Problem 34 - Advanced Engineering Mathematics

Question

For \(f(z)=1 /(\overline{z-1}), \overline{f(z)}=1 /(z-1),\) so on \(z-1=2 e^{i t}, d z=2 i e^{i t} d t,\) and \(\oint_{C} \overline{f(z)} d z=\int_{0}^{2 \pi} \frac{1}{2 e^{i t}} \cdot 2 i e^{i t} d t=i \int_{0}^{2 \pi} d t=2 \pi i\). Thus circulation \(=\operatorname{Re}\left(\oint_{C} \overline{f(z)} d z\right)=0,\) and net \(\operatorname{flux}=\operatorname{Im}\left(\oint_{C} \overline{f(z)} d z\right)=2 \pi\).

Step-by-Step Solution

Answer

The net flux is \( 2\pi \) and the circulation is 0.

Step 1: Expressing \( \overline{f(z)} \) in terms of \( z \)

Given \( f(z) = \frac{1}{\overline{z - 1}} \), we can write its conjugate as \( \overline{f(z)} = \frac{1}{z-1} \).

Step 2: Parametrize the contour \( z-1 = 2 e^{it} \)

To perform a contour integral, we set \( z - 1 = 2 e^{it} \) so that we traverse a circle of radius 2 centered at \( z = 1 \).

Step 3: Deriving \( dz \)

Since \( z = 1 + 2 e^{it} \), the differential \( dz \) is given by \( dz = 2i e^{it} dt \).

Step 4: Setting up the contour integral

To evaluate \( \oint_C \overline{f(z)} dz \), substitute \( \overline{f(z)} = \frac{1}{z-1} = \frac{1}{2 e^{it}} \) and \( dz = 2i e^{it} dt \).

Step 5: Simplifying the expression

The integral becomes \( \int_0^{2\pi} \frac{1}{2 e^{it}} \cdot 2i e^{it} dt = i \int_0^{2\pi} dt \). The \( e^{it} \) terms cancel out.

Step 6: Calculate the integral

Evaluate \( i \int_0^{2\pi} dt \), which is \( i \times 2\pi \) since the integral of \( dt \) over \( [0, 2\pi] \) equals \( 2\pi \).

Step 7: Determine the real part (circulation)

The real part of \( i \times 2\pi \) is zero, so \( \operatorname{Re}(\oint_C \overline{f(z)} dz) = 0 \).

Step 8: Determine the imaginary part (net flux)

The imaginary part of \( i \times 2\pi \) is \( 2\pi \), so the net flux is \( \operatorname{Im}(\oint_C \overline{f(z)} dz) = 2\pi \).

Key Concepts

Contour Integration

Contour integration is a powerful tool in complex analysis used to evaluate integrals of complex functions over a specified contour or path in the complex plane. This method leverages the concept of a contour, which is a closed, continuous path along which the integral is evaluated. In this exercise, we are interested in integrating a complex function of the form \( \overline{f(z)} \), over a circular contour centered at \( z=1 \).Key ideas in contour integration involve:

• Parameterizing the contour to express complex functions in a form suitable for integration.
• Evaluating the integral by simplifying the expression, often making use of symmetries or cancellations inherent in the function or the contour.

Contour integrals are unique as they often reveal properties such as residues and poles, which provide insight into the behavior of complex functions.

Complex Conjugate

The complex conjugate is a fundamental concept in complex analysis. For any complex number \( z = x + yi \), its conjugate is expressed as \( \overline{z} = x - yi \). The conjugate essentially "flips" the sign of the imaginary part while keeping the real part unchanged.In this problem, the function \( f(z) = \frac{1}{\overline{z-1}} \) prompts us to work with \( \overline{f(z)} \), which under the manipulation of complex conjugates, becomes \( \overline{f(z)} = \frac{1}{z-1} \). This transformation is crucial as it simplifies the integral, making the mathematical operation more tractable.Conjugates are often used to determine the real and imaginary components of functions and play a critical role in resolving equations involving complex numbers.

Parametrization

In complex analysis, parametrization is a technique used to transform complex expressions into a function of a single parameter, which simplifies integration. During the contour integration process, functions over complex paths are often represented in terms of a parameter.For the given contour \( z - 1 = 2 e^{it} \), it parameterizes a circle of radius 2 centered at \( z = 1 \). Here, \( t \) runs from 0 to \( 2\pi \), which defines a full circle in the complex plane. By parameterizing \( z \), the differential \( dz \) is expressed as \( dz = 2i e^{it} dt \), accommodating the circular nature of the contour.Parametrization allows the complex function to be expressed in terms of a real variable \( t \), which aids in evaluating the integral over the contour.

Imaginary and Real Parts

In complex analysis, every complex number can be broken down into a real part and an imaginary part. For instance, \( z = x + yi \) where \( x \) is the real part and \( yi \) is the imaginary part.For a contour integral, we are usually interested in these components separately because they often represent different physical or geometric interpretations:

• The real part of the integral is often associated with circulation or as a measure of rotation around a point.
• The imaginary part is linked to flux, representing the net flow across the boundary of the contour.

In this exercise, when evaluating \( \oint_C \overline{f(z)} dz \), the real part \( \operatorname{Re}(i \times 2\pi) \) is zero, indicating no circulation. Conversely, the imaginary part \( \operatorname{Im}(i \times 2\pi) \) is \( 2\pi \), representing the net flux through the contour.