Problem 7 - Advanced Engineering Mathematics

Question

Using \(z=e^{i t}=\cos t+i \sin t, d z=(-\sin t+i \cos t) d t\) and \(x=\cos t\), \(\begin{aligned} \oint_{C} \operatorname{Re}(z) d z &=\int_{0}^{2 \pi} \cos t(-\sin t+i \cos t) d t=-\int_{0}^{2 \pi} \sin t \cos t d t+i \int_{0}^{2 \pi} \cos ^{2} t d t \\ &=-\frac{1}{2} \int_{0}^{2 \pi} \sin 2 t d t+\frac{1}{2} i \int_{0}^{2 \pi}(1+\cos 2 t) d t=\pi i \end{aligned}\).

Step-by-Step Solution

Answer

The integral evaluates to \(\pi i\).

Step 1: Express \( z \) in terms of \( t \)

Given \( z = e^{it} = \cos t + i \sin t \), we identify the expression for \( z \) as a complex number based on its trigonometric components.

Step 2: Compute \( dz \)

Differentiate \( z \) with respect to \( t \) to get \( dz \). This gives \( dz = (-\sin t + i \cos t) dt \).

Step 3: Integrate using parametrization

Substitute \( \operatorname{Re}(z) = \cos t \) and \( dz = (-\sin t + i \cos t) dt \) into the line integral expression: \( \int_{0}^{2\pi} \cos t (-\sin t + i \cos t) dt \).

Step 4: Separate into real and imaginary components

Separate the integral into real and imaginary parts: \(-\int_{0}^{2\pi} \sin t \cos t dt + i \int_{0}^{2\pi} \cos^2 t dt \).

Step 5: Solve real part integral

The real part is \(-\int_{0}^{2\pi} \sin t \cos t dt\), which using the identity \(\sin 2t = 2\sin t\cos t\) becomes \(-\frac{1}{2}\int_{0}^{2\pi} \sin 2t dt = 0\) since the integral of \(\sin 2t\) over one period is zero.

Step 6: Solve imaginary part integral

The imaginary part is \(\int_{0}^{2\pi} \cos^2 t dt\). Use the identity \(\cos^2 t = \frac{1 + \cos 2t}{2}\) to rewrite this as \(\frac{1}{2} \int_{0}^{2\pi} 1 dt + \frac{1}{2} \int_{0}^{2\pi} \cos 2t dt\). The first integral is \( \pi \), and the second integral is 0, giving a result of \( \pi \).

Step 7: Combine real and imaginary parts

Combine both results: the real part contributes 0, and the imaginary part contributes \(i \pi\). Thus, the integral evaluates to \(\pi i\).

Key Concepts

Line Integrals

A line integral, like the one in our example, allows you to integrate a function over a curve. In complex analysis, you'll often deal with line integrals involving complex numbers. Consider it as the generalization of a simple integral, extending to functions along a path.Understanding line integrals requires recognizing their different components:

• Your path or curve, denoted as \( C \).
• The function you're integrating, which in this example is \(\operatorname{Re}(z)\), or the real part of a complex function \( z \).
• The differential element \( dz \), which represents a tiny segment of your path in the complex plane.

For the integral \( \oint_C \operatorname{Re}(z) dz \), this involves evaluating the real part of \( z \) along the curve \( C \). By following the curve from start to end and summing the values of the function, you get the integral's value.

Complex Numbers

Complex numbers are a fundamental concept in complex analysis and play a crucial role in the exercise. A complex number is typically expressed as \( z = x + yi \), where \( x \) is the real part, and \( yi \) is the imaginary part. Here, \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).In our scenario, the complex number \( z = e^{it} \) is expressed using Euler's formula: \( e^{it} = \cos t + i \sin t \). This shows how complex numbers can represent points in the complex plane with an angle \( t \) around the origin. This form is particularly useful when dealing with problems involving rotation or oscillation. When calculating differntials like \( dz \), you differentiate each component of \( z \), resulting in a new expression that can be used in line integrals.

Trigonometric Identities

Trigonometric identities serve as powerful tools in simplifying the line integral. They help transform complex trigonometric expressions into more manageable forms for integration.In the provided solution, we used the identity \( \sin 2t = 2\sin t \cos t \) to simplify the real part of the integral, \( -\int_{0}^{2\pi} \sin t \cos t dt \). This identity breaks the product of sine and cosine into a single sine term, which is easier to work with.Similarly, the identity \( \cos^2 t = \frac{1 + \cos 2t}{2} \) transformed the imaginary part, \( \int_{0}^{2\pi} \cos^2 t dt \), into integrals involving constants and double angle terms. These identities allow you to compute integrals more gracefully by reducing complexity and avoiding cumbersome multiplications. Recognizing and applying these identities is key to mastering trigonometric integrals.