Problem 24 - Advanced Engineering Mathematics

Question

On \(C, x=\sin t, y=\cos t, 0 \leq t \leq \pi / 2\) or \(z=i e^{-i t}, d z=e^{-i t} d t\), \(\int_{C}\left(z^{2}-z+2\right) d z=\int_{0}^{\pi / 2}\left(-e^{-2 i t}-i e^{-i t}+2\right) e^{-i t} d t=\int_{0}^{\pi / 2}\left(-e^{-3 i t}-i e^{-2 i t}+2 e^{-i t}\right) d t\) \(=-\frac{1}{3} i e^{-3 \pi i / 2}+\frac{1}{2} e^{-\pi i}+2 i e^{-\pi i / 2}+\frac{1}{3} i-\frac{1}{2}-2 i=\frac{4}{3}-\frac{5}{3} i\).

Step-by-Step Solution

Answer

The value of the integral is \( \frac{4}{3} - \frac{5}{3}i \).

Step 1: Understand the Given Functions

The exercise involves two representations: one is parameterized by trigonometric functions, \( x = \sin t \) and \( y = \cos t \), while the other uses complex exponential forms with \( z = ie^{-it} \) and \( dz = -ie^{-it}dt \). This complex representation is suited for integration in the complex plane.

Step 2: Simplify and Arrange the Integral

Rewrite the integral with the given function of \( z \): \( \int_{0}^{\pi/2}(-e^{-3it} - ie^{-2it} + 2e^{-it}) \, dt \). The problem statement provides a formula to evaluate directly, noting each term has a direct exponential form.

Step 3: Evaluate Each Exponential Integral

Separate the terms: - \( \int_{0}^{\pi/2} -e^{-3it} \, dt = \frac{i}{3}(e^{-3it}) \bigg|_{0}^{\pi/2} = -\frac{1}{3}ie^{-3\pi i/2} + \frac{1}{3}i \).- \( \int_{0}^{\pi/2} -ie^{-2it} \, dt = \frac{1}{2}(e^{-2it}) \bigg|_{0}^{\pi/2} = \frac{1}{2}e^{-\pi i} - \frac{1}{2} \).- \( \int_{0}^{\pi/2} 2e^{-it} \, dt = 2i(e^{-it}) \bigg|_{0}^{\pi/2} = 2ie^{-\pi i/2} - 2i \).

Step 4: Sum the Parts

Add the results from step 3 to obtain the total:\(\begin{align*} -\frac{1}{3} & i e^{-3\pi i / 2} + \frac{1}{3}i \ + & \frac{1}{2}e^{-\pi i} - \frac{1}{2} \ + & 2ie^{-\pi i/2} - 2i = \frac{4}{3} - \frac{5}{3}i\end{align*}\)

Step 5: Check the Solution

Confirm the computations match the provided solution: The value \( \frac{4}{3} - \frac{5}{3}i \) agrees with the answer in the problem after simplification.

Key Concepts

Complex Exponentials

In the realm of complex numbers, complex exponentials provide a powerful way to express functions and solve integrals. They leverage Euler's formula, which states that \[ e^{i\theta} = \cos \theta + i\sin \theta \]. This bridges the gap between trigonometric functions and exponential forms.
The given exercise simplifies processes by using complex exponentials, turning trigonometric parameterizations into more manageable forms.

• In the exercise, the expression \( z = i e^{-it} \) is used. This involves the complex exponential \( e^{-it} \) representing a point on the unit circle in the complex plane.
• By utilizing these exponentials, derivative operations translate into simple multiplication or division, easing the evaluation of integrals.

The key advantage here is the ability to transform complicated trigonometric identities into a form that is linear and more straightforward when integrating over a contour on the complex plane.

Trigonometric Parameterization

Trigonometric parameterization is a technique where trigonometric identities describe a curve in the calculus of complex functions. It is often employed when the curve is circular or elliptic in nature.

The problem uses this technique by setting \( x = \sin t \) and \( y = \cos t \), which map onto the circular segment from \( t = 0 \) to \( t = \pi/2 \).

• Here, \( t \) acts as a parameter that describes a path along the curve, specifically a quarter of a unit circle in this context.
• This parameterization offers a geometric intuition, simplifying complex integration when calculating areas or other properties of the curve.

By equating trigonometric functions to coordinates in complex integration, it allows the transition from geometric to algebraic expressions efficiently.

Complex Numbers

Complex numbers extend the traditional real number system to include all numbers of the form \( a + bi \), where \( i \) is the imaginary unit satisfying \( i^2 = -1 \).

This exercise involves these numbers because it occurs in the complex plane, which is pivotal in expressing functions of multiple variables with a single expression.

• In our solution, \( z \) represents a complex number \( ie^{-it} \), emphasizing its pathway traced from the origin in the complex plane.
• The use of complex numbers enables more comprehensive methods of differentiation and integration, crucial for evaluating integrals like the ones in this exercise.

All calculations, therefore, rely on the foundation laid by complex numbers, ensuring the mathematical technique aligns with the geometric intuition.

Definite Integrals

Definite integrals hold the key to calculating areas under curves, solutions to differential equations, and in this exercise, segments of curves in the complex plane.

The integral \( \int_{0}^{\pi/2}(-e^{-3it} - ie^{-2it} + 2e^{-it}) \, dt \) demonstrates how definite integrals can be computed over intervals defined by limits.

• Each distinct term in the expression represents a segment of the path in the complex plane, conforming to rules of additivity in line integrals.
• By solving such integrals, one can determine the total 'accumulation' along the path from start to finish, given the parameters and bounds.

Definite integrals facilitate a mathematical 'sum' over continuous parts of the curve, offering the final solution that merges both real and imaginary components seamlessly.