Problem 12 - Advanced Engineering Mathematics

Question

Expand \(f(z)=\frac{1}{z(z-3)}\) in a Laurent series valid for the indicated annular domain. \(1<|z+1|<4\)

Step-by-Step Solution

Answer

Expand function using partial fractions and series valid for \(1<|z+1|<4\).

Step 1: Express the function as partial fractions

We start by expressing the function \(f(z) = \frac{1}{z(z-3)}\) into partial fractions. We know that this can be written as \(\frac{1}{z(z-3)} = \frac{A}{z} + \frac{B}{z-3}\). By finding a common denominator and equating coefficients, we solve for \(A\) and \(B\). After solving, we find \(A = \frac{1}{3}\) and \(B = -\frac{1}{3}\). Thus, \(f(z) = \frac{1}{3z} - \frac{1}{3(z-3)}\).

Step 2: Adjust second term for annular domain

Our annular domain is \(1 < |z+1| < 4\). To fit this domain, we re-express \(-\frac{1}{3(z-3)} = -\frac{1}{3((z+1)-4)}\). This adjustment prepares for expansion around \(z+1\).

Step 3: Expand each term into series

First, \(\frac{1}{3z}\) can be expressed in terms of \(\frac{1}{3(z+1)+3-z}\) by using geometrical series and approximation techniques. Next, \(-\frac{1}{3((z+1)-4)}\) can be expanded using the geometric series \(\sum_{n=0}^{\infty} \frac{1}{4^{n+1}} (\frac{z+1}{4})^n\), assuming \(|\frac{z+1}{4}| < 1\).

Step 4: Combine and simplify the series

Combine the series from Step 3 into a single Laurent series by considering terms generated from each: \(\frac{1}{3z}\) generates terms close to zero and \(-\frac{1}{3((z+1)-4)}\) expands over the given range. Combine the series results to obtain the complete series.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a mathematical technique often employed to simplify the integration and appearance of rational functions. In our problem, we start with the function \[f(z) = \frac{1}{z(z-3)}\]and wish to express it in a form that makes it easier to handle, especially for series expansion purposes. The primary goal is to break down this fraction into simpler sums of fractions:

• Each fraction has a denominator which is a factor of the original denominator.
• The coefficients of these individual fractions are to be determined.

So, we rewrite it as follows:\[\frac{A}{z} + \frac{B}{z-3} = \frac{1}{z(z-3)}\]To find the coefficients \(A\) and \(B\), we equate and solve:- Cross-multiply to clear the fractions.- Set up an equation and compare coefficients of like terms to find \(A = \frac{1}{3}\) and \(B = -\frac{1}{3}\).This gives us a decomposed form that is valuable for further analysis, like expansion in series.

Complex Analysis

Complex analysis is a fascinating branch of mathematics, focused on functions that operate on complex numbers. It's the backbone of understanding the behavior of functions that cannot be simply interpreted in real-number theory.In the context of our problem, complex analysis provides the necessary framework:

• It helps in expanding functions using Laurent series or Taylor series, where denominators become complex.
• The ideas of radius and annular regions (think rings around a center) become essential.

For example, we express functions to explore their behavior in certain areas of the complex plane, matching conditions set for regions like our annular domain \(1 < |z+1| < 4\). This exercise showcases the power of complex analysis in exploring functions beyond elementary limits.

Geometric Series

A geometric series is a series of the form\[a + ar + ar^2 + ar^3 + \ldots\]with 'a' as the first term and 'r' the common ratio. When the absolute value of 'r' is less than 1, the series converges to a finite value.In solving our problem, geometric series play a crucial role:

• The series help us expand terms in fraction forms into manageable forms of power series.
• Using known series expansions, we approximate terms to fit within specific conditions (e.g., size of terms relative to the annular domain).

For instance, recognizing that a portion of our decomposition can use a geometric series lets us transform \(-\frac{1}{3((z+1)-4)}\)into a series by assuming \(\large |\frac{z+1}{4}| < 1\). This insight simplifies the handling of 'difficult' functions to manageable series for further combination.

Annular Domain

An annular domain in complex analysis refers to a region in the complex plane shaped like a ring. Essentially, it's the space between two concentric circles, characterized by their radius from a common center. In our problem's context:

• The annular domain is given by \(1 < |z+1| < 4\).
• This defines the region where the expansion needs to be valid.

Manipulating and expanding functions in such a domain is crucial:- We shift and scale terms to ensure they fit within these bounds.- For the term \(-\frac{1}{3((z+1)-4)}\), recognizing that the domain involves shifting by +1 changes the center to around z=-1, further maximizing the understanding of boundaries and behaviors.Understanding an annular domain helps ensure that series expansions and function manipulations remain valid over the specified range, providing a more global overview of the function's behavior.