Problem 7 - Advanced Engineering Mathematics

Question

(a) By Theorem 18.9 with \(f(z)=\frac{z^{2}}{z+2 i}\) \\[ \oint_{C} \frac{\frac{z^{2}}{z+2 i}}{\frac{z+2 i}{z-2 i}} d z=2 \pi i\left(-\frac{4}{4 i}\right)=-2 \pi \\] (b) By Theorem 18.9 with \(f(z)=\frac{z^{2}}{z-2 i}\) \\[ \oint_{C} \frac{\frac{z^{2}}{z-2 i}}{z-(-2 i)} d z=2 \pi i\left(\frac{-4}{-4 i}\right)=2 \pi \\]

Step-by-Step Solution

Answer

(a) Integral is \(-2\pi\); (b) Integral is \(2\pi\).

Step 1: Identify the Function and Contour

The exercise requires evaluating contour integrals using two different functions, both centered around the theorem 18.9. The contour integral given is around a closed path, denoted as \( C \). For each function, this involves solving a contour integral around \( C \). The functions are \( \frac{z^2}{z+2i} \) for part (a) and \( \frac{z^2}{z-2i} \) for part (b).

Step 2: Understanding Theorem 18.9

Theorem 18.9 suggests the utilization of the Cauchy's residue theorem, which calculates contour integrals based on the residues of the poles inside the contour. The formula used is \( \oint_C f(z) \, dz = 2\pi i \sum \text{Residues} \).

Step 3: Calculate the Residue for Part (a)

For the function \( \frac{z^2}{z+2i} \), the pole at \( z = -2i \) is considered. The residue is calculated by finding the limit: \[ \text{Res}_{z=-2i} \left( \frac{z^2}{z+2i} \right) = \lim_{z \to -2i} \frac{z^2}{z+2i} = (-2i)^2 = -4. \]

Step 4: Apply Theorem 18.9 to Part (a)

Plug the residue into the formula: \( 2\pi i \times (-\frac{4}{4i}) = -2\pi \). Thus, the integral for part (a) evaluates to \( -2\pi \).

Step 5: Calculate the Residue for Part (b)

For the function \( \frac{z^2}{z-2i} \), the pole at \( z = 2i \) is considered. The residue is computed through: \[ \text{Res}_{z=2i} \left( \frac{z^2}{z-2i} \right) = \lim_{z \to 2i} \frac{z^2}{z-2i} = (2i)^2 = -4. \]

Step 6: Apply Theorem 18.9 to Part (b)

Use the residue for \( z = 2i \): \( 2\pi i \times (-\frac{4}{-4i}) = 2\pi \). Thus, the integral for part (b) is \( 2\pi \).

Key Concepts

Contour Integrals

Contour integrals are a fundamental concept in complex analysis. They involve integrating a complex-valued function along a specific path or contour in the complex plane. This process is an extension of the idea of line integrals, which you might be familiar with from calculus.

• A contour is a curve or path, typically a closed path, that is traversed when performing the integration.
• Contour integrals are often denoted by the integral sign with a circle, \(\oint\), indicating that the path is closed.
• The function to be integrated, called the integrand, can have singularities — points where the function is not defined or behaves poorly.

When computing contour integrals, one important theorem that comes into play is the Cauchy's Residue Theorem. This theorem provides a way to evaluate the integral by examining the singularities (or poles) of the function inside the contour.

Complex Analysis

Complex analysis is the branch of mathematics that explores functions of complex numbers. These functions have both a real and an imaginary part, which opens up a fascinating world of possibilities and applications.

• One of the core ideas in complex analysis is that complex functions are often more tractable due to analytic properties, like being differentiable everywhere in their domain.
• This branch of math uses complex numbers, where each number has the form \(a + bi\), with real part \(a\) and imaginary part \(b\).
• Complex analysis investigates different types of singularities or special points of functions, such as poles.

A significant result in complex analysis is Cauchy's Residue Theorem, which allows us to compute complex integrals by looking just at these singular points (poles). It's particularly powerful because it simplifies otherwise complicated integrals.

Poles and Residues

Poles and residues are crucial concepts when evaluating contour integrals in complex analysis. They help simplify the integration process immensely.

Understanding Poles

A pole is a type of singularity of a complex function where the function goes to infinity. They are classified by their order, with a second order pole indicating a steeper climb to infinity than a first order pole.

• A simple pole is where the function behaves like something divided by \(z - a\).
• Poles are essential in calculating contour integrals, especially when using the Cauchy's Residue Theorem.

Exploring Residues

The residue of a function at a given pole is a number that represents the behavior or contribution of the pole to the integral.

• To find a residue at a simple pole, you can use the limit formula:
• \[ \text{Res}_{z=a} \left(f(z)\right) = \lim_{z \to a} (z-a)f(z) \]
• Residues are used in the residue theorem to calculate integrals efficiently. Only the residues of poles inside the contour contribute to the integral.

In the exercises provided, residues at poles are used to evaluate the contour integrals, illustrating the power and elegance of the residue theorem.