Problem 13 - Advanced Engineering Mathematics

Question

Evaluate the given integral along the indicated contour. \(\int_{C} \operatorname{Im}(z-i) d z\), where \(C\) is the polygonal path consisting of the circular arc along \(|z|=1\) from \(z=1\) to \(z=i\) and the line segment from \(z=i\) to \(z=-1\)

Step-by-Step Solution

Answer

The integral evaluates to zero.

Step 1: Analyze the integral expression

The integral we need to evaluate is \(\int_{C} \operatorname{Im}(z-i) dz\), where \(\operatorname{Im}(z-i)\) represents the imaginary part of \(z-i\).

Step 2: Break down the contour into segments

The contour \(C\) is composed of two parts: (1) a circular arc from \(z=1\) to \(z=i\) on \(|z|=1\), and (2) a line segment from \(z=i\) to \(z=-1\). We will analyze these parts separately.

Step 3: Evaluate the integral over the circular arc

For the first segment, parameterize the circular arc from \(z=1\) to \(z=i\). Using the parameterization \(z=e^{i\theta}\) where \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\), we have: \(\int_{0}^{\frac{\pi}{2}} \operatorname{Im}(e^{i\theta} - i)i e^{i\theta} d\theta\). This simplifies to \(\int_{0}^{\frac{\pi}{2}}(\sin\theta - 1)(ie^{i\theta}) d\theta\). Compute this integral.

Step 4: Evaluate the integral over the line segment

For the second segment, parameterize the line from \(z=i\) to \(z=-1\) using \(z = i - (1+t)i\), where \(t\) ranges from \(0\) to \(1\). This gives the integral \(\int_{0}^{1}\operatorname{Im}(i - (1+t)i)(-i) dt\). Simplify this to \(\int_{0}^{1}(-t+1) (-i) dt\) and compute it.

Step 5: Sum the results of each segment

Sum the results of the integrals evaluated over the two segments to find the total value of the contour integral. Given that the individual integrals result in real values due to symmetry, we focus on their sum to determine the final value.

Key Concepts

Complex analysis

Complex analysis is a fascinating area of mathematics that deals with functions of complex numbers. The field extends many ideas of calculus to the complex plane, which provides new insights and techniques.
The complex plane is a two-dimensional plane where each point represents a complex number. The horizontal axis represents the real part, while the vertical axis represents the imaginary part. This geometric interpretation is crucial when exploring integrals of complex functions.

• Complex numbers can be expressed in the form \(z = x + yi\), where \(x\) is the real part and \(y\) is the imaginary part.
• Functions of complex numbers may return parts of the number, such as the real or imaginary component. In the problem provided, the expression \(\operatorname{Im}(z-i)\) represents the imaginary part of \(z-i\).

With complex analysis, we can transform problems related to real-valued functions into more manageable forms by considering them in a two-dimensional complex framework. This transformation opens up a wide range of techniques and theorems that simplify and solve many seemingly difficult problems.

Line integrals

Line integrals are a type of integral that generalize the notion of integrating over a straight line to integrating along a path in space. In complex analysis, the paths are often in the complex plane.
Line integrals can be thought of as adding up contributions from a field along a certain path:

• A line integral in the complex plane involves integrating a complex-valued function over a contour, which is a path within the complex plane.
• For example, in our exercise, the contour \(C\) consists of a combination of a circular arc and a line segment.
• The integral calculates the cumulative effect of the field, represented by the function, as you move along the given path.

In solving a line integral over a complex contour, it's common to break it into simpler pieces corresponding to simpler physical or geometric paths, just as we see in the solution to the provided problem. This approach allows us to handle complex paths by focusing on more manageable segments.

Parameterization

Parameterization is a mathematical technique used to specify a function using parameters. It provides a way to represent complex shapes and paths using simpler expressions. This technique is especially useful in complex analysis when dealing with integral contours.
Parameterization involves:

• Assigning a parameter to a variable that describes a curve or path. This allows for handling shapes that aren't easily expressed in the standard \(x, y\) coordinate system.
• In the given exercise, we used parameterization to express the path along the arc and the line segment with respect to a parameter \(\theta\) and \(t\), respectively.
• This simplifies the integral because we convert a potentially complex expression into something more tractable, often modifying the integral limits to suit the parameter range.

By using parameterization, integrals that might seem difficult initially can be rewritten into forms where standard integration techniques can be easily applied, turning complex paths into simpler, parameter-defined trajectories.