Laurent Series Expansion for 1/ (z^2 - 4) Centered at 2i: Step-by-Step Guide

Laurent series, a fundamental concept in complex analysis, enable us to expand complex functions in a series of terms that capture both the positive and negative powers of the variable. This guide will walk through the process of computing the Laurent series for the function ( f(z) frac{1}{z^2 - 4} ) centered at ( z 2i ). Understanding this process will not only provide a valuable tool for further analysis but also deepen the understanding of complex function behavior in various applications.

1. Introduction to the Function and Its Singularities

Consider the function ( f(z) frac{1}{z^2 - 4} ). To compute its Laurent series centered at ( z 2i ), the first step is to identify the singularities of the function. These are the points where the function is undefined or diverges – in this case, the denominator is zero. Solving the equation ( z^2 - 4 0 ), we find the following singularities:

( z 2 ) ( z -2 )

Since we are centering the series at ( z 2i ), which is a different point, we note that ( z -2 ) is another singularity. This means the function has singularities at both ( z 2 ) and ( z -2 ), and we need to address how these singularities affect our expansion.

2. Rewriting the Function in Terms of ( z - 2i )

To begin the expansion, we rewrite the function ( f(z) ) in terms of the center of the series ( z 2i ):

[ f(z) frac{1}{z^2 - 4} frac{1}{(z - 2i)(z 2i)} ]

This representation allows us to separate the function into two parts, each with its own behavior around the singularity at ( z 2i ).

3. Partial Fraction Decomposition

Next, we use partial fraction decomposition to decompose the function into simpler terms:

[ frac{1}{z^2 - 4} frac{A}{z - 2i} frac{B}{z 2i} ]

By solving for ( A ) and ( B ), we get:

Setting ( z 2i ):

[ 1 A(2i 2i) Rightarrow 1 4Ai Rightarrow A -frac{i}{4} ]

Setting ( z -2i ):

[ 1 B(-2i - 2i) Rightarrow 1 -4Bi Rightarrow B -frac{i}{4} ]

Thus, we can write:

[ f(z) -frac{i}{4(z - 2i)} - frac{i}{4(z 2i)} ]

Given that ( z 2i (z - 2i) ), the second term can be expanded using a geometric series.

4. Expanding the Second Term Using a Geometric Series

To expand the second term, note that:

[ frac{1}{z 2i} frac{1}{2i (z - 2i)} frac{1}{2i}cdotfrac{1}{1 frac{z - 2i}{2i}} ]

This can be expanded as a geometric series, valid for ( left| frac{z - 2i}{2i} right| [ frac{1}{1 frac{z - 2i}{2i}} sum_{n0}^{infty} left( -frac{z - 2i}{2i} right)^n ]

Thus, the expanded term becomes:

[ frac{1}{z 2i} frac{1}{2i} sum_{n0}^{infty} left( -frac{z - 2i}{2i} right)^n ]

Since ( frac{1}{2i} -frac{i}{2} ), the expansion simplifies to:

[ frac{1}{z 2i} -frac{i}{2} sum_{n0}^{infty} left( -frac{z - 2i}{2i} right)^n ]

Substituting this back into our expression for ( f(z) ), we get:

[ f(z) -frac{i}{4(z - 2i)} - frac{i}{4}cdot -frac{i}{2} sum_{n0}^{infty} left( -frac{z - 2i}{2i} right)^n ]

Simplifying the coefficient:

[ f(z) -frac{i}{4(z - 2i)} frac{1}{16} sum_{n0}^{infty} left( -frac{z - 2i}{2i} right)^n ]

The first term is ( -frac{i}{4(z - 2i)} ), while the second term provides a series expansion for ( left| z - 2i right|

5. Final Laurent Series

The Laurent series for ( f(z) frac{1}{z^2 - 4} ) centered at ( z 2i ) is given by:

[ f(z) -frac{i}{4(z - 2i)} frac{1}{16} sum_{n0}^{infty} left( -frac{z - 2i}{2i} right)^n ]

This series converges for ( left| z - 2i right|

Keywords: Laurent series, complex analysis, function expansion, singularity