Problem 19 - Advanced Engineering Mathematics

Question

Expand \(f(z)=\frac{z}{(z+1)(z-2)}\) in a Laurent series valid for the indicated annular domain. . \(1<|z|<2\)

Step-by-Step Solution

Answer

Laurent series is \(\sum_{n=0}^{\infty} \left( \frac{1}{3} \left(-\frac{1}{z}\right)^n - \frac{2}{3} \left(\frac{z}{2}\right)^n \right)\).

Step 1: Partial Fraction Decomposition

To expand the function using a Laurent series, we start with a partial fraction decomposition. Given \( f(z) = \frac{z}{(z+1)(z-2)} \), we want to express it as:\[\frac{z}{(z+1)(z-2)} = \frac{A}{z+1} + \frac{B}{z-2}\]Multiplying through by \((z+1)(z-2)\), we get:\[z = A(z-2) + B(z+1)\]Expanding and equating coefficients gives the following equations:\[A + B = 1\]\[-2A + B = 0\]Solving these simultaneous equations, we find \(A = \frac{1}{3}\) and \(B = \frac{2}{3}\). Substituting back, we get:\[\frac{z}{(z+1)(z-2)} = \frac{1}{3(z+1)} + \frac{2}{3(z-2)}\]

Step 2: Expand Individual Fractions

For the Laurent series valid in \(1 < |z| < 2\), we need to expand each fraction appropriately:- Fraction 1: \(\frac{1}{3(z+1)}\) Since \(|z| > 1\), we can write: \[ \frac{1}{z+1} = \frac{1}{z} \cdot \frac{1}{1 + \frac{1}{z}} = \frac{1}{z} \sum_{n=0}^{\infty} \left(-\frac{1}{z}\right)^n = \sum_{n=0}^{\infty} \left(-\frac{1}{z}\right)^n \] Hence, \[ \frac{1}{3(z+1)} = \frac{1}{3} \sum_{n=0}^{\infty} \left(-\frac{1}{z}\right)^n \]

Step 3: Continue Expansion for Second Fraction

- Fraction 2: \(\frac{2}{3(z-2)}\)Since \(|z| < 2\), rewrite and expand using: \[ \frac{1}{z-2} = -\frac{1}{2} \cdot \frac{1}{1 - \frac{z}{2}} = -\frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n = -\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n \]Thus, \[ \frac{2}{3(z-2)} = \frac{2}{3} \cdot -\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n = -\frac{2}{3} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n \]

Step 4: Combine Series

Now we combine the expansions obtained from Steps 2 and 3:- From Fraction 1: \(\frac{1}{3} \sum_{n=0}^{\infty} \left(-\frac{1}{z}\right)^n\)- From Fraction 2: \(-\frac{2}{3} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\)The Laurent series is therefore:\[\sum_{n=0}^{\infty} \left( \frac{1}{3} \left(-\frac{1}{z}\right)^n - \frac{2}{3} \left(\frac{z}{2}\right)^n \right)\]

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to express a complex rational function as a sum of simpler fractions. This helps in expanding the function into a series, making complex analysis problems more manageable. In the given problem, we need to decompose the function \( f(z) = \frac{z}{(z+1)(z-2)} \). The idea is to rewrite it in the form \( \frac{A}{z+1} + \frac{B}{z-2} \). This step simplifies the process of finding the Laurent series for an annular domain.

To find the coefficients \( A \) and \( B \), we multiply both sides by the denominator \((z+1)(z-2)\), leading to the equation \( z = A(z-2) + B(z+1) \). By equating coefficients, we solve the resulting system of equations \( A + B = 1 \) and \(-2A + B = 0 \). Solving gives \( A = \frac{1}{3} \) and \( B = \frac{2}{3} \).

Partial fraction decomposition lays the groundwork for transforming our complex function into simpler components, making it easier to work with in complex analysis.

Complex Analysis

Complex analysis is the study of functions of a complex variable. It is a profound field of mathematics that extends methods of calculus to the complex plane, focusing on functions that are differentiable in the complex sense everywhere in their domain. A key concept here is the expansion of functions into power series, including Laurent series, particularly useful for functions with singularities.

The function \( f(z) = \frac{z}{(z+1)(z-2)} \) has singularities at \( z = -1 \) and \( z = 2 \). When dealing with complex functions in specific regions of the complex plane, knowing how to expand using series like Laurent series is crucial. These expansions allow us to represent functions around certain points or inside annular regions in the complex plane.

Complex analysis techniques, like power series expansions and partial fraction decomposition, provide tools to study and solve complex problems dealing with analytic properties, singularities, and integration over complex paths.

Annular Domain

An annular domain refers to a ring-shaped region in the complex plane. It is defined by two concentric circles, producing a region where a function can be expressed using series expansions suitable for annular nature. For the problem given, the annular domain is specified as \( 1 < |z| < 2 \). This means we are looking at a region outside the circle of radius 1 and inside the circle of radius 2 in the complex plane.

In the context of Laurent series, this annular domain indicates that we need a series expansion that works around the singularities and is valid only within this region. Each part of the function (resulting from partial fraction decomposition) must be expanded to fit within this domain. For example:

• \( \frac{1}{z+1} \) is expanded considering \( |z| > 1 \),
• \( \frac{1}{z-2} \) is expanded considering \( |z| < 2 \).

Working with annular domains requires careful consideration of how singularities affect function behavior in complex planes and demands suitable expansions that align with these specific regions.