Question

Find the (unique) Laurent expansion of each of the following functions about the origin for its entire region of analyticity. (a) \(\frac{1}{(z-2)(z-3)} ;\) (b) \(z \cos \left(z^{2}\right)\) (c) \(\frac{1}{z^{2}(1-z)}\); (d) \(\frac{\sinh z-z}{z^{4}}\); (e) \(\frac{1}{(1-z)^{3}}\); (f) \(\frac{1}{z^{2}-1}\); (g) \(\frac{z^{2}-4}{z^{2}-9} ;\) (h) \(\frac{1}{\left(z^{2}-1\right)^{2}}\); (i) \(\frac{z}{z-1}\).

Short answer

This exercise involved applying series expansions, mainly Maclaurin and geometric series in order to convert the given functions into their Laurent series representations, focusing on identifying and handling the singularities of each function.

Step-by-step solution

Function (a) \(\frac{1}{(z-2)(z-3)}\)

Since this function is analytic everywhere except at z=2 and z=3, it can be rewritten using a partial fraction expansion as: \(\frac{1}{(z-2)(z-3)} = \frac{1}{z-3} - \frac{1}{z-2}\).

Function (b) \(z \cos \left(z^{2}\right)\)

Expanding the cosine term into its Maclaurin series, the function becomes: \(z(1 - \frac{(z^2)^2}{2!} + \frac{(z^2)^4}{4!} - ... )\). From this series, we can clearly read off the Laurent series centered at zero.

Function (c) \(\frac{1}{z^{2}(1-z)}\)

The function has singularities at z=0 and z=1. The Laurent expansion of \(1/(1-z)\) is straightforward since it's a geometric series. Therefore the function can be written as: \(\frac{1}{z^2}(1 + z + z^2 + ...)\).

Function (d) \(\frac{\sinh z-z}{z^{4}}\)

Write the function \(\sinh z\) in its power series form. Then simplify the expression obtaining: \( \frac{(z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!} - ...) - z}{z^4} = -\frac{1}{6} + \frac{z^2}{120} - ...\).

Function (e) \(\frac{1}{(1-z)^{3}}\)

Here, the function can be written as: \(-(-z)^2(1-z)^{-3} = -(-1)(1+ z + z^2), which is already in a power series form.

Function (f) \(\frac{1}{z^{2}-1}\)

This function has singularities at z=±1. The function can be rewritten as \(\frac{1}{z-1} - \frac{1}{z+1}.\)

Function (g) \(\frac{z^{2}-4}{z^{2}-9}\)

The function has singularities at z=±3. It can be rewritten as \(\frac{1}{3} - \frac{(z-1)}{3 (z+3)}.\)

Function (h) \(\frac{1}{(z^{2}-1)^{2}}\)

The function can be written as \(\frac{1}{(z-1)^2 (z+1)^2}\). Using geometric series expansion again, the power series form can be obtained.

Function (i) \(\frac{z}{z-1}\)

The function has a singularity at z=1. It can be expressed as \(1+ z + z^2 + ... \), which is a power series.

Key concepts

Analytic Function

An analytic function is one that is differentiable at every point in its domain. This means the function can be represented as a power series around these points. For example, \(\frac{1}{(z-2)(z-3)}\) is analytic everywhere except at points \(z=2\) and \(z=3\). Analytic functions have the vital property that they can be uniquely expressed as a series of terms involving powers of \(z\), which allows for various techniques in finding the series representation, including Laurent expansions in complex analysis.

It's important to note that singularities, which are points where the function is not analytic, play a crucial role in determining the form of the series. In the given exercise, careful attention is needed to discover where the functions are not analytic, as in the case of singularities at \(z=2\) and \(z=3\) for function (a).

Partial Fraction Expansion

Partial fraction expansion is a method used to decompose a complex rational function into simpler fractions that are easier to work with, especially while finding integrals or series expansions. This technique is particularly useful when dealing with Laurent series expansions. For instance, function (a) \(\frac{1}{(z-2)(z-3)}\) can be broken down into \(\frac{1}{z-3} - \frac{1}{z-2}\), where each term represents a partial fraction.

This simplification allows us to expand the function into a Laurent series around the given singularities. It is an invaluable tool when we have functions with multiple singularities that are not necessarily at the origin.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series, expanded at the point \(z=0\). It is a way to represent an analytic function as an infinite sum of terms calculated from the derivatives of the function. For example, the cosine function in the given exercise \(\cos(z^{2})\) has a well-known Maclaurin series expansion. This is applied in function (b) by expanding \(\cos(z^{2})\) into its Maclaurin series and then multiplying by \(z\) to obtain the Laurent series centered at the origin.

Understanding the Maclaurin series is crucial because many functions encountered in analysis and applied mathematics can be expressed in this form, which is also the beginning step towards finding their Laurent expansions.

Geometric Series

A geometric series is a series with a constant ratio between successive terms. For our purpose, it is often represented as \(1 + z + z^2 + z^3 + ...\), which converges when \(|z| < 1\). In the context of Laurent expansions, geometric series can be used to express functions with singularities, as seen in functions (c) and (h), where \(1/(1 - z)\) and its powers provide the respective Laurent expansions for different regions of analyticity.

Knowing the geometric series allows for easy expansion around singularities, especially when the singularity appears as a term like \(1 - z\) in the denominator of a function. These expansions are particularly helpful when considering the domains where the function is analytic.

Power Series

A power series is an infinite sum of terms in the form \(a_n z^n\), where \(a_n\) represents the coefficients, and \(z\) is the variable raised to the power \(n\). The concept of power series is fundamental in representing analytic functions and in the process of finding the Laurent expansion. For instance, function (e) \(\frac{1}{(1-z)^3}\) can be seen as a power series through the binomial expansion.

Being able to rewrite a function in a power series format is key when tackling problems in complex analysis because it reveals the behavior of the function near different points, particularly near singularities. Power series are also used when functions have removable singularities, enabling one to find the Laurent series directly.

Singularities

Singularities are points at which a function ceases to be analytic and can no longer be described with a power series. They are essential in determining the Laurent expansion of a function. There are different types of singularities: removable, poles, and essential. For example, the singularities at \(z=2\) and \(z=3\) in function (a) are poles, where the function goes to infinity.

Understanding singularities is crucial when working with Laurent expansions because they divide the complex plane into regions where different series expansions are valid. Identifying and classifying singularities allows students to apply the correct expansion method, such as partial fraction decomposition or geometric series, to find the corresponding Laurent series.