Problem 13 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The Laurent series for \( f(z) \) valid in \( 1 < |z| < 2 \) is \( - \sum_{n=0}^{\infty} z^{-(n+1)} + \sum_{n=0}^{\infty} \left(\frac{z}{2} \right)^n \).

Step 1: Factor the Expression

The given function is \( f(z) = \frac{1}{(z-1)(z-2)} \). Start by factoring this expression in a form that we can use to identify a Laurent series. This is already in a factored form, but we will manipulate to fit into partial fractions for easier expansion.

Step 2: Apply Partial Fraction Decomposition

Use partial fraction decomposition to rewrite the expression as sums of simpler fractions. Assume: \( \frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2} \). \( 1 = A(z-2) + B(z-1) \). Solve for A and B by substituting convenient values of \(z\).

Step 3: Solve for Coefficients A and B

Set \( z = 1 \): \( 1 = A(1-2) + B(1-1) \Rightarrow A = -1 \). Set \( z = 2 \): \( 1 = A(2-2) + B(2-1) \Rightarrow B = 1 \). The original expression becomes \( \frac{1}{(z-1)(z-2)} = \frac{-1}{z-1} + \frac{1}{z-2} \).

Step 4: Expand Each Fraction

The series expansion needs to be made valid within the annular region \(1 < |z| < 2\). For \( \frac{-1}{z-1} \), expand using \( |z-1| < 1 \) and for \( \frac{1}{z-2} \), expand using \( |z| < 2 \). For the term \( \frac{-1}{z-1} \): this requires rewriting \( \frac{1}{z-1} = -1/(1 - (z-1)) \) and for the term \( \frac{1}{z-2} \): rewrite into \( \frac{-1}{1 - (z/2)} \).

Step 5: Express Terms as Geometric Series

Recognizing \( \frac{1}{1-w} = 1 + w + w^2 + \ldots \), rewrite the expressions: \( \frac{1}{z-1} = - \sum_{n=0}^{\infty} (z-1)^n \) (for convergence, \(|z-1| < 1\)) and \( \frac{1}{z-2} = \sum_{n=0}^{\infty} \left( \frac{z}{2} \right)^n \) (for convergence, \(|z| < 2\)).

Step 6: Combine Series for Region 1 < |z| < 2

The valid region \( 1 < |z| < 2 \) implies we must combine the geometric series such that the net result converges in this region. After adjusting the terms, the combined series is: \(- \sum_{n=0}^{\infty} \left(\frac{1}{z^{n+1}} \right) + \sum_{n=0}^{\infty} \frac{z^n}{2^n}\).

Step 7: Write Final Laurent Series

Combine terms into a Laurent series valid for \(1 < |z| < 2\): \(f(z) = - \sum_{n=0}^{\infty} z^{-(n+1)} + \sum_{n=0}^{\infty} \left(\frac{z}{2} \right)^n \).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a critical technique when working with complicated rational functions, particularly for functions composed of non-linear terms. The goal is to express a given rational function as a sum of simpler fractions, which can then be individually manipulated and expanded easily.

Consider a rational function like \( f(z) = \frac{1}{(z-1)(z-2)} \). We can't easily expand this directly, so we decompose it into partial fractions. We assume a set of simpler fractions that would sum up to the original function, like \( \frac{A}{z-1} + \frac{B}{z-2} \). This assumption allows us to solve for constants \( A \) and \( B \) by multiplying both sides by the common denominator and comparing coefficients.

For \( f(z) = \frac{1}{(z-1)(z-2)} \), setting \( z = 1 \) and \( z = 2 \) helps calculate \( A \) and \( B \). From this, you find \( A = -1 \) and \( B = 1 \). This simplifies the original function to \( \frac{-1}{z-1} + \frac{1}{z-2} \).

Having simpler terms allows us to apply other mathematical techniques, such as geometric series expansions, to further simplify or solve the expression.

Geometric Series

A geometric series is one of the simplest types of infinite series, where each term is a constant multiple of the previous term. It is very useful for series expansion in complex analysis, especially when dealing with rational functions expanded within a given region.

When you have a function like \( \frac{1}{1-w} \), it can be expanded to the infinite series \( 1 + w + w^2 + \ldots \), provided \(|w| < 1\). This is critical when expanding the functions obtained from partial fraction decomposition.

For example, to expand \( \frac{-1}{z-1} \) in the region \( |z-1| < 1 \), we set \( w = z-1 \), giving us \( \frac{1}{1-(z-1)} = -\sum_{n=0}^{\infty} (z-1)^n \).

Similarly, \( \frac{1}{z-2} \) transforms with \( w = \frac{z}{2} \) into the series \( \sum_{n=0}^{\infty} \left( \frac{z}{2} \right)^n \) for \(|z| < 2\). These expansions are essential for finding a power series valid in the specific annular domain of \( 1 < |z| < 2 \).

Complex Analysis

Complex analysis studies functions of a complex variable and offers powerful techniques for solving problems in mathematics and engineering. The Laurent series is a fundamental tool in this field, particularly for functions with singularities, or points where they aren't well-defined.

In complex analysis, the Laurent series allows expansion of functions not just into powers of \( z \) (like with Taylor series), but also negative powers, accommodating more complex functional behaviors.

The solution for \( f(z) = \frac{1}{(z-1)(z-2)} \) in the annular domain \( 1 < |z| < 2 \) illustrates this. Partial fraction decomposition is applied, followed by geometric series expansions. These expansions are then combined to result in a full Laurent series valid for the specified region:

• \( - \sum_{n=0}^{\infty} \left( \frac{1}{z^{n+1}} \right) \) results from \( \frac{-1}{z-1} \)
• \( + \sum_{n=0}^{\infty} \frac{z^n}{2^n} \) results from \( \frac{1}{z-2} \)

This process highlights the power of combining algebraic manipulation with series expansions in complex analysis for developing solutions that cover intricate regions and characteristics.