Problem 14 - A First Course in Complex Analysis with Applications

Question

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

Step-by-Step Solution

Answer

The Laurent series is \( \sum_{n=0}^{\infty} \frac{2^n - 1}{z^{n+1}} \)."

Step 1: Understanding the Problem

We need to expand \( f(z) = \frac{1}{(z-1)(z-2)} \) as a Laurent series valid in the domain \(|z| > 2\). This involves expressing \(f(z)\) in terms of powers of \(z^{-1}\) since \(|z| > 2\).

Step 2: Partial Fraction Decomposition

First, perform partial fraction decomposition on \( \frac{1}{(z-1)(z-2)} \). This can be expressed as \( \frac{A}{z-1} + \frac{B}{z-2} \). Solve for \(A\) and \(B\) by equating coefficients or substituting convenient values for \(z\).

Step 3: Finding Coefficients in Partial Fractions

Solving \( 1 = A(z-2) + B(z-1) \) by substitution: let \( z = 1 \), then \( 1 = A(1-2) \Rightarrow A = -1 \). Let \( z = 2 \), then \( 1 = B(2-1) \Rightarrow B=1 \). So, \( \frac{1}{(z-1)(z-2)} = \frac{-1}{z-1} + \frac{1}{z-2} \).

Step 4: Representing Negative Powers

Since \(|z| > 2\), for \( \frac{1}{z-1} \), multiply numerator and denominator by \(\frac{1}{z}\), giving \(\frac{-1/z}{1-1/z} = -\frac{1}{z}\sum_{n=0}^{\infty} \left( \frac{1}{z} \right)^n\).

Step 5: Representing Negative Powers for Second Term

Similarly, for \( \frac{1}{z-2} \), multiply numerator and denominator by \(\frac{1}{z}\), giving \(\frac{1/z}{1-2/z} = \frac{1}{z}\sum_{n=0}^{\infty} \left( \frac{2}{z} \right)^n\).

Step 6: Construct the Laurent Series

Combine the series expansions for the two terms to get the Laurent series: \(\sum_{n=0}^{\infty} \left( -\frac{1}{z^{n+1}} + \frac{2^n}{z^{n+1}} \right)\), which simplifies to \(\sum_{n=0}^{\infty} \left( \frac{2^n - 1}{z^{n+1}} \right)\).

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex numbers. Complex numbers are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, defined as \(i^2 = -1\). In complex analysis, we explore how these functions behave, including limits, continuity, derivatives, and integrals in the complex plane.

One core concept of complex analysis is the field's deep connection with power series and complex functions.

• Complex functions: Just like real functions, but they map complex numbers to complex numbers.
• Analytic functions: If a function is differentiable at a point in its domain, it is considered analytic.
• Laurent series: A representation that extends power series to include terms with negative exponents. This is particularly useful for handling complex functions that have singularities (points where a function is not defined).

Complex analysis is integral to solving the exercise at hand because it involves finding a Laurent series, which is closely tied to the singularities and behavior of complex functions beyond traditional domains.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down a complex rational function into simpler fractions that are more manageable for further analysis. This method is especially useful when integrating functions and finding power series expansions.

To decompose a rational function, you express it as a sum of fractions with simpler denominators. It's essential to find constants for these fractions.

• Identify the factors in the denominator: For instance, consider \(f(z) = \frac{1}{(z-1)(z-2)}\).
• Express as a sum of fractions: Write as \(\frac{A}{z-1} + \frac{B}{z-2}\).
• Solve for constants: Use substitution or algebraic manipulation to find \(A\) and \(B\).

For the given exercise, the decomposition helped transform the function into simpler terms, allowing us to perform further expansion using power series.

Power Series Expansion

Power series expansion is a technique for expressing a function as an infinite sum of terms in a series, each involving a power of a variable. In complex analysis, power series allow us to approximate complex functions and even handle infinite domains.

The key steps of power series expansion are:

• Identify the function to be expanded: Here, it’s the terms obtained from partial fraction decomposition.
• Express each term in the form of a power series: Instead of dealing with a fraction, rewrite it as a series involving powers of \(z^{-1}\).
• Simplify and combine series: Combine series from different fractions, in this case, using the Laurent series form for a specific annular domain.

For our exercise, this involved converting \(\frac{1}{z-1}\) and \(\frac{1}{z-2}\) into expressions involving \(z^{-1}\), which are more convenient given \(|z| > 2\). This forms the basis of the final Laurent series.

Annular Domain

An annular domain refers to a ring-shaped region in the complex plane, defined between two concentric circles. It's a crucial concept because it extends the usual path of operation in complex analysis from a simple disk to this more complex domain.

In our context, the annular domain is defined as \(|z| > 2\), which means we are working in the area outside the circle of radius 2. Understanding the concept of an annular domain is key for determining the validity of Laurent series expansions, especially as these series often converge only within specific ranges.

• Outer annular region: The area outside a circle of given radius.
• Inner annular region: The space inside a circle but larger than another, possibly smaller circle.
• Factors of convergence: The nature of the product's singularities within the annular domain affects the series outcome.

By focusing on \(|z| > 2\), the exercise narrows down where the Laurent series can converge, providing insights into how the function behaves in that particular area.