Problem 452 - Advanced Calculus

Question

Evaluate \(^{1} \int_{-1} z^{1 / 2} d z\) where \(z^{1 / 2}=\sqrt{\mid z} \mid e^{i \theta / 2}\) where \(0<\theta<2 \pi\) along any contour lying in the upper half plane and also along any contour lying in the lower half plane. Use indefinite integrals to do this.

Step-by-Step Solution

Answer

The given integral can be expressed as an indefinite integral involving polar form of complex numbers and separated into real and imaginary parts: \(\int z^{1 / 2} d z = \int r^{1/2} dr e^{i3\theta / 2} + \int ir^{3/2} d\theta e^{i5\theta / 2}\). Evaluating along contours in the upper and lower half-planes, we have: Upper half-plane: \(\int_{-1}^1 r^{1/2} dr e^{i3\theta / 2}\Big|_0^\pi + \int_0^\pi ir^{3/2} d\theta e^{i5\theta / 2}\) Lower half-plane: \(\int_{-1}^1 r^{1/2} dr e^{i3\theta / 2}\Big|_\pi^{2\pi} + \int_\pi^{2\pi} ir^{3/2} d\theta e^{i5\theta / 2}\) Evaluate these integrals using standard integration techniques and compare the results along the contours in the upper and lower half-planes.

Step 1: Express the square root function in a familiar form

The square root function is given as: \( z^{1 / 2}=\sqrt{\mid z} \mid e^{i \theta / 2} \) where \(0< \theta < 2\pi\). Rewrite this expression using the polar form of a complex number. Let \(z = re^{i\theta}\), where \(r = \mid z \mid\) and \(0< \theta < 2\pi\). Then, the square root function becomes: \( z^{1 / 2} = \sqrt{r}e^{i\theta / 2} \)

Step 2: Express the given integral using indefinite integrals

Now, express the integral as an indefinite integral: \( \int z^{1 / 2} d z = \int \sqrt{r}e^{i\theta / 2} d(re^{i \theta}) \) We'll need to separate the real and imaginary parts of \(dz\). Recall that if \(z = re^{i\theta}\), then: \( dz = dre^{i\theta} + ire^{i\theta}d\theta \) Thus, the integral becomes: \( \int \sqrt{r}e^{i\theta / 2} (dre^{i\theta} + ire^{i\theta}d\theta) \)

Step 3: Evaluate the indefinite integral in terms of \(r\) and \(\theta\)

Now we can evaluate the indefinite integral for each component \(dr\) and \(d\theta\). First, consider the \(dr\) component: \( \int \sqrt{r}e^{i\theta / 2} dre^{i\theta}= \int r^{1/2} dr e^{i3\theta / 2} \) Now, evaluate the \(d\theta\) component: \( \int \sqrt{r}e^{i\theta / 2} ire^{i\theta}d\theta = \int ir^{3/2} d\theta e^{i5\theta / 2} \)

Step 4: Combine the two integrals

We now have two integrals for the problem, one involving \(dr\) and the other involving \(d\theta\). Now, we'll combine these two integrals to solve the problem: \(\int z^{1 / 2} d z = \int r^{1/2} dr e^{i3\theta / 2}+ \int ir^{3/2} d\theta e^{i5\theta / 2}\)

Step 5: Evaluate along the contours

Now, evaluate this integral along contours in the upper half-plane and lower half-plane. In both cases, the limits of integration for \(r\) are \(1\) and \(-1\). In the upper half-plane, the limits of integration for \(\theta\) are \(0\) and \(\pi\), while in the lower half-plane, they are \(\pi\) and \(2\pi\). Upper half-plane: \(\int_{-1}^1 r^{1/2} dr e^{i3\theta / 2}\Big|_0^\pi + \int_0^\pi ir^{3/2} d\theta e^{i5\theta / 2}\) Lower half-plane: \(\int_{-1}^1 r^{1/2} dr e^{i3\theta / 2}\Big|_\pi^{2\pi} + \int_\pi^{2\pi} ir^{3/2} d\theta e^{i5\theta / 2}\) Now, evaluate these integrals using standard integration techniques and compare the results along the contours in the upper and lower half-planes.

Key Concepts

Contour Integration

Contour integration is a powerful method in complex analysis for evaluating integrals along a path in the complex plane. It involves integrating a complex function along a contour, which is a curve on the complex plane. Unlike real integrals, contour integrals take into account the behavior of the complex function around singularities and the path taken.

In our exercise, we considered the contour integrations in the upper and lower half-planes. Contours in these planes entail different ranges of the angle \(\theta\). For the upper half-plane, \(\theta\) ranges from \(0\) to \(\pi\); for the lower half-plane, it ranges from \(\pi\) to \(2\pi\).

While solving contour integrals, the

• choice of contour path matters significantly
• complex functions and expressions derive from the path
• singularities within the path can contribute significantly to the integral's value

Thus, knowing how to properly set up and evaluate contour integrals allows us to solve complex problems efficiently.

Indefinite Integrals

An indefinite integral represents a family of functions that differ by a constant. In calculus, it is represented by integrating a function without specific bounds. The result is expressed as an antiderivative plus an arbitrary constant C.

In complex integration, indefinite integrals play a crucial role as they help in decomposing the problem into simpler parts to tackle. During the solution of the exercise, the integral was expressed as:

• \( \int z^{1 / 2} d z \)

This involves finding the antiderivative of \( z^{1/2} \), which was further split into real and imaginary components for ease of calculation.

Working with indefinite integrals in complex numbers means:

• Solving the real and imaginary parts separately
• Understanding how functions behave without fixed limits
• Allowing flexibility for boundaries determining overall path integration

Employing indefinite integrals provides a foundation for setting up the contour integration later in the exercise.

Polar Form of Complex Numbers

The polar form of complex numbers allows us to express complex numbers in terms of magnitude and direction, making certain calculations more straightforward. A complex number \(z\) can be written as \(z = re^{i\theta}\), where \(r\) is the magnitude of \(z\) and \(\theta\) is the angle with the positive real axis.

In this problem, understanding the polar form was vital to express the square root \(z^{1/2}\) as \( \sqrt{r}e^{i\theta/2}\). When dealing with roots or powers of complex numbers, the polar form provides clarity:

• Magnitude \(r\) is easily computed
• Angle \(\theta\) helps adjust the phase correctly

Knowing that this form aids in simplifying multiplication and division operations, as well as integration, it streamlines the involved process making it less prone to algebraic errors.

Upper and Lower Half-Plane Analysis

The analysis of expressions in the upper and lower half-planes is significant in contour integration. The complex plane is split into two by the real axis:

• The upper half-plane (where the imaginary part \(\text{Im}(z) > 0\))
• The lower half-plane (where \(\text{Im}(z) < 0\))

In the exercise, varying the contour through these different regions required adjustments in the \(\theta\) limits. For example:

• The upper half-plane ran from \(\theta = 0\) to \(\theta = \pi\)
• The lower half-plane ran from \(\theta = \pi\) to \(\theta = 2\pi\)

Why is this crucial?

• Each half-plane can influence the functional behavior of the contour paths.
• Understanding the changes in \(\theta\) ensures correct direction and value computation associated with paths.
• Analysis of both halves can reveal symmetries or unique behaviors within the complex function being integrated.

Grasping these concepts solidifies the ability to evaluate and interpret integrals accurately across diverse contours.