Problem 11 - Complex Analysis for Mathematics and Engineering

Question

Let \(f\) be analytic at the points \(0, \pm 1, \pm 2, \ldots\). If \(g(z)=\pi f(z) \cot \pi z\), then show that \(\operatorname{Res}[g, n]=f(n)\) for \(n=0, \pm 1, \pm 2, \ldots .\)

Step-by-Step Solution

Answer

\(\operatorname{Res}[g, n]=f(n)\) for \(n = 0, \pm 1, \pm 2, \ldots\).

Step 1: Understanding the Function Setup

We have the function \(g(z) = \pi f(z) \cot \pi z\). The \(\cot \pi z\) function, being \(\frac{\cos \pi z}{\sin \pi z}\), has simple poles at all integers, specifically \(z = 0, \pm 1, \pm 2, \ldots\). We recognize that these are also the points where \(f\) is analytic.

Step 2: Locating and Identifying Residues of Poles

For each integer \(n = 0, \pm 1, \pm 2, \ldots\), \(\cot \pi z\) has a simple pole with residue 1. Since \(f(z)\) is analytic at these points, it does not contribute to the singularity, thus the residue of the product \(f(z) \cot \pi z\) at \(z = n\) is \(f(n)\).

Key Concepts

Residue Theorem

The residue theorem is a powerful tool in complex analysis, particularly useful when evaluating complex integrals around closed contours. It states that:

• The integral of a function around a closed contour is equal to \(2\pi i\) times the sum of residues at all poles inside the contour.

This theorem simplifies many complex curve integrals, making the evaluation straightforward when you know the poles and their residues.

In the original exercise, we have a function \(g(z)\) formed by \(\pi f(z) \cot \pi z\), where \(f(z)\) is analytic at integers. The residue theorem helps us find the contribution at each simple pole, which matches the setup perfectly to extract the residue and evaluate the function at these critical points.

Analytic Functions

Analytic functions are central to complex analysis. A function \(f(z)\) is analytic at a point if it is differentiable at that point and in a neighborhood surrounding it.

Analytic functions boast several intriguing properties:

• Their derivative exists everywhere in the neighborhood.
• They can be represented as a power series.
• They are infinitely differentiable.

In the original exercise, \(f(z)\) is analytic at points \(0, \pm 1, \pm 2, \ldots\). This analiticity ensures that there are no poles introduced by \(f(z)\) at these integer points, allowing the focus to remain on the properties of \(\cot \pi z\). Therefore, the residue at these points depends entirely on the properties of the pole introduced by \(\cot \pi z\), leading us to a clear path in finding the residues as \(f(n)\).

Poles and Residues

Poles are essential features to identify when studying complex functions. A pole of a function is a point where the function's magnitude becomes unbounded, but it can be characterized by a Laurent series expansion with a finite principal part.

Residues are coefficients of the \(1/(z - a)\) term in a function's Laurent series, and they play a vital role in calculating integrals involving complex functions. Key insights about residues include:

• If the pole is simple, the residue at a point \(a\) is \(\lim_{z \to a} (z - a)f(z)\).
• Residues compute the contribution of each pole to a contour integral.

In our specific setup, \(\cot \pi z\) exhibits poles at integer points \(n\), each with a residue of 1. Since \(f(z)\) is analytic, it does not affect the residue extraction process at these points. Therefore, the residue of \(g(z) = \pi f(z) \cot \pi z\) at each pole becomes straightforwardly determined as \(f(n)\), showcasing the method's elegance and efficiency.