Problem 12 - Advanced Engineering Mathematics

Question

\(\int_{c} \sin z d z\), where \(C\) is the polygonal path consisting of the line segments from \(z=0\) to \(z=1\) and from \(z=1\) to \(z=1+i\)

Step-by-Step Solution

Answer

The integral evaluates to \(-\cos(1)\cosh(1) + i\sin(1)\sinh(1) + 1.\)

Step 1: Understanding the Path C

The given path \(C\) consists of two line segments: the first segment goes from \(z=0\) to \(z=1\), and the second segment goes from \(z=1\) to \(z=1+i\). We need to evaluate the integral on each segment separately and then add the results.

Step 2: Parametrize the First Segment

For the first segment from \(z=0\) to \(z=1\), we can parametrize this segment as \(z = t\), where \(t\) ranges from \(0\) to \(1\). Then, \(dz = dt\). The integral over this segment becomes \(\int_0^1 \sin(t) \, dt\).

Step 3: Calculate the Integral on the First Segment

The integral on the first segment is \(\int_0^1 \sin(t) \, dt\). The antiderivative of \(\sin(t)\) is \(-\cos(t)\). Evaluating the integral, we have:\[ \int_0^1 \sin(t) \, dt = [-\cos(t)]_0^1 = [-\cos(1) + \cos(0)] = -\cos(1) + 1. \]

Step 4: Parametrize the Second Segment

For the second segment from \(z=1\) to \(z=1+i\), we can parametrize this segment as \(z = 1 + it\), where \(t\) ranges from \(0\) to \(1\). Then, \(dz = i\, dt\). The integral over this segment becomes \(\int_0^1 \sin(1+it) \, i\, dt\).

Step 5: Calculate the Integral on the Second Segment

The integral on the second segment is \(\int_0^1 i \sin(1+it) \, dt\). Using the identity \(\sin(a+bi) = \sin(a)\cosh(b) + i\cos(a)\sinh(b)\), we have:\[ \sin(1+it) = \sin(1)\cosh(t) + i\cos(1)\sinh(t). \]Therefore, the integral becomes:\[ i \int_0^1 \sin(1)\cosh(t) + i\cos(1)\sinh(t) \, dt = i\sin(1)\int_0^1 \cosh(t) \, dt - \cos(1)\int_0^1 \sinh(t) \, dt. \]Evaluating these integrals:\[ \int_0^1 \cosh(t) \, dt = [\sinh(t)]_0^1 = \sinh(1) - \sinh(0) = \sinh(1), \]\[ \int_0^1 \sinh(t) \, dt = [\cosh(t)]_0^1 = \cosh(1) - \cosh(0) = \cosh(1) - 1. \]Substituting back, we get:\[ i\sin(1)\sinh(1) - \cos(1)(\cosh(1) - 1). \]

Step 6: Combine the Integrals from Both Segments

Add the results from the integrals of the two segments:1. From the first segment, we have \(-\cos(1) + 1.\)2. From the second segment, we have \(i\sin(1)\sinh(1) - \cos(1)(\cosh(1) - 1).\)Combine these to get the overall integral:\(-\cos(1) + 1 + i\sin(1)\sinh(1) - \cos(1)\cosh(1) + \cos(1).\)Simplify to final answer:\(-\cos(1)\cosh(1) + i\sin(1)\sinh(1) + 1.\)

Key Concepts

Polygonal Path

In complex integration, understanding the path along which integration occurs is crucial. The term "polygonal path" refers to a path consisting of a series of straight line segments connected in a piecewise manner. For the integral \( \int_{c} \sin z \, dz \), the polygonal path \( C \) is composed of two line segments.
The first segment goes from the point \( z=0 \) to \( z=1 \), creating a horizontal line segment on the real axis. The second segment continues vertically from \( z=1 \) to \( z=1+i \). These segments form a path that resembles an L-shape when plotted in the complex plane.
This construction is often used because it allows for sequential, segmented evaluation of more complex paths through separate integrals over each segment. This step-by-step breakdown simplifies the overall calculation.

Parametrization

When dealing with integrals over complex paths, parametrization is key. It involves expressing each part of a path as a function of a parameter. For our polygonal path, each segment is parametrized separately.
The first segment, from \( z=0 \) to \( z=1 \), is parametrized by setting \( z = t \) where \( t \) ranges from \( 0 \) to \( 1 \). This straightforward parametrization allows us to describe the linear relationship of the segment since \( dz = dt \).
The second segment, from \( z=1 \) to \( z=1+i \), is parametrized by \( z = 1 + it \), where \( t \) also ranges from \( 0 \) to \( 1 \). The derivative \( dz = i \, dt \), shows the imaginary unit \( i \) involved, indicating a path perpendicular to the real axis.
Breaking down the path into segments and parametrizing each one allows for easier integral evaluation, as each parameterization transforms the segment into a form suitable for calculus.

Cosh and Sinh Functions

The cosh and sinh functions, also known as hyperbolic cosine and hyperbolic sine, play a vital role in complex analysis. These functions are used to describe certain complex exponential growth and decay behaviors.
In the context of our exercise, we use the identity \( \sin(a+bi) = \sin(a)\cosh(b) + i\cos(a)\sinh(b) \) during the evaluation of the integral over the second segment. Here, cosh \( (t) \) and sinh \( (t) \) function descriptions help in transforming a complex trigonometric expression into parts that are easier to integrate over.

• \( \cosh(t) \) is defined as \( \frac{e^t + e^{-t}}{2} \).
• \( \sinh(t) \) is defined as \( \frac{e^t - e^{-t}}{2} \).

To evaluate the segments:

• For \( \cosh(t) \), the integral becomes \( \int_0^1 \cosh(t) \, dt = \sinh(1) \).
• For \( \sinh(t) \), we calculate \( \int_0^1 \sinh(t) \, dt = \cosh(1) - 1 \).

These hyperbolic functions bridge trigonometric functions with complex numbers.

Line Integral

Line integrals extend the concept of integrating functions along straight paths to curved paths in complex analysis. Here, the integral \( \int_{c} \sin z \, dz \) is calculated over the polygonal path \( C \).
In simpler terms, a line integral helps us understand the total accumulation of the function \( \sin z \) as \( z \) traces the path. Each segment of the path contributes to the final integral value.
For each segment:

• The first line segment, \[ \int_0^1 \sin(t) \, dt \], integrates the function straight along the real line.
• The second segment, \[ i \int_0^1 \sin(1 + it) \, dt \], involves a more complex evaluation using the identity for sine of a complex number.

By adding the results of these integrals, we capture the holistic effect of \( \sin z \) over the entire polygonal path \( C \). Such calculations are crucial in various fields like fluid dynamics and electromagnetic theory, where understanding the influence along a path in space is essential.