Problem 3 - Advanced Engineering Mathematics

Question

Use a Laurent series to find the indicated residue. \(f(z)=\frac{4 z-6}{z(2-z)} ; \operatorname{Res}(f(z), 0)\)

Step-by-Step Solution

Answer

The residue of \( f(z) \) at \( z = 0 \) is \(-3\).

Step 1: Identify the Singularities

The function is given by the fraction \( f(z) = \frac{4z - 6}{z(2-z)} \). The singularities occur where the denominator is zero, that is, \( z = 0 \) and \( z = 2 \). For this problem, we are interested in the singularity at \( z = 0 \).

Step 2: Expand Denominator around Singularity

Express \( \frac{1}{2-z} \) using a geometric series expansion around \( z=0 \):\[\frac{1}{2-z} = \frac{1}{2\left(1 - \frac{z}{2}\right)} = \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\]This expansion is valid for \( |z| < 2 \).

Step 3: Derive Laurent Series for \( f(z) \)

Substitute the expanded form of \( \frac{1}{2-z} \) into \( f(z) \):\[f(z) = \frac{4z-6}{z} \cdot \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n = \left(4 - \frac{6}{z}\right) \cdot \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\]Simplify this to get the Laurent series:\[2 \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n - 3 \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n \cdot \frac{1}{z}\]Which simplifies further to:\[\sum_{n=0}^{\infty} 2 \left(\frac{z}{2}\right)^n - \sum_{n=0}^{\infty} 3 \frac{z^n}{2^n z}\]

Step 4: Solve for the Residue

The residue is the coefficient of the \( \frac{1}{z} \) term in the Laurent series. In our expanded series:\[-\sum_{n=0}^{\infty} 3 \frac{z^n}{2^n z} = -3\]Thus, the residue of \( f(z) \) at \( z = 0 \) is \( -3 \).

Key Concepts

Complex Analysis

Complex analysis is a fascinating branch of mathematics that studies complex numbers and the functions that depend on them. Here are a few key ideas:

• Complex numbers are numbers of the form \(a + bi\), where \(i\) is the imaginary unit, equal to the square root of \(-1\).
• Functions of complex variables can exhibit behaviors quite different than their real counterparts.
• Complex analysis includes many powerful techniques, such as the use of complex integrals, series expansions, and analytic continuation.

In the context of the given exercise, complex analysis helps us understand the behavior of functions like \(f(z) = \frac{4z - 6}{z(2-z)}\) around certain points called singularities, using tools like Laurent series.

Singularities

A singularity in complex analysis is a point where a function ceases to be analytic or well-behaved. For our function \(f(z) = \frac{4z - 6}{z(2-z)}\), singularities are where the denominator equals zero:

• \(z = 0\) and \(z = 2\) are the singular points of the function.
• At these points, the function is undefined or infinite.
• Singularities can be classified as removable, poles, or essential depending on their nature.

In this exercise, we focus on the singularity at \(z = 0\). This is a point where a pole exists, more specifically a simple pole, which plays a crucial role when calculating the residue.

Residue Theorem

The residue theorem is a central result in complex analysis used for evaluating integrals of complex functions around closed contours. Here's a quick summary:

• The residue of a function at a singular point is the coefficient of the \(\frac{1}{z}\) term in its Laurent series expansion about that point.
• The residue theorem allows for transformation of contour integrals of a function into a summation of residues inside the contour.
• This theorem simplifies complex integral calculations significantly.

In the given solution, we used the residue at \(z = 0\) found in the Laurent series expansion to determine that the residue is \(-3\), which directly leads to the result once the series is known.

Geometric Series

A geometric series is a sum of terms, each which is a fixed multiple of the previous one. In mathematical terms, a geometric series with a common ratio \(r\) and first term \(a\) is described as:

• \(a + ar + ar^2 + ar^3 + \ldots\)
• The series converges if \(|r| < 1\).
• Summing an infinite geometric series when it converges is done using the formula: \(\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\).

In our solution, a geometric series is employed to simplify the expression for \(\frac{1}{2-z}\). Knowing how to expand terms into a geometric series is crucial for deriving the Laurent series, which helps isolate singular behavior like poles.