Problem 4 - Complex Analysis

Question

Assume that \(f\) has in \(\infty\) an isolated singularity. We define $$ \begin{aligned} \operatorname{Res}(f ; \infty) &:=-\operatorname{Res}(\tilde{f} ; 0), \quad \text { where we set } \\ \tilde{f}(z):=\frac{1}{z^{2}} \widehat{f}(z)=\frac{1}{z^{2}} f\left(\frac{1}{z}\right) \end{aligned} $$ The factor \(z^{-2}\) is natural, as it will become transparent from the following computation rules, especially exercise \(5 .\) (a) Show: $$ \operatorname{Res}(f ; \infty)=-\frac{1}{2 \pi i} \oint_{\alpha_{R}} f(\zeta) d \zeta $$ where \(\alpha_{R}(t)=R \exp (i t), t \in[0,2 \pi]\), and \(R\) is chosen large enough to insure that \(f\) is analytic in the complement of the closed disk centered in 0 with radius \(R\) (b) The function $$ f(z)= \begin{cases}1 / z, & \text { if } z \neq \infty \\ 0, & \text { if } z=\infty\end{cases} $$ has in \(\infty\) a removable singularity, but \(\operatorname{Res}(f ; \infty)=-1\) (not zero !!). It looks like \infty plays a special role. The sensation of discomfort immediately disappears, if we (re)define the notion of "residue" by using the differential \(f(z) d z\), this being the more structural definition. The notion of residue of \(f\) is related to the differential \(f(z) d z\), the notion of order of \(f\) to the function \(f\) itself.

Step-by-Step Solution

Answer

To show \( \operatorname{Res}(f; \infty) = -\frac{1}{2\pi i} \oint_{\alpha_R} f(\zeta) d\zeta \), transform \( f \) to \( \tilde{f}(z) \) and find the residue at 0. For \( f(z) = \frac{1}{z} \), the residue at infinity is \(-1\).

Step 1: Define the Residue at Infinity

The residue at infinity for a function \( f \) is given by \( \operatorname{Res}(f; \infty) = -\operatorname{Res}(\tilde{f}; 0) \). This means that to find the residue at infinity, we transform the function \( f \) into \( \tilde{f}(z) = \frac{1}{z^2} f\left(\frac{1}{z}\right) \) and find the residue of this new function at 0.

Step 2: Apply the Residual Theorem

Using the definitions provided, we apply the theory from complex analysis that relates the contour integrals and residues. The theorem involves integrating \( f(\zeta) \) over the contour \( \alpha_R \), which is a large circle of radius \( R \). The formula given in the problem is \( \operatorname{Res}(f; \infty) = -\frac{1}{2\pi i} \oint_{\alpha_R} f(\zeta) d\zeta \).

Step 3: Evaluate the Integral

For large enough \( R \), the function \( f \) becomes analytic outside a disk centered at zero, except possibly at infinity if \( f \) has a singularity there. The contour integral \( \oint_{\alpha_R} f(\zeta) d\zeta \) captures this contribution from \( f \). By the given theorem from complex analysis, this integral equals \(-2\pi i\) times the residue at infinity.

Step 4: Interprete (b) with Example Function

For the function \( f(z) = \frac{1}{z} \), evaluate its behavior as it approaches infinity. At infinity, \( f(z) = \frac{1}{z} \) has a removable singularity, but the residue is not zero. Applying our previously derived expression, it can be shown that the residue at infinity is \(-1\), as the integral from step 3 would resolve to this value based on evaluations of asymptotic limits.

Key Concepts

Complex Analysis

Complex Analysis is a fascinating area of mathematics that explores functions, particularly those that are defined using complex numbers. It provides powerful tools to analyze these functions by exploring their properties and behavior in the complex plane. Complex functions can be expressed as powers of complex numbers, and thus can exhibit unique properties not always present in real-valued functions.
- Complex analysis introduces the concept of holomorphic functions, which are complex functions that are differentiable at every point in their domain. - A significant aspect of complex analysis is the study of analytic functions through their singular points, using ideas such as residues and contour integrals.
The idea is to represent complex functions in a simpler form and study these through the introduction of techniques like **Laurent series** and **Cauchy's integral formulas**. By examining these series, we gain insight into the behavior of functions around singularities. These methods allow mathematicians and engineers to solve complex integrals and predict the behavior of physical systems modeled using complex variables.

Singularities

In the realm of Complex Analysis, singularities are points in the complex plane where a function ceases to be well-behaved in some manner. This can include points where a function is not defined or is not differentiable. - **Isolated Singularities:** Singularities can be isolated, meaning a function may have a singular point in a region surrounded by points where the function is analytic. - There are three types of isolated singularities: **removable, poles, and essential singularities**. Understanding these is crucial to applying complex analysis techniques. For example: - **Removable singularities** are points where the function can be redefined so it becomes analytic. - **Poles** are singularities where a function’s magnitude becomes infinite, yet these have a manageable behavior. - **Essential singularities** have more complicated behavior, making the function’s value unpredictable around these points. Identifying the type of singularity can enable mathematicians to determine the appropriate methods for analysis, like contour integration, to handle complex functions effectively.

Contour Integration

Contour Integration is a fundamental technique in Complex Analysis used to evaluate complex integrals over specified contours or paths in the complex plane. This technique connects complex function theory to integration, offering a powerful method for calculating integrals in engineering and physics. - Contours are described as paths along which the integral of the complex function is to be evaluated. - The **Cauchy Integral Theorem** and **Cauchy Integral Formula** are central in contour integration. They offer valuable insights into finding integrals of functions that are complex differentiable across certain contours. The steps in contour integration involve:

• Identifying the singularities enclosed by the contour.
• Applying Cauchy's integral theorem which states the integral of a function over a closed contour can be zero if no singularities are enclosed.
• Utilizing the residue theorem, which simplifies the calculation by focusing on residues at these singularities.

Logical and critical understanding of contour integration can greatly simplify problems of evaluating complex integrals, especially in regions with singularities.

Residue at Infinity

The concept of Residue at Infinity allows for the evaluation of residue not just at finite points, but at the point at infinity, which might seem a bit abstract. However, this is vital in understanding the global behavior of a complex function.To compute the residue at infinity:- Transform the original function into a related function by substituting the variable, effectively flipping the analysis point to zero. This transformation is given by \( \tilde{f}(z) = \frac{1}{z^2}f\left( \frac{1}{z} \right) \).- Apply the residue theorem, converting the contour integral around infinity into a contour integral at zero. This relates to finding residues in finite parts of the complex plane.The **Residue Theorem** expands the applicability of complex analysis to contours that encircle infinity by creatively redefining the problem. By understanding the residue at infinity:

• We simplify complex contour integrals that involve infinite boundaries.
• It provides insight into the asymptotic behavior of functions.

This understanding serves as a crucial tool in complex analysis, facilitating the calculation of integrals over closed contours that span infinite regions.