Problem 5 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Write the Laurent series definition

A Laurent series is a representation of a complex function that includes terms with both positive and negative powers. The general form is: \[ f(z) = rac{a_{-n}}{z^n} + rac{a_{-(n-1)}}{z^{n-1}} + rac{a_{-(n-2)}}{z^{n-2}} + ext{...} + a_0 + a_1z + a_2z^2 + ext{...} \]We need to find the coefficient of \(\frac{1}{z}\), which is \(a_{-1}\), as the residue at a point is this particular coefficient.

Step 2: Express the function in terms of a series

Given the function is \(f(z) = e^{-2/z^2}\). First, write the exponential term as a series: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \]Substitute \(x = -\frac{2}{z^2}\):\[ e^{-2/z^2} = 1 - \frac{2}{z^2} + \frac{4}{2!z^4} - \frac{8}{3!z^6} + \dots \]

Step 3: Identify the terms and find the residue

In the series expansion from Step 2, observe that:\[ e^{-2/z^2} = 1 - \frac{2}{z^2} + \frac{2}{z^4} - \frac{8}{3!z^6} + \dots \]None of these terms have \(\frac{1}{z}\), meaning \(a_{-1} = 0\).

Step 4: Conclusion about the residue

Since there is no \(\frac{1}{z}\) term, the residue at \(z=0\) is 0. The function \(e^{-2/z^2}\) does not have a simple pole at \(z=0\), so the residue is 0.

Key Concepts

Residue Calculation

The residue is a crucial part of complex analysis, simplifying the evaluation of integrals around singular points. It is the coefficient of the term \(\frac{1}{z}\) in a Laurent series expansion. To calculate the residue of a complex function at a certain point, first, express the function as a Laurent series.
Laurent series are unique because they include both positive and negative powers of \(z\). This dual-power series allows for expressing complex functions even at points where they are not analytic.
Identifying the coefficient \(a_{-1}\) of \(\frac{1}{z}\) is exactly what "finding the residue" is all about. For the function \(f(z) = e^{-2/z^2}\), the calculated Laurent series doesn't include \(\frac{1}{z}\), proving the residue at \(z=0\) is 0.
This lack of a \(\frac{1}{z}\) term means there is no pole at this point, and thus the residue is found to be zero. Reside calculations help in determining the behavior of integrals and solutions of differential equations involving complex functions.

Complex Functions

Complex functions form a foundation for advanced mathematical analysis and encompasses a wide range of applications in engineering and physics. Understanding how they behave, particularly around peculiar points, is crucial. This makes exploring the functions via expansions, like Laurent series, essential.
Complex functions can include both real components and imaginary components: represented as \( f(z) = u(x, y) + iv(x, y) \) where \( u \) and \( v \) are real-valued functions. They extend the idea of functions of a single real variable, providing a richer analysis.
The power of complex functions lies in their ability to encapsulate challenging concepts into manageable mathematical expressions. Evaluating these functions involves attending to their singular points, zeros, and poles. By transforming these functions into series, such as Laurent series, you can gain a deeper understanding of their properties and behavior.

Series Expansion

Series expansion is a method to express complex functions as the sum of an infinite series of terms. The Laurent series, in particular, is significant because it accommodates functions with singularities. It incorporates terms of both positive and negative powers of \(z\), unlike the Taylor series which only has positive powers.
Learners need to recognize that Laurent series is particularly useful in finding residues. By directly evaluating each term, you can identify poles and calculate residues effectively.
For the function \(f(z) = e^{-2/z^2}\), using the series expansion helped clarify the absence of a \(\frac{1}{z}\) term. Consequently, it confirmed that the residue at \(z=0\) is 0. This specific insight is valuable when working with complex analysis problems related to evaluating complex integrals.
Series expansions, such as the Laurent series, provide invaluable tools in both theoretical assessments and practical computations in complex analysis.