Problem 15 - Advanced Engineering Mathematics

Question

We have \(\oint_{C} z e^{z} d z=\int_{C_{1}} z e^{z} d z+\int_{C_{2}} z e^{z} d z+\int_{C_{3}} z e^{z} d z+\int_{C_{4}} z e^{z} d z\) \(\mathrm{On} C_{1}, y=0,0 \leq x \leq 1, z=x, d z=d x\), \(\int_{C_{1}} z e^{z} d z=\int_{0}^{1} x e^{x} d x=x e^{x}-\left.e^{x}\right|_{0} ^{1}=1\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}} z e^{z} d z=i \int_{0}^{1}(1+i y) e^{1+i y} d y=i e^{i+1}\). On \(C_{3}, y=1,0 \leq x \leq 1, z=x+i, d z=d x\), \(\int_{C_{3}} z e^{z} d z=\int_{1}^{0}(x+i) e^{x+i} d x=(i-1) e^{i}-i e^{1+i}\). On \(C_{4}, x=0,0 \leq y \leq 1, z=i y, d z=i d y\), \(\int_{C_{4}} z e^{z} d z=-\int_{1}^{0} y e^{i y} d y=(1-i) e^{i}-1\). Thus \(\oint_{C} z e^{z} d z=1+i e^{i+1}+(i-1) e^{i}-i e^{1+i}+(1-i) e^{i}-1=0\).

Step-by-Step Solution

Answer

The integral over the closed contour \( C \) is 0.

Step 1: Analyze Contour C1

For contour \( C_1 \), where \( y = 0 \) and \( 0 \leq x \leq 1 \), the function \( z = x \) and \( dz = dx \). This makes the integral \( \int_{C_1} z e^z dz = \int_0^1 x e^x dx \). Using integration by parts, we evaluate this to get \( xe^x - e^x \big|_0^1 = 1 \).

Step 2: Evaluate Contour C2

On contour \( C_2 \), we have \( x = 1 \), \( 0 \leq y \leq 1 \), which translates to \( z = 1 + iy \) and \( dz = i dy \). Thus \( \int_{C_2} z e^z dz = i \int_0^1 (1 + iy) e^{1 + iy} dy \). This integral evaluates to \( ie^{1+i} \), as given.

Step 3: Solve Contour C3

For contour \( C_3 \), where \( y = 1 \), \( 0 \leq x \leq 1 \), the integration involves \( z = x + i \), \( dz = dx \), giving the integral \( \int_{C_3} z e^z dz = \int_1^0 (x+i) e^{x+i} dx \). After evaluating, \( (i-1) e^i - i e^{1+i} \) is the result.

Step 4: Evaluate Contour C4

On contour \( C_4 \), where \( x = 0 \), \( 0 \leq y \leq 1 \), we have \( z = iy \), \( dz = i dy \), and inversion of limits leads to \( \int_{C_4} z e^z dz = -\int_1^0 y e^{iy} dy \). This evaluates to \( (1-i)e^i - 1 \).

Step 5: Combine Results

Combine the results of each contour: \( 1 + ie^{1+i} + (i-1)e^i - ie^{1+i} + (1-i)e^i - 1 \). Simplify the expression to find the total result. The terms cancel each other yielding \( 0 \), confirming that the integral around the closed contour \( C \) is zero.

Key Concepts

Contour Integration

Contour integration is a fundamental technique in complex analysis that involves evaluating a complex integral over a contour or path in the complex plane. This technique is often used to integrate complex-valued functions, where the contour can be a straight line, a curve, or any closed path.

In the given exercise, the contour is divided into four parts: \(C_1, C_2, C_3,\) and \(C_4\). Each part of the contour is specified by different equations that dictate the path of integration. These contours are summed to evaluate the integral around the entire closed path \(C\).

Contour integration is especially handy because of the Cauchy-Goursat theorem, which states that if a function is analytic inside and on a closed contour, the contour integral is zero. This is reflected in the exercise as the integration around the closed contour \(C\) resulted in zero.

When performing contour integration, it's important to:

• Identify the path correctly.
• Express the differential \(dz\) in terms of the path's differential elements (such as \(dx\) or \(dy\)).
• Apply the limits specific to each path.
• Combine all parts of the path for a complete solution.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is expressed by the formula:

\[\int u \, dv = uv - \int v \, du\]This tool is useful in scenarios where simple integration is complicated by the presence of two multiplicative functions.

In the exercise, integration by parts was applied to the contour \(C_1\), where the function \(z e^z\) was differentiated into parts: program \(z\) as \(u\) and \(e^z\, dz\) as \(dv\). Upon integration, it was evaluated within given limits to derive the integral along this contour as 1.

When using integration by parts:

• Identify the functions to assign as \(u\) and \(dv\).
• Differentiate \(u\) to get \(du\), and integrate \(dv\) to obtain \(v\).
• Substitute back into the formula carefully using limits.

This process smoothly handles the product \(z e^z\) resulting in simpler components for integration.

Complex Integration

Complex integration extends the concept of integration to complex-valued functions along a path in the complex plane. It incorporates the properties and rules similar to real integration but is applied to functions of complex numbers.

In the exercise, each of the integrals along \(C_1, C_2, C_3,\) and \(C_4\) required integrating \( z e^z\) over specific paths defined by the contours. Each part was calculated separately, considering the path geometry and the complex nature of the function.

Key aspects in complex integration include:

• Paths are defined using complex numbers \(z = x + iy\), involving both real and imaginary components.
• Differential elements \(dz\) are expressed accordingly, considering the specific path, such as \(dz = i \, dy\) for vertical paths.
• Complex integration can result in complex numbers, reflecting both the magnitude and direction-like behavior due to the complex plane nature.

Path Independence

In complex analysis, path independence is a characteristic of functions that can be integrated without concern for the specific path taken, provided the endpoints remain constant, and the function is analytic on and between the paths.

This exercise showcases the important principle that when a function is analytic and the path is closed, the integral around the path is zero, as seen in the zero result of the contour integral \( \oint_{C} z e^{z} dz = 0 \).

Path independence relies on:

• The function must be analytic (holomorphic) inside the region and on the path.
• The contour must be closed, forming a continuous loop without gaps.
• The result of the contour integral should equal zero, indicating path independence according to the Cauchy-Goursat theorem.

This property helps simplify complex calculations, as the evaluation can often be avoided if certain conditions are met. It also emphasizes the power of contour integration in complex analysis.