Problem 10 - Advanced Engineering Mathematics

Question

Using \(z=e^{i t}, \pi \leq t \leq 2 \pi, d z=i e^{i t} d t, x=\cos t=\left(e^{i t}+e^{-i t}\right) / 2, y=\sin t=\left(e^{i t}-e^{-i t}\right) / 2 i\), \(\begin{aligned} \int_{C}\left(x^{3}-i y^{3}\right) d z &=\frac{1}{8} i \int_{\pi}^{2 \pi}\left(e^{3 i t}+3 e^{i t}+3 e^{-i t}+e^{-3 i t}\right) e^{i t} d t+\frac{1}{8} i \int_{\pi}^{2 \pi}\left(e^{3 i t}-3 e^{i t}+3 e^{-i t}-e^{-3 i t}\right) e^{i t} d t \\ &=\frac{1}{8} i \int_{\pi}^{2 \pi}\left(2 e^{4 i t}+6\right) d t=\frac{3 \pi}{4} i \end{aligned}\).

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{3\pi}{4} i \).

Step 1: Define the Complex Variable

Given the complex variable \(z = e^{i t}\), with the condition \(\pi \leq t \leq 2\pi\), recognize that this represents a semi-circular path in the complex plane.

Step 2: Relate \(dz\) to \(dt\)

We are given \(dz = i e^{i t} dt\). This indicates how \(z\) changes with respect to \(t\).

Step 3: Express \(x\) and \(y\) in Terms of \(e^{i t}\)

From the provided equations, \(x = \cos t = \frac{e^{i t} + e^{-i t}}{2}\) and \(y = \sin t = \frac{e^{i t} - e^{-i t}}{2i}\). These represent the real and imaginary parts of \(e^{i t}\).

Step 4: Compute \(x^3 - i y^3\)

To calculate, start by expressing \(x^3\) as \(\left(\frac{e^{i t} + e^{-i t}}{2}\right)^3\) and \(\left(i y\right)^3\) as \(i \left(\frac{e^{i t} - e^{-i t}}{2i}\right)^3\). Combine these to simplify into trigonometric terms.

Step 5: Simplify the Integral

Combine the two integrals: \[\frac{1}{8} i \int_{\pi}^{2\pi}(e^{3it} + 3e^{it} + 3e^{-it} + e^{-3it} + e^{3it} - 3e^{it} + 3e^{-it} - e^{-3it}) e^{it} dt\].

Step 6: Final Integral Evaluation

Sum the terms under the integral \[\frac{1}{8} i \int_{\pi}^{2\pi}(2e^{4it} + 6) dt\]. The integrals split into two parts: the integral of \(2e^{4it}\) which evaluates to zero upon periodic function integration over its period, and the integral of 6 which gives \(6(t)\) evaluated from \(\pi\) to \(2\pi\).

Step 7: Calculate the Result

Compute the evaluated integral: \[\frac{1}{8} i \left(6(2\pi) - 6(\pi)\right) = \frac{6\pi}{8} i = \frac{3\pi}{4} i\].

Key Concepts

Complex Integration

Complex integration is a key concept in complex analysis. It involves integrating functions whose variables are complex numbers. In this context, we use contour integrals, which integrate over a path or a curve in the complex plane. When integrating, it's essential to choose an appropriate path that fits the given function. For instance, in the exercise, we use a semi-circular path. This choice simplifies integration using periodic functions over one complete period or half-period.
Complex integration is similar to real integration but deals with two components, real and imaginary. Here, the path of integration, especially if it's circular or semi-circular, affects how we compute the integral. The exercise integrates a function over a parameterized path in the complex plane, making the calculations more tractable by using trigonometric identities and properties of the exponential function.
Some aspects of complex integration include:

• Choosing a contour based on the function's nature.
• Using parameterization to define the path explicitly, like using \(z = e^{i t}\) to describe the path.
• Using known identities to simplify the integral's complexity.

Trigonometric Identities

Trigonometric identities are formulas involving the trigonometric functions, like sine and cosine, that are true for all values of the variable. In complex analysis, these identities are often related to the unit circle, tied to the complex exponential function. One primary identity used here is Euler's formula, \(e^{i t} = \cos t + i \sin t\).
In the step-by-step solution provided, trigonometric identities help transform complex expressions for easier integration. For \(x = \cos t\) and \(y = \sin t\), the exercise translates these into terms involving \(e^{i t}\). Such conversions are possible through identities like:

• \(\cos t = \frac{e^{i t} + e^{-i t}}{2}\)
• \(\sin t = \frac{e^{i t} - e^{-i t}}{2i}\)

Using these rows simplifies complex expressions and plays a crucial role in solving integrals in the context of a complex variable.
These identities bridge the trigonometric functions and complex exponential functions, offering a richer viewpoint of periodic functions, especially beneficial for solving integrals over circular paths.

Complex Variable

In mathematics, a complex variable is simply a variable that can represent complex numbers. Complex numbers extend the idea of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part.
The exercise's variable, \(z = e^{i t}\), is a function of a real parameter \(t\), highlighting this concept as it maps real numbers to a semi-circular path in the complex plane. This is because \(e^{i t}\) traces out a circle on the complex plane as \(t\) varies from \(\pi\) to \(2\pi\).
Working with complex variables often involves:

• Parameterization: Defining complex variables as functions of real parameters to describe paths or curves.
• Recognizing the real and imaginary components: For \(z = a + bi\), \(a\) is the real part, and \(b\) is the imaginary part.
• Utilizing trigonometric identities to simplify expressions that involve these variables.

Complex variable techniques are foundational in complex analysis and lead to many exciting insights into calculus when extended to complex functions.