Problem 8 - A First Course in Complex Analysis with Applications

Question

Expand \(f(z)=\frac{1}{z(z-3)}\) in a Laurent series valid for the indicated annular domain. $$ |z|>3 $$

Step-by-Step Solution

Answer

The Laurent series for \(f(z)\) in \(|z| > 3\) is \( \sum_{n=1}^{\infty} \frac{3^n}{3z^{n+1}} \).

Step 1: Express the Function in Partial Fraction Form

First, we decompose the given function into partial fractions. We have:\[f(z) = \frac{1}{z(z-3)} = \frac{A}{z} + \frac{B}{z-3}\]To find the constants \(A\) and \(B\), we multiply through by the denominator \(z(z-3)\) to obtain:\[1 = A(z - 3) + Bz\]

Step 2: Solve for Constants A and B

We expand and solve the equation:\[1 = Az - 3A + Bz\]Equate the coefficients:- For \(z\): \(A + B = 0\)- Constant term: \(-3A = 1\)From \(-3A = 1\), we find \(A = -\frac{1}{3}\). Substituting \(A\) into the equation \(A + B = 0\), we get \(-\frac{1}{3} + B = 0\), so \(B = \frac{1}{3}\).

Step 3: Rewrite the Function with Constants

Substitute the values for \(A\) and \(B\) back into the partial fraction decomposition:\[f(z) = \frac{-\frac{1}{3}}{z} + \frac{\frac{1}{3}}{z-3}\]So, it becomes:\[f(z) = -\frac{1}{3z} + \frac{1}{3(z-3)}\]

Step 4: Express \(\frac{1}{z-3}\) as a Geometric Series

To express \(\frac{1}{3(z-3)}\) for \(|z| > 3\), note that\[\frac{1}{z-3} = \frac{1}{z} \cdot \frac{1}{1 - \frac{3}{z}}\]For \(|z| > 3\), \(\left|\frac{3}{z}\right| < 1\), so we use the geometric series expansion:\[= \frac{1}{z} \sum_{n=0}^{\infty} \left(\frac{3}{z}\right)^n\]

Step 5: Combine the Series to form the Laurent Series

Substitute the geometric series back into the function:\[f(z) = -\frac{1}{3z} + \frac{1}{3} \left( \frac{1}{z} \sum_{n=0}^{\infty} \left(\frac{3}{z}\right)^n \right)\]This becomes:\[= -\frac{1}{3z} + \frac{1}{3z} \sum_{n=0}^{\infty} \left(\frac{3}{z}\right)^n\]Thus, the expression:\[= \frac{1}{3z} \left( -1 + \sum_{n=0}^{\infty} \left(\frac{3}{z}\right)^n \right)\]Expressing this using a single summation, we find:\[f(z) = \sum_{n=1}^{\infty} \frac{3^n}{3z^{n+1}}\]

Key Concepts

Partial Fraction Decomposition

Understanding partial fraction decomposition is crucial when dealing with complex rational functions. It allows us to split a complex function into simpler parts. For the function \( f(z) = \frac{1}{z(z-3)} \), partial fraction decomposition can make our life easier.
We separate \( f(z) \) into simpler fractions:

• \( \frac{A}{z} \)
• \( \frac{B}{z-3} \)

Here, the goal is to find \( A \) and \( B \) that satisfy this equation. We do this by multiplying the equation by the denominator \( z(z-3) \), simplifying, and then comparing coefficients. This helps isolate constants \( A \) and \( B \) effectively. Such decomposition is particularly helpful for finding Laurent series, which brings simplicity to the solution.

Geometric Series

A geometric series is a series of numbers with a constant ratio between consecutives. In complex analysis, this concept helps us express parts of functions for series expansions. To use a geometric series, the value \( \left| \frac{3}{z} \right| \) must be less than 1, which is true in our domain \( |z| > 3 \). This leads to the series:

• \( \frac{1}{z} \cdot \frac{1}{1 - \frac{3}{z}} \), expressed as \( \frac{1}{z} \sum_{n=0}^{\infty} \left( \frac{3}{z} \right)^n \)

It allows expressing seemingly complex functions in a form where every term is a power of \( z \), which is critical for building Laurent series. Such series help evaluate and simplify functions over specified domains.

Complex Analysis

Complex analysis deals with functions of complex variables and provides tools for manipulating and studying these functions. In expanding \( f(z) = \frac{1}{z(z-3)} \) into a Laurent series, we apply complex analysis concepts, including:

• Understanding domains of convergence like \( |z| > 3 \).
• Utilizing analytical methods to separate and transform functions into usable forms, like partial fraction decomposition and geometric series.

Complex analysis not only aids in finding solutions but also in understanding the behavior of functions. It helps us manage and simplify expressions that appear intricate at first glance. This insight is invaluable in deeper mathematical fields.

Annular Domain

An annular domain is a ring-shaped region in the complex plane, important for determining where a function behaves analytically. For our function \( f(z) = \frac{1}{z(z-3)} \) and the condition \( |z| > 3 \), it refers to the region outside a circle of radius 3.
This domain affects how we construct the Laurent series:

• We ensure that the series converges within the specified region.
• Deciding the form and validity of the series over this annular space matters.

It’s crucial to understand such domains because they tell us where our function representations are accurate. Working within specified annular domains ensures that our solutions remain valid within the intended scope.