral remains the same as in part (a), since the pole \(z=2\) is outside the contour and does not contribute to the integral.

Step 3: Analyze Part (c)

For part (c), Theorem 18.9 is again applied with a different function \( f(z) = \frac{2z+5}{z} \). The integral is given as \( \oint_{C} \frac{\frac{2z+5}{z}}{\frac{z}{z-2}} \, dz \). Simplifying the integrand gives \( \oint_{C} \frac{2z+5}{z-2} \, dz \), indicating the pole at \( z=2 \) is enclosed by the contour \( C \). Using the Residue Theorem again, the residue at \( z=2 \) is \( \frac{9}{2} \), leading to the result \( 9\pi i \).

Step 4: Evaluate Part (d)

In part (d), the given circle \(|z+2i|=1\) encloses neither \( z=0 \) nor \( z=2 \). According to the Cauchy-Goursat Theorem (Theorem 18.4), which states that the integral of a holomorphic function around a closed contour enclosing no singularities is zero, this results in \( \oint_{C} \frac{2z+5}{z(z-2)} \, dz = 0 \).

Key Concepts

Complex Integration

Complex integration is a fundamental concept in the study of complex analysis. It involves integrating complex-valued functions over a specific path or contour in the complex plane. Complex integration is analogous to conventional real-variable integration but has unique properties and applications.

• Paths: The path of integration is crucial. It determines which singularities or poles are included within the contour, affecting the value of the integral.
• Complex Function: The function being integrated should be defined and holomorphic, except at a finite number of singularities within the path.

Complex integration is not just about finding areas under curves but also understanding the mappings performed by complex functions. It aligns closely with the idea of summing infinitesimal changes of the function as you move along a path in the complex plane.

Cauchy-Goursat Theorem

The Cauchy-Goursat Theorem is a pivotal result in complex analysis. It states that if a function is analytical or holomorphic everywhere on and inside a closed curve in the complex plane, then the contour integral around that curve equals zero. This theorem simplifies many complex integrals and serves as a cornerstone to understand more advanced theorems.
Key aspects include:

• Holomorphic Function: The theorem applies to functions that are holomorphic, meaning they have a derivative at every point in their domain.
• Closed Curve: The path must be closed, forming a loop in the complex plane.
• No Singularities: The function should not have singularities within the contour.

When conditions are met, the Cauchy-Goursat Theorem guarantees zero for the integral, significantly simplifying complex integration problems by reducing unnecessary calculations.

Contour Integral

Contour integrals are an extension of complex integration where the integral is evaluated along a path or contour in the complex plane. This technique is essential in complex analysis and often involves the use of parameterization to describe the path.

• Parameterization: A contour is typically parameterized by a complex function, which describes how the path is traversed in the complex plane.
• Independence of Path: If a function is holomorphic within a region, the contour integral does not depend on the specific path taken between two points, just the endpoints.
• Residue Theorem: Contour integrals also tie into the Residue Theorem, which uses residues at poles within the contour for evaluating integrals.

Understanding contour integrals enables the evaluation of complex integrals by focusing on paths and the singularities they enclose.

Poles and Residues

In complex analysis, poles and residues provide powerful tools for evaluating integrals. A pole is a type of singularity where a complex function behaves like the reciprocal of a monomial. Residues are used to calculate contour integrals via the Residue Theorem.

• Identifying Poles: Poles are identified by examining the points where the function's denominator approaches zero and defines the order of the pole based on the power of zeroes.
• Residue Calculation: The residue of a function at a pole is the coefficient of \( \frac{1}{z-a} \) in the function's Laurent series expansion. It can also be calculated by specific formulas for simple poles.
• Residue Theorem Application: The theorem states that the integral around a closed contour is \( 2\pi i \) times the sum of the residues within that contour, making it crucial for solving integral problems efficiently.

By understanding poles and residues, students can effectively evaluate complex integrals, especially those involving closed contours and singularities.