Problem 26 - Advanced Engineering Mathematics

Question

Expand \(f(z)=\frac{7 z-3}{z(z-1)}\) in a Laurent series valid for the indicated annular domain. \(0<|z-1|<1\)

Step-by-Step Solution

Answer

The Laurent series is \( f(z) = \frac{3}{z} + 4\sum_{n=0}^{\infty}(z-1)^n \) for \(0<|z-1|<1\).

Step 1: Decompose the Function into Partial Fractions

The first step is to express the given function \(f(z) = \frac{7z-3}{z(z-1)}\) in terms of partial fractions. The partial fraction decomposition is: \[\frac{7z-3}{z(z-1)} = \frac{A}{z} + \frac{B}{z-1}\]where \(A\) and \(B\) are constants. To find \(A\) and \(B\), equate coefficients in the identity:\[7z - 3 = A(z-1) + Bz\]By considering the limits and solving the resulting system of equations, we find that \(A = 3\) and \(B = 4\). Thus:\[\frac{7z - 3}{z(z-1)} = \frac{3}{z} + \frac{4}{z-1}\]

Step 2: Expand Each Partial Fraction into Laurent Series

Now, expand each term into a Laurent series valid for the annular domain \(0 < |z-1| < 1\):1. The term \(\frac{3}{z}\) is already a Laurent series as it is.2. The term \(\frac{4}{z-1}\) can be expanded using the geometric series formula. We manipulate the denominator: - Let \(w = 1 - z\) so \(z = 1 - w\). Then, \(\frac{1}{z-1} = -\frac{1}{w}\). - For \(0 < |1-z| < 1\), \(w = 1 - z\) suggests \(0 < |w| < 1\), and the series expansion becomes: \[\frac{4}{z-1} = -4 \sum_{n=0}^{\infty} (z-1)^n\] which simplifies in terms of \(w\) to \(\sum_{n=0}^{\infty} (1 - z)^{-n-1}\), reflected in the correct variable bounds.

Step 3: Write the Complete Laurent Series

Combine the series expansions:\[\frac{7z - 3}{z(z-1)} = \frac{3}{z} + 4\sum_{n=0}^{\infty}(z-1)^n\]We are interested in the region where \(0 < |z-1| < 1\). The series representation is:- The term \(\frac{3}{z}\) has a simple pole at \(z=0\).- The series \(4\sum_{n=0}^{\infty}(z-1)^n\) provides the contributions for each \(n\).Thus, the Laurent series representation valid for \(0<|z-1|<1\) is:\[ f(z) = \frac{3}{z} + 4\left(\frac{1}{z-1} + 1 + (z-1) + (z-1)^2 + \cdots\right) \]

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down a complex rational function into simpler fractions that are easier to work with. In the function given in the exercise, we are dealing with \[ f(z) = \frac{7z-3}{z(z-1)}. \]Breaking it down into partial fractions involves expressing it as: \[ \frac{A}{z} + \frac{B}{z-1}. \]This form allows us to manage each fraction separately using simpler operations.To find the constants \(A\) and \(B\), you equate terms of the polynomial, producing equations to solve:- Equating coefficients for terms with the same power. - Solving for constants: When you do this for the given problem, we find \(A = 3\) and \(B = 4\).This method is particularly useful for integrating complex functions or in cases, like this, where we need a series expansion.

Annular Domain

The concept of an annular domain is foundational in complex analysis. It refers to the region between two concentric circles in the complex plane, not touching either the inner or the outer boundary.In our exercise, the annular domain is specified as \(0 < |z-1| < 1. \)This notation means 'an open ring' around \(z = 1,\) with radius extending from just outside 0 to up to 1.Why is it important? - Because series expansions depend on these domains in complex analysis.- Laurent series, in contrast to Taylor series, can handle poles, and this annular domain signifies a region excluding a pole precisely at \(z=1.\)Understanding domains allows us to use the appropriate series for a specific region of interest, avoiding singularities directly on the boundaries and concentrating on the functional behavior within the region itself.

Geometric Series Expansion

Geometric series are infinite series that are straightforward to work with because they follow a simple rule for their coefficients.For any number \( r\) in the valid range, the series \(1 + r + r^2 + r^3 + \cdots\) sums up to \(\frac{1}{1-r}.\)This is the technique we use when expanding \(\frac{4}{z-1}.\)First, we manipulate it by letting \(w = 1 - z,\)which gives us \(-\frac{1}{w}.\)The transformation helps us express it as a geometric series:- \(\frac{4}{z-1} = -4 \sum_{n=0}^{\infty} (z-1)^n.\)- This expansion is valid in the circle defined by \(0 < |z - 1| < 1.\)- It forms the foundation of the exponential elements that make the Laurent series.By breaking complex fractions into sums of geometric series, we capture each component's contribution clearly and manage the region specified in the annular domain effectively.