Problem 7 - Complex Analysis for Mathematics and Engineering

Question

Evaluate \(\int_{C}\left(4 z^{2}+4 z-3\right)^{-1} d z=\int_{C}(2 z-1)^{-1}(2 z+3)^{-1} d z\) for (a) the circle \(C=C_{1}^{+}(0)\). (b) the circle \(C=C_{1}^{+}\left(-\frac{2}{3}\right)=\left\\{z:\left|z+\frac{2}{3}\right|=1\right\\}\). (c) the circle \(C=C_{3}^{+}(0)\).

Step-by-Step Solution

Answer

(a) \(-\frac{\pi i}{2}\); (b) \(\frac{2\pi i}{5}\); (c) \(-\frac{\pi i}{10}\)."

Step 1: Transform the Integral

We begin by transforming the integral expression into a simpler form: \( \int_{C} \frac{1}{(2z-1)(2z+3)} \, dz \). This form allows us to apply the residue theorem as it highlights the poles of the integrand: \( z = \frac{1}{2} \) and \( z = -\frac{3}{2} \).

Step 2: Part (a) Evaluate Over Circle \( C=C_{1}^{+}(0) \)

For the circle \( C \) with center at 0 and radius 1, check the positions of the poles. Both poles \( z = \frac{1}{2} \) and \( z = -\frac{3}{2} \) relative to \( C \): \( z = \frac{1}{2} \) is inside the circle, but \( z = -\frac{3}{2} \) is outside. We only need to consider the pole inside. The residue at \( z = \frac{1}{2} \) is \( -\frac{1}{4} \), so the integral evaluates to \( 2\pi i \times \left(-\frac{1}{4}\right) = -\frac{\pi i}{2} \).

Step 3: Part (b) Evaluate Over Circle \( C=C_{1}^{+}\left(-\frac{2}{3}\right) \)

This circle has center \( -\frac{2}{3} \) and radius 1. The pole \( z = \frac{1}{2} \) is outside while \( z = -\frac{3}{2} \) is inside. Therefore, we consider only the pole \( z = -\frac{3}{2} \). The residue at this pole is \( \frac{1}{5} \), resulting in the integral being \( 2\pi i \times \frac{1}{5} = \frac{2\pi i}{5} \).

Step 4: Part (c) Evaluate Over Circle \( C=C_{3}^{+}(0) \)

The circle \( C \) with center 0 and radius 3 includes both poles \( z = \frac{1}{2} \) and \( z = -\frac{3}{2} \). The integral evaluates by considering residues at both poles. The sum of residues is \( -\frac{1}{4} + \frac{1}{5} = -\frac{1}{20} \). The integral evaluates to \( 2\pi i \times \left(-\frac{1}{20}\right) = -\frac{\pi i}{10} \).

Key Concepts

Contour Integration

Contour integration is a technique used in complex analysis to evaluate integrals of complex functions along a certain path or contour in the complex plane. This method is particularly useful when dealing with functions that have singularities, or poles, within the path of integration. The idea is to take advantage of the properties of complex functions, particularly their behavior near these singularities.

In our specific exercise, we dealt with contours in the form of circles. To understand the process better, consider contour integration as a way of walking along a path (the contour) around these points of interest (the poles).

• You start with an integral that might seem challenging to evaluate directly.
• Transform it into a form that shows off its poles.
• Identify the contours based on the area you want to traverse.
• Determine if these poles lie within these contours.

By doing this, we can apply powerful theorems such as Cauchy’s integral theorem and the residue theorem to compute the integral effectively. These steps emphasize the general outline of tackling complex integrals through contour integration.

Residue Theorem

The residue theorem is a key tool in complex analysis, offering a way to evaluate contour integrals by using residues at poles inside the contour. It simplifies the process of integration considerably when you have functions with isolated singularities, or poles.

The residue theorem states that if a function is analytic inside and on some closed contour except for a few isolated singularities, then the integral of the function over the contour is given by:\[ \oint_{C} f(z) \, dz = 2 \pi i \sum_{ \text{Residues inside } C} \text{Residue}(f, z_k)\]
This theorem was pivotal in solving the given exercise. By identifying the poles within the various contours and calculating residues at these points, we could determine the value of the integrals:

• For a simple contour with one pole inside, like in part (a), the result was a straightforward integration using the residue at that pole.
• In cases with only one relevant pole inside such as part(b), the theorem still directly allowed us to sum up the contributions from that pole only.
• In more complex scenarios involving multiple poles, the residues at each significant pole are summed, as seen in part (c).

The residue theorem makes what could be a daunting issue into a much simpler exercise in evaluation.

Complex Poles

Complex poles are particular points in the complex plane where a complex function approaches infinity. These poles play a crucial role in complex analysis because they dictate the behavior and characteristics of functions when performing contour integrations.

Identifying poles is the first step in utilizing residue calculus. To illustrate, the decomposition of our given integral revealed two poles at: \[ z = \frac{1}{2} \text{ and } z = -\frac{3}{2} \]
Once we've identified these poles, it’s about finding which ones lie within the contour of integration. The position of these poles relative to the contour determines whether they contribute to the integral’s value via the residue theorem.

• A pole inside the contour means it will affect the integral’s value.
• If outside the contour, it has no impact and can be ignored for that particular integration path.

Calculating the residue at these poles involves finding the coefficient of the \( \frac{1}{z - z_0} \) term in the Laurent series expansion of the function around the pole. The residues then plug directly into the residue theorem, showing their utility in simplifying complex integrals.