Problem 18 - Advanced Engineering Mathematics

Question

$$\begin{aligned} f(z) &=\frac{1}{3(z+1)}+\frac{2}{3} \cdot \frac{1}{(z+1)-3}=\frac{1}{3(z+1)}+\frac{2}{3(z+1)} \cdot \frac{1}{1-\frac{3}{z+1}} \\ &=\frac{1}{3(z+1)}+\frac{2}{3(z+1)}\left(1+\frac{3}{z+1}+\frac{3^{2}}{(z+1)^{2}}+\frac{3^{3}}{(z+1)^{3}}+\cdots\right) \\\ &=\frac{1}{z+1}+\frac{2}{(z+1)^{2}}+\frac{2 \cdot 3}{(z+1)^{3}}+\frac{2 \cdot 3^{2}}{(z+1)^{4}}+\cdots \end{aligned}$$

Step-by-Step Solution

Answer

The function simplifies to an infinite series starting with \(\frac{1}{z+1} + \frac{2}{(z+1)^2} + \cdots\).

Step 1: Identify Initial Terms

First, identify the terms of the function given: \(f(z) = \frac{1}{3(z+1)} + \frac{2}{3} \cdot \frac{1}{(z+1) - 3}\). Separate the two parts: the first fraction \(\frac{1}{3(z+1)}\) and the product of the second term.

Step 2: Simplify Second Term

Simplify \(\frac{2}{3} \cdot \frac{1}{(z+1) - 3}\). It becomes \(\frac{2}{3(z+1)} \cdot \frac{1}{1 - \frac{3}{z+1}}\). This represents a geometric series expansion.

Step 3: Apply Geometric Series Expansion

Use the formula \(\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \) for the geometric series, where \(x = \frac{3}{z+1}\). Thus, \(\frac{1}{1-\frac{3}{z+1}} = 1 + \frac{3}{z+1} + \frac{3^2}{(z+1)^2} + \cdots \).

Step 4: Multiply Series by \(\frac{2}{3(z+1)}\)

Multiply each term of the series by \(\frac{2}{3(z+1)}\). This gives: \(\frac{2}{3(z+1)}\cdot1 + \frac{2\cdot3}{3(z+1)^2} + \frac{2\cdot 3^2}{3(z+1)^3} + \cdots\).

Step 5: Add Initial Term and Series Expansion

Combine the \(\frac{1}{3(z+1)}\) with the series obtained in the previous step: \(\frac{1}{3(z+1)} + \frac{2}{3(z+1)} + \frac{2\cdot 3}{3(z+1)^2} + \cdots\). Simplifying, we get \(\frac{1}{z+1} + \frac{2}{(z+1)^2} + \frac{2\cdot 3}{(z+1)^3} + \cdots\).

Key Concepts

Geometric Series

A geometric series is a series of numbers with a constant ratio between successive terms. In our exercise, the series expansion of \(\frac{1}{1-x}\) plays a major role. This formula states that for \(|x| < 1\), the series will converge to \(1 + x + x^2 + x^3 + \cdots\). Here, the constant ratio is \(x\) and each term is multiplied by this ratio to generate the subsequent term.
Whenever you encounter a function like \(\frac{1}{1-x}\), it’s like finding a secret lever that unlocks an infinite series, connecting complex numbers with a simple pattern. This concept is crucial when dealing with problems in complex analysis because it helps to simplify complicated expressions into an infinite sum of terms that are much easier to manage.

• The geometric series converges only if \(|x| < 1\).
• It provides a straightforward path to represent functions as sums, which can be extraordinarily powerful in analysis.

Understanding how to apply geometric series lets you tackle complex problems and aids in spotting patterns within different functions.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. This method is incredibly handy for handling integrals and series expansions. In our function, we see this in action when the expression \(\frac{2}{3} \cdot \frac{1}{(z+1)-3}\) is separated into parts.
The idea here is to express the given function in terms of simpler fractions that can be more easily manipulated. This often involves taking a complex fraction and splitting it into a sum of simpler fractions with linear denominators. For instance, a rational function can often be expressed as a sum of fractions like \(\frac{A}{x+B}\).

• This method simplifies complex expressions, making them easier to integrate or differentiate.
• Decomposition is particularly useful in both real and complex analysis to solve integrals more efficiently.

Being able to decompose fractions allows for a more in-depth understanding of how functions behave, as well as facilitating the integration process when tackling empirical or theoretical problems.

Series Expansion

Series Expansion refers to expressing a function as the sum of terms in a series. This is essential in complex analysis, allowing complicated functions to be expressed in terms of their series. In this problem, after applying the geometric series expansion for \(\frac{1}{(z+1)-3}\), we get an infinite series: \(\frac{1}{3(z+1)} + \frac{2}{3(z+1)} \left(1 + \frac{3}{z+1} + \frac{3^2}{(z+1)^2} + \cdots \right)\).
By multiplying the calculated series with \(\frac{2}{3(z+1)}\), we expand the function further. You merge initial terms with the newly expanded series, turning it into: \(\frac{1}{z+1} + \frac{2}{(z+1)^{2}} + \frac{2 \cdot 3}{(z+1)^{3}} + \cdots\).

• Series expansion allows for precise approximations and expressions for functions over an interval.
• It is particularly useful in deducing the behavior of functions in limits, differentiability, and integration.

The beauty of series expansion is in its ability to provide insight into the inner workings of complex functions, turning them into manageable expressions that reveal their core properties.