Problem 20 - Advanced Engineering Mathematics

Question

\(\oint_{C} \frac{1}{z^{3}+2 i z^{2}} d z ;|z|=1\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi i}{2} \).

Step 1: Simplify the Integrand

First, factor the denominator of the integrand. We have \( z^3 + 2iz^2 = z^2(z + 2i) \). This factorization will help identify the poles of the integrand.

Step 2: Identify the Poles and Check if They Lie Inside the Contour

The integrand has poles at \( z = 0 \) (a double pole) and \( z = -2i \) (a simple pole). Since the contour is the unit circle \(|z| = 1\), check which of these poles lie inside the contour. \( -2i \) has a modulus of 2, so it lies outside the contour, while \( 0 \) lies inside since its modulus is 0.

Step 3: Apply the Residue Theorem

Since the only pole inside the contour is at \( z = 0 \), we will apply the Residue Theorem: \[\oint_{C} f(z) \, dz = 2 \pi i \sum \text{Res}(f, a_k),\]where \(a_k\) are the poles inside \(C\). Calculate the residue at \( z = 0 \).

Step 4: Calculate the Residue at Pole \( z = 0 \)

The residue at a pole of order 2 (double pole) at \( z = 0 \) is given by the formula: \[\text{Res}(f, 0) = \lim_{z \to 0} \frac{d}{dz}\left[ (z - 0)^2 \frac{1}{z^2(z + 2i)} \right].\] Simplify the expression: \[(z^2)(z^2 + 2iz)^{-1} = \frac{1}{z + 2i}.\]Differentiate \( \frac{1}{z + 2i} \) with respect to \(z\), and then take the limit as \( z \to 0 \): \[\frac{d}{dz}\left( \frac{1}{z + 2i} \right) = -\frac{1}{(z + 2i)^2},\]evaluate at \(z = 0\) gives \(-\frac{1}{(2i)^2} = -\frac{1}{-4} = \frac{1}{4}.\)

Step 5: Evaluate the Contour Integral Using Residue

With the residue found, apply it to the integral using the Residue Theorem:\[\oint_{C} \frac{1}{z^3 + 2iz^2} \, dz = 2 \pi i \cdot \frac{1}{4} = \frac{\pi i}{2}.\]Hence, the value of the integral is \( \frac{\pi i}{2} \).

Key Concepts

Contour Integration

In complex analysis, one powerful tool for evaluating integrals is contour integration. This method is particularly useful for integrals of complex functions over a specific path, called a contour. Think of a contour as a path you can draw on the complex plane.
The concept revolves around integrating a complex function along a designated path, usually a curve or a closed loop. This path is known as the contour. When working with contour integrals, it's important to consider the function's singularities (or poles) and how they interact with the chosen path.

• Contour integration helps in simplifying integrals by transforming them into something more manageable.
• Understanding the path or contour is crucial, as it determines which singularities affect the outcome.
• It's especially useful in complex analysis, where traditional methods may fail.

Contour integration is often paired with other powerful tools, such as the Residue Theorem and Cauchy's Integral Theorem, to solve complex integrals.

Residue Theorem

The Residue Theorem is a cornerstone of complex analysis, particularly in evaluating complex integrals. It provides a powerful technique to find the value of contour integrals by considering the residues of a function's poles that lie inside the contour.
Essentially, the theorem states that for a function with isolated singularities (poles) inside a closed contour, the integral can be calculated by summing up the residues of these singularities. Mathematically, it's expressed as:\[\oint_{C} f(z) \, dz = 2 \pi i \sum \text{Res}(f, a_k),\]where \(a_k\) are the poles within the contour.

• Residues provide vital information for evaluating integrals over closed paths.
• The theorem simplifies complex integration tasks by converting them into a simple sum of residues.
• It allows us to ignore poles that lie outside the chosen path, reducing unnecessary complexity.

This makes Residue Theorem invaluable for tackling tricky integrals and is an indispensable tool for anyone learning complex analysis.

Poles and Residues

Within complex analysis, poles and residues are crucial concepts essential for understanding how functions behave around singular points.

• A pole is a point where a function takes on an infinite value, essentially a location where the function "blows up." These are classified as simple (order 1), double (order 2), or higher order poles.
• The residue of a function at a pole gives the coefficient of \((z-a)^{-1}\) in its Laurent series expansion around that pole \(a\).

In the context of solving integrals, residues are like shortcuts. They allow us to compute the value at poles without manually evaluating complex integrals over contours.
For instance, in the given exercise, understanding that you have a double pole at \(z=0\) impacts how you evaluate the contour integral using the Residue Theorem.

Cauchy Integral Theorem

The Cauchy Integral Theorem is one of the foundational results in complex analysis and serves as the basis for several other important theorems, including the Residue Theorem.
This theorem states that if a function is holomorphic (meaning it is complex differentiable everywhere) within and on some closed contour \(C\), then the integral of that function over this contour is zero:\[\oint_{C} f(z) \, dz = 0.\]
This property makes working with holomorphic functions incredibly neat as it simplifies the process of determining the outcome of integrals involving these functions around closed paths. The function's differentiability ensures all its derivatives exist, leading to smooth behavior without any sudden infinity points, except possibly at isolated singularities that the theorem helps to bypass.
Understanding this theorem is crucial because it underpins why and how contour integration works for holomorphic functions and sets the stage for knowing when and how to apply the Residue Theorem effectively.