Question

Find the Laurent series for the following functions about the indicated points; hence find the residue of the function at the point. (Be sure you have the Laurent series which converges near the point.) $$\frac{1}{z(z+1)}, z=0$$

Short answer

The Laurent series is \(\frac{1}{z} - \sum_{n=0}^{\infty} (-1)^n z^n\) and the residue at \(z=0\) is 1.

Step-by-step solution

Express the function

Rewrite the given function to identify its singularity at the point of interest. \[\frac{1}{z(z+1)}\] The singularities are at \(z=0\) and \(z=-1\).

Decompose the function

Use partial fraction decomposition to break down the function. \[\frac{1}{z(z+1)} = \frac{A}{z} + \frac{B}{z+1}\] To find A and B, solve the equation: \[1 = A(z+1) + Bz\] Setting \(z=0\): \[1 = A \Rightarrow A = 1\] Setting \(z=-1\): \[1 = -B \Rightarrow B = -1\] Therefore, \[\frac{1}{z(z+1)} = \frac{1}{z} - \frac{1}{z+1}\]

Expand the second term in a series

Expand \(\frac{1}{z+1}\) in a Laurent series about \(z=0\). Since \(\frac{1}{z+1} = \frac{1}{1+z}\), we can use the geometric series expansion: \[\frac{1}{1+z} = \sum_{n=0}^{\infty} (-1)^n z^n\] Thus, \[\frac{1}{z+1} = \sum_{n=0}^{\infty} (-1)^n z^n\]

Construct the Laurent series

Now combine the terms to construct the Laurent series: \[\frac{1}{z(z+1)} = \frac{1}{z} - \sum_{n=0}^{\infty} (-1)^n z^n\]

Identify the residue

The residue is the coefficient of \(\frac{1}{z}\) in the Laurent series. Here, it is: 1

Key concepts

Laurent series

The Laurent series is a representation of a complex function as a series expansion which includes both positive and negative powers of the variable. This series is particularly useful when dealing with functions that have singularities, like the point at which the function becomes undefined or infinite.
In the context of our problem, we want to find the Laurent series for \frac{1}{z(z+1)} at the point \(z=0\). A Laurent series differs from a Taylor series because it allows for terms with negative exponents, which means it can accommodate functions with singularities.
The steps to finding the Laurent series usually involve:

• Identifying singularities.
• Decomposing the function, often using partial fractions.
• Expanding each part in a series around the point of interest.

In our problem, we decomposed \frac{1}{z(z+1)} into \frac{1}{z} - \frac{1}{z+1} and then expanded \frac{1}{z+1} as a geometric series.

Complex functions

Complex functions are functions that involve complex numbers, where a complex number is any number of the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, defined by \(i^2 = -1\).
Functions of complex variables have rich structures and properties which differ in certain ways from functions of a real variable.
For example, the function \frac{1}{z(z+1)} is defined everywhere in the complex plane except at the points where the denominator is zero, which are \(z = 0\) and \(z = -1\). These points are known as singularities. Understanding these singularities is crucial when working with complex functions.
This problem required us to understand the behavior of the given function around its singularity at \(z = 0\), and hence, find a series that helps in analyzing its behavior near this point. Using partial fractions translation and series expansion, we can easily manage such complex functions.

Residue

The residue of a complex function at a particular point is a key concept in complex analysis. It is especially valuable in the computation of complex integrals.
Residue theorem states that for a simple closed curve that encloses several singular points, the integral of a function over this curve is \(2\pi i\) times the sum of residues at these singular points. Finding the residue involves isolating the term with the \frac{1}{z}\ component in the Laurent series.
For our function \frac{1}{z(z+1)} expanded around \(z=0\), the Laurent series is \frac{1}{z} - \sum_{n=0}^{\infty} (-1)^n z^n\, which simplifies identifying the residue quickly. The coefficient of \frac{1}{z}\ term in this series is just 1. Hence, the residue at \(z=0\) for this function is 1.
Knowing how to find the residue not only helps with this kind of problem but is a powerful tool in broader contexts within complex function theory.