Problem 19 - Advanced Engineering Mathematics

Question

$$ \text { In Problems 17-20, expand } f(z)=\frac{z}{(z+1)(z-2)} \text { in a Laurent } $$ $$ 1<|z|<2 $$

Step-by-Step Solution

Answer

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

Step 1: Factor Denominator

Write the denominator of the function as \(f(z) = \frac{z}{(z+1)(z-2)}\) and recognize that we need to break it into partial fractions for the expansion.

Step 2: Partial Fraction Decomposition

Express the function as a sum of simpler terms: \(\frac{z}{(z+1)(z-2)} = \frac{A}{z+1} + \frac{B}{z-2}\). Solve for \(A\) and \(B\). By equating \(z = A(z-2) + B(z+1)\), solve the system: \(Az - 2A + Bz + B = z\). This results in \(A + B = 1\) and \(-2A + B = 0\). Solving gives \(A = \frac{1}{3}\) and \(B = \frac{2}{3}\).

Step 3: Rewrite Partial Fractions

Insert the values from the partial fraction decomposition into the function: \(f(z) = \frac{1/3}{z+1} + \frac{2/3}{z-2}\).

Step 4: Series Expansion for Each Term

For \(\frac{1/3}{z+1}\), express it as a geometric series within the given annulus \(1<|z|<2\). Rewriting as \(\frac{1/3}{z(1+1/z)}\), this gives \(\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n(z^{-n})\). Similarly, express \(\frac{2/3}{z-2} = \frac{2/3}{z(1-2/z)}\) as \(\frac{2}{3z}\sum_{n=0}^{\infty}(2/z)^n \) for the given region.

Step 5: Combine Series

Combine the series from each term to form the Laurent series: \(\frac{1}{3z} \sum_{n=0}^{\infty} (-1)^n \, z^{-n} + \frac{2}{3z} \sum_{n=0}^{\infty} \left(\frac{2}{z}\right)^n = \sum_{n=0}^{\infty} \left(\frac{(-1)^n}{3z^{n+1}} + \frac{2}{3}\cdot\frac{2^n}{z^{n+1}}\right)\)

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex fractions into simpler components. This technique is especially useful in calculus and complex analysis when dealing with rational expressions.

• To decompose fractions, we express a single complicated fraction as a sum of simpler fractions.
• For instance, take the function \( \frac{z}{(z+1)(z-2)} \). By expressing it as \( \frac{A}{z+1} + \frac{B}{z-2} \), we transform it into parts that are easier to handle.

In the original problem, the system of equations \( A + B = 1 \) and \( -2A + B = 0 \) is solved to find the values of \( A \) and \( B \), which are \( \frac{1}{3} \) and \( \frac{2}{3} \) respectively. This decomposition is the first step in simplifying the function into a form suitable for further analysis.

Geometric Series

A geometric series is a series with a constant ratio between successive terms. In complex analysis, geometric series become a powerful tool for expressing functions as infinite sums.

• A geometric series has the form \( a + ar + ar^2 + ar^3 + \ldots \), where each term is multiplied by a constant "ratio" \( r \).
• This series converges when the absolute value of \( r \) is less than one, allowing us to express functions in a series form over a given domain.

To use geometric series in the original exercise, each decomposed fraction was rewritten so it could be expressed in this form. For the term \( \frac{1/3}{z+1} \), it can be expanded into \( \frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n(z^{-n}) \) when \( 1 < |z| < 2 \), following the properties of geometric series. This transforms our function into an infinite sum that suits the given region.

Complex Analysis

Complex analysis is a field of mathematics focusing on functions of complex numbers. It involves studying the properties and behaviors of these complex functions.

• One main feature of complex analysis is the expansion of functions into power series, such as Taylor or Laurent series.
• Laurent series are like Taylor series but allow for negative powers, enabling the expansion of functions with singularities within an annular region.

The function \( f(z) = \frac{z}{(z+1)(z-2)} \) is expanded using a Laurent series because the series expansion includes negative powers to accommodate the domain \( 1 < |z| < 2 \). This expansion is crucial in analyzing functions within specific boundaries, making complex analysis a vital tool in understanding the behavior of these functions across different regions in the complex plane.