Problem 2 - Advanced Engineering Mathematics

Question

In Problems 1-6, use a Laurent series to find the indicated residue. $$ f(z)=\frac{1}{z^{3}(1-z)^{3}} ; \operatorname{Res}(f(z), 0) $$

Step-by-Step Solution

Answer

The residue of \( f(z) \) at \( z = 0 \) is 6.

Step 1: Identify the function and the singularity

The function given is \( f(z) = \frac{1}{z^3(1-z)^3} \), and we are tasked with finding the residue at \( z = 0 \). The singularity at \( z = 0 \) is a pole of order 3 due to the \( z^3 \) term in the denominator.

Step 2: Expand \((1-z)^{-3}\) using a binomial series

Since \( (1-z)^{-3} \) is part of the function, we can expand it as a binomial series:\[(1-z)^{-3} = \sum_{n=0}^{\infty} \binom{n+2}{2} z^n = 1 + 3z + 6z^2 + 10z^3 + \ldots\]This expansion will help us find the Laurent series of \( f(z) \).

Step 3: Write the Laurent series for \( f(z) \)

Substitute the series expansion of \( (1-z)^{-3} \) into the expression for \( f(z) \):\[f(z) = \frac{1}{z^3} \left(1 + 3z + 6z^2 + 10z^3 + \ldots\right)\]Expanding this, we find:\[f(z) = \frac{1}{z^3} + \frac{3}{z^2} + \frac{6}{z} + 10 + \ldots\]

Step 4: Identify the residue

The residue of \( f(z) \) at \( z = 0 \) is the coefficient of \( \frac{1}{z} \) in its Laurent series expansion. From step 3, we see that this coefficient is 6.

Key Concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis used to evaluate complex integrals along closed contours. It leverages residues, which are specific coefficients in the Laurent series expansion of a function. When finding a complex integral, the theorem states: If a function is analytic inside and on some closed contour, except for a finite number of singularities inside, then the integral over the contour can be calculated using the residues of the function at these singularities.

• Residues are the coefficients of the \ \( \frac{1}{z-a} \ \) term in the Laurent series expansion.
• The theorem simplifies the evaluation of integrals in the complex plane.
• It reduces complex integral computations to a sum of residues.

In our exercise, the residue theorem helps us find the residue at \( z = 0 \), which plays a crucial role in the expansion and evaluation.

Complex Analysis

Complex analysis deals with functions of a complex variable, offering a profound way to extend concepts of calculus to the complex plane. It studies things like transformations, mappings, and properties of complex functions. Several aspects of complex analysis are crucial for understanding mathematical concepts such as the residue, especially when functions can be expanded into series.

• Functions may have singularities, places where they are not defined or not analytic.
• The Laurent series offers a way to represent these functions around singularities.
• Complex analysis includes evaluating functions around these singularities with tools like the residue theorem.

This exercise highlights the importance of complex analysis, using concepts such as power series and singular points to find residues.

Singularities

In complex functions, singularities are points where a function fails to be analytic. Understanding singularities is key to analyzing complex functions and is essential for applying the residue theorem. This analysis revolves around expanding functions into series like the Laurent series. Some types of singularities include poles and essential singularities.

• A pole is a type of singularity where a function goes to infinity.
• Poles have different orders, determined by the leading term's power as \( z \to a \).
• The residue allows us to understand the behavior near these singular points.

In the problem, \( z=0 \) is a pole of order 3, meaning the function \( f(z) \) has terms up to \( \frac{1}{z^3} \), underscoring why expansion and residue calculation are necessary.