Problem 18 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The total contour integral \( \oint_{C}(2 z-1) \, dz \) is 0.

Step 1: Understanding the Problem

We are given a contour integral \( \oint_{C}(2 z-1) \, dz \) which is split into three parts: \( C_1, C_2, \) and \( C_3 \). Each contour segment has its own integral to solve, and we need to calculate the total by summing the results of these integrals.

Step 2: Solving Integral on C1

The contour \( C_1 \) is described by \( y=0 \) over \( 0 \leq x \leq 1 \) with \( z=x \) and \( dz=dx \). The integral is calculated as: \[ \int_{C_{1}}(2 z-1) \, dz = \int_{0}^{1}(2 x-1) \, dx = 0. \]

Step 3: Solving Integral on C2

For \( C_2 \), \( x=1 \) and \( 0 \leq y \leq 1 \), so \( z=1+i y \) and \( dz=i \, dy \). The integral calculation is: \[ \int_{C_{2}}(2 z-1) \, dz = \int_{0}^{1} (2(1+i y)-1) \, i \, dy = -2 \int_{0}^{1} y \, dy + i \int_{0}^{1} dy = -1+i. \]

Step 4: Solving Integral on C3

The contour \( C_3 \) follows \( y=x \), giving \( z=x+i x \) and \( dz=(1+i) \, dx \). The calculation is done from \( x=1 \) to \( x=0 \): \[ \int_{C_{3}}(2 z-1) \, dz = (1+i) \int_{1}^{0}(2 x-1+2 i x) \, dx = 1-i. \]

Step 5: Summing the Integrals

Now, sum the results from \( C_1, C_2, \) and \( C_3 \): \( \int_{C_{1}}(2 z-1) \, dz + \int_{C_{2}}(2 z-1) \, dz + \int_{C_{3}}(2 z-1) \, dz = 0 - 1 + i + 1 - i = 0. \)

Step 6: Conclusion

The total contour integral \( \oint_{C}(2 z-1) \, dz \) evaluates to \( 0 \). This indicates that the path integral along the closed curve is zero, confirming that the vector field described is conservative along this path.

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics that explores complex numbers and their functions. It deals with numbers of the form \(z = x + iy\), where \(x\) and \(y\) are real numbers and \(i\) is the imaginary unit, satisfying \(i^2 = -1\). This field is fundamental when analyzing complex functions, derivatives, and integrals. Complex analysis provides tools like Cauchy's Integral Theorem and the Residue Theorem, which simplify the process of solving integrals of complex functions over specific paths or contours in the complex plane. Understanding these concepts is crucial for executing contour integrals, which often evaluate to simply manageable values like zero, just as shown in the given problem, owing to specific properties of the functions involved.

Path Integral

A path integral, often used in complex analysis, is the method by which you compute the integral of a complex function along a certain path or contour within the complex plane. Unlike regular integrals, which evaluate over an interval on the real line, path integrals consider a trajectory in space. This exercise features path integrals across different paths \(C_1, C_2,\) and \(C_3\) to calculate the function \((2z-1)dz\). It shows the integration process over straight and diagonal paths in the plane where different expressions for \(z\) and \(dz\) are used based on the defined path. These integrals contribute to the understanding of how different directional movements in the complex plane affect the resulting values from the integration.

Conservative Vector Field

In vector calculus, a conservative vector field is one where line integrals over a closed loop path return zero. That suggests the potential function from which the field is derived does not depend on the path taken between two points, only on the endpoints themselves. For the given exercise, the contour integral equalling zero confirms the field is conservative. Cauchy's Integral Theorem plays a big part here. It states that if a function is holomorphic (i.e., differentiable in the complex sense) within and on some closed curve, then the integral of that function over the curve is zero. This simplification occurs because holomorphic functions retain many properties similar to real differentiability, such as existence and smoothness. These properties ensure a conservative nature.

Integration of Complex Functions

Integration of complex functions involves calculating the integral of a complex-valued function over a certain contour in the complex plane. The process involves splitting the integral into real and imaginary components and solving related real-valued integrals. In the given exercise, we see this practice applied to the function \((2z-1)\). Each segment, \(C_1\), \(C_2\), and \(C_3\), needs separate treatment depending on its direction and limits, resulting in expressions such as \((2x-1)dx\) and \((2(1+iy)-1)idy\). The successful integration over these paths and summing them to a total of zero shows how paths influence the integral and demonstrate key results like path independence and the properties of analytic functions over closed paths. These details are the heart of complex integral solutions.