Problem 4 - Advanced Engineering Mathematics

Question

In Problems 1-6, use a Laurent series to find the indicated residue. $$ f(z)=(z+3)^{2} \sin \frac{2}{z+3} ; \operatorname{Res}(f(z),-3) $$

Step-by-Step Solution

Answer

The residue of \( f(z) \) at \( z = -3 \) is \( -\frac{4}{3} \).

Step 1: Understand the Function

We are given the function \( f(z) = (z+3)^2 \sin \left( \frac{2}{z+3} \right) \) and need to find the residue at \( z = -3 \). This function has a singularity at \( z = -3 \) due to the term \( \frac{2}{z+3} \).

Step 2: Simplify Around the Singularity

To deal with the singularity at \( z = -3 \), set \( w = z+3 \), so \( z = w - 3 \). Then \( f(z) = w^2 \sin \left( \frac{2}{w} \right) \) and the singularity now appears at \( w = 0 \).

Step 3: Expand \( \sin \frac{2}{w} \) Using Series

Use the Taylor series expansion for sine:\[\sin \left( \frac{2}{w} \right) = \frac{2}{w} - \frac{\left( \frac{2}{w} \right)^3}{6} + \frac{\left( \frac{2}{w} \right)^5}{120} - \ldots\]Thus:\[\sin \left( \frac{2}{w} \right) = \frac{2}{w} - \frac{8}{6w^3} + \frac{32}{120w^5} - \ldots\]

Step 4: Multiply and Simplify Series

Multiply the series with \( w^2 \):\[w^2 \left( \frac{2}{w} - \frac{8}{6w^3} + \frac{32}{120w^5} - \ldots \right) = 2w - \frac{4}{3w} + \frac{8}{15w^3} - \ldots\]

Step 5: Identify the Residue

The residue is the coefficient of \( \frac{1}{w} \) in the Laurent series expansion. From Step 4, the coefficient of \( \frac{1}{w} \) is \( -\frac{4}{3} \). So, the residue at \( z = -3 \) is \( -\frac{4}{3} \).

Key Concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis that allows us to evaluate complex integrals, especially over closed curves. This theorem states that if a function is holomorphic (complex differentiable) within and on some closed contour except at a finite number of singularities, the integral of that function around the contour can be expressed in terms of the residues at those singularities.

Here's a simple rundown of the process:

• Identify the singular points of the function within the contour.
• Determine the residues at these singularities.
• Sum the residues and multiply by the constant factor \(2\pi i\).

Residues are especially useful because they help simplify the computation of difficult integrals. In our problem, the focus is on finding the residue of a function at a particular singular point. By calculating this residue, you can predict the outcome of the integral around that point without directly computing it. For our given function, the residue at \(z = -3\) provides the important contribution to our integral over any contour enclosing the singularity.

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex numbers. Analysts focus on concepts such as differentiation and integration in the complex plane. Functions in complex analysis exhibit behaviors that can be dramatically different from functions defined in real analysis.

A crucial aspect of complex analysis is dealing with functions that are not differentiable everywhere in the plane, often due to points known as singularities. These singularities are locations where the function fails to be holomorphic.

• The function \(f(z) = (z+3)^2 \sin(\frac{2}{z+3})\) has a singularity at \(z = -3\).
• This singularity is essential to finding the Laurent series expansion, affecting how we handle the function mathematically.

By converting cases with singular points into simpler expressions, complex analysis offers insightful tools, like the residue theorem, to work around traditional difficulties in integration. Understanding these underlying concepts equips students and analysts to handle complex functions with more confidence and to derive results otherwise deemed complex through real analysis alone.

Singularity Expansion

Singularity expansion involves representing functions using series that take into account the peculiarities around singular points. One such representation is the Laurent series, which is particularly useful in expressing functions around isolated singularities.

Unlike the Taylor series, which only represents power series of non-negative integer terms, the Laurent series includes terms with negative powers. This inclusion is crucial when dealing with functions that have singular points, as it allows the incorporation of terms that describe behavior near the singularity. Here’s how:

• The Laurent series specifically separates the function into a principal part (terms with negative powers) and a regular part.
• In our example, after rewriting \(f(z)\) at the singularity, we express it at \(w = 0\) with negative powers using, \[ w^2 \left( \frac{2}{w} - \frac{8}{6w^3} + \frac{32}{120w^5} - \ldots \right) = 2w - \frac{4}{3w} + \ldots \] The term \( -\frac{4}{3w}\) indicates the residue after identifying it as part of a Laurent expansion.

Expansion around singularity thus reveals significant traits of a function's behavior nearby, leading to the efficient application of the residue theorem and further insights into complex integrals within a broader region.