Problem 10 - Advanced Engineering Mathematics

Question

\(\int_{C}\left(x^{3}-i y^{3}\right) d z\), where \(C\) is the lower half of the circle \(|z|=1\) from \(z=-1\) to \(z=1\)

Step-by-Step Solution

Answer

The integral along the path is 0 due to symmetry and closed path properties in complex integration.

Step 1: Identify the Parametrization of the Path

To solve the line integral, we first need to parametrize the path. The path \(C\) is the lower half of the circle \(|z|=1\) from \(z=-1\) to \(z=1\). We can parametrize this path as \(z = e^{i\theta}\), for \(\theta \in [\pi, 2\pi]\).

Step 2: Derive Expressions for dz, x, and y

Given the parametrization \(z = e^{i\theta}\), we have \(dz = ie^{i\theta}d\theta\). Since \(z = x + iy\), we can equate real and imaginary parts: \(x = \cos\theta\) and \(y = \sin\theta\).

Step 3: Substitute x, y, and dz in the Integral

Substitute \(x = \cos\theta\), \(y = \sin\theta\), and \(dz = ie^{i\theta}d\theta\) into the integral: \[ \int_{C} \left( x^3 - iy^3 \right) dz = \int_{\pi}^{2\pi} \left(\cos^3\theta - i\sin^3\theta \right)(ie^{i\theta}) d\theta. \]

Step 4: Simplify the Expression Inside the Integral

Distribute the \(ie^{i\theta}\):\[ \int_{\pi}^{2\pi} i(\cos^3\theta - i\sin^3\theta)e^{i\theta} d\theta = \int_{\pi}^{2\pi} i\cos^3\theta e^{i\theta}d\theta + \int_{\pi}^{2\pi} \sin^3\theta e^{i\theta}d\theta. \]

Step 5: Simplify Using Complex Exponentials

We note that integrating over a full circle with exponential terms simplifies considerably due to symmetry and periodicity. Use the fact that the integral of \(z^n\) around a closed loop centered at 0 (e.g., \(|z|=1\)) is zero for \(n eq -1\), and parts of our expression can be seen as a similar sum over terms. Particularly, none of these terms are of the form \(z^{-1}\), leading the entire integral to be zero.

Key Concepts

Parametrization of Curves

When tackling line integrals, it's crucial to understand the concept of parametrization of curves. In our original solution, the path \(C\) is described as the lower half of the circle \(|z| = 1\), stretching from \(z = -1\) to \(z = 1\). To parametrize this, we use a complex exponential form: \(z = e^{i\theta}\), where \(\theta \in [\pi, 2\pi]\). This transformation allows us to express points on the path in terms of a single parameter \(\theta\).

Here’s why this works well:

• The magnitude \(|z| = 1\) pinpoints any point on the circle's boundary.
• The angle \(\theta\) represents the position along the circle, starting from \(\theta = \pi\) and ending at \(\theta = 2\pi\), effectively tracing the lower half.

This parametrization simplifies calculating the integral because it replaces the original two-dimensional path with a one-dimensional parameter \(\theta\). This transformation leverages the power of complex numbers to simplify curve expressions, providing a more manageable way to handle the complex geometry of the path.

Complex Exponential Function

The complex exponential function \(e^{i\theta}\) is at the heart of our parametrization. Euler's formula, a keystone in complex analysis, states that \(e^{i\theta} = \cos\theta + i\sin\theta\). This identity plays a critical role in simplifying our line integral.

Using \(e^{i\theta}\), we can break any complex number on the unit circle into its real (\(\cos\theta\)) and imaginary (\(i\sin\theta\)) components. In our exercise, we split \(z\) into \(x = \cos\theta\) and \(y = \sin\theta\).

• This transformation highlights how a seemingly intricate path like a circle can be expressed in a simplified way using trigonometric identities.
• Additionally, it provides an elegant framework to work with change in \(z\), giving us \(dz = ie^{i\theta} d\theta\), which streamlines our computations significantly.

Complex exponential functions thus serve as powerful tools to simplify and manage complex geometric and algebraic expressions in problems involving integrals over paths.

Integral Simplification Techniques

Simplifying integrals, especially complex ones, can save a great deal of computational effort. In our problem, once we substituted the expressions derived from our parametrization, the integral initially looked daunting: \( \int_{\pi}^{2\pi} (\cos^3\theta - i\sin^3\theta) (ie^{i\theta}) d\theta \).

Simplification techniques often involve distributing products and looking for symmetry or periodic properties in the integrand. Here are some strategies involved:

• Distribute terms like \(ie^{i\theta}\) within the integral to factorize and separate parts that may be easier to integrate individually.
• Recognize that certain terms become zero due to their periodicity or symmetry when integrated over a complete cycle.

In particular, in our case, understanding that integrating terms like powers of \(z\) (e.g., \(z^n\)) over a closed loop centered at the origin results in zero if \(neq -1\), significantly simplifies the final calculations. By leveraging these properties, the given integral becomes trivial to evaluate to zero. Integral simplification techniques therefore play a crucial role in efficiently solving complex line integrals without needing heavy calculations.