Problem 15 - Advanced Engineering Mathematics

Question

Expand \(f(z)=\frac{1}{(z-1)(z-2)}\) in a Laurent series valid for the indicated annular domain. \(0<|z-1|<1\)

Step-by-Step Solution

Answer

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

Step 1: Partial Fraction Decomposition

The function \( f(z) = \frac{1}{(z-1)(z-2)} \) can be expressed using partial fraction decomposition as follows:\[ f(z) = \frac{A}{z-1} + \frac{B}{z-2} \].To find \( A \) and \( B \), equate \( 1 = A(z-2) + B(z-1) \). Solving this, set \( z = 1 \) to find \( A \):\[ A(1-2) = 1 \Rightarrow A = -1. \]Set \( z = 2 \) to find \( B \): \[ B(2-1) = 1 \Rightarrow B = 1. \]Thus, \( f(z) = \frac{-1}{z-1} + \frac{1}{z-2} \).

Step 2: Expand Terms into Series

For the domain \( 0<|z-1|<1 \), the point \( z = 2 \) is outside the circle. We need to expand the term \( \frac{1}{z-2} \) in terms of \( z-1 \).This can be done by manipulation:\[ \frac{1}{z-2} = \frac{1}{(z-1) - 1} = \frac{1}{-(1 - (z-1))} = - \frac{1}{1 - (z-1)}. \]This can be expanded using the geometric series formula for \(|z-1|<1\):\[ \frac{1}{z-2} = -\sum_{n=0}^{\infty} (z-1)^n = -1 - (z-1) - (z-1)^2 - \cdots. \]

Step 3: Combine Series

Combine the series expansions from the partial fraction decomposition:\[ f(z) = \frac{-1}{z-1} + \left( -1 - (z-1) - (z-1)^2 - \cdots \right). \]This simplifies to:\[ f(z) = \frac{-1}{z-1} - 1 - (z-1) - (z-1)^2 - \cdots. \]Rewriting this in terms of a single series:\[ f(z) = -\sum_{n=0}^{\infty} (z-1)^n - \frac{1}{z-1}. \]

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions. This method makes it easier to analyze or integrate the function. Imagine you have a complex fraction that seems too complicated to work with directly. Partial fraction decomposition helps by expressing it as a sum of smaller, easier-to-handle fractions.

Let's see how we used this method with the given expression:

• The function is initially given as \(f(z) = \frac{1}{(z-1)(z-2)}\).
• The goal is to express it in the form \(\frac{A}{z-1} + \frac{B}{z-2}\).
• By matching coefficients, we can solve for \(A\) and \(B\). We set up the equation \(1 = A(z-2) + B(z-1)\), then solve it by substituting appropriate values for \(z\) to isolate each constant separately.
• Setting \(z = 1\) gives us \(A = -1\) and setting \(z = 2\) gives us \(B = 1\).

This splits our function into simpler terms, namely \(\frac{-1}{z-1} + \frac{1}{z-2}\), making it easier to manipulate, especially for expansions.

Geometric Series

A geometric series is a series with a constant ratio between successive terms. It is a powerful tool in various fields of mathematics, including complex analysis, because of its straightforward formula.

In our exercise, we encounter a situation where we need to expand a fraction using a geometric series. Here’s how:

• We focus on the expression \(\frac{1}{z-2}\), where we set up a substitution to transform this into a geometric series.
• We treat it as \(\frac{1}{(z-1)-1}\), which can then be rewritten to fit the form \(\frac{1}{1 - (z-1)}\).
• This aligns nicely with the geometric series formula \(\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\), for \(|x| < 1\).
• Therefore, in this context, \(\frac{1}{z-2}\) expands into the series \(-\sum_{n=0}^{\infty} (z-1)^n\).

Understanding geometric series allows us to convert our transformed algebraic expressions into infinite series that are simpler to comprehend and compute.

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex numbers. It’s essential in many advanced areas of mathematics and engineering.

When dealing with complex functions, we often explore their properties using different series and decompositions. Here’s how complex analysis plays into our solution:

• We start by considering the Laurent series, a tool used for expressing such functions as a series that includes negative powers.
• In our solution, we desire a series valid on an annular region around \(z = 1\), specifically \(0<|z-1|<1\).
• The process involved combining the series we've expanded from partial fractions, particularly focusing on how each term behaves in the specified domain.
• In this case, the function \(f(z) = \frac{-1}{z-1} - \sum_{n=0}^{\infty} (z-1)^n\) provides the desired representation in the given annulus.

In essence, complex analysis gives us the tools to understand and manipulate complex functions, allowing us to convert problems into more manageable forms via series like the Laurent series.