Problem 28 - A First Course in Complex Analysis with Applications

Question

Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C} \frac{\cot \pi z}{z^{2}} d z, C:|z|=\frac{1}{2} $$

Step-by-Step Solution

Answer

The integral evaluates to \(-2\pi^2 i\).

Step 1: Identify the Pole of the Integrand

The integrand \( \frac{\cot \pi z}{z^2} \) has a singularity at \( z = 0 \) because of the \( \frac{1}{z^2} \) term. This is a pole of order 2. In addition, \( \cot \pi z = \frac{\cos \pi z}{\sin \pi z} \) has simple poles at \( z = n, n \in \mathbb{Z} \). However, within the contour \( C: |z|=\frac{1}{2} \), only \( z = 0 \) is relevant.

Step 2: Express \( \cot \pi z \) as a Series

\( \cot \pi z \) can be expressed as a Laurent series expansion: \( \cot \pi z = \frac{1}{\pi z} + \sum_{n=1}^{\infty} (-1)^{n} (2 \pi z)^{2n-1} \). This helps identify the behaviour of the integrand around \( z = 0 \).

Step 3: Find the Residue at \( z = 0 \)

We know \( Res(f,z_0) = \lim_{z \to z_0} \frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}}((z-z_0)^mf(z)) \). For a pole of order 2 at \( z = 0 \), \( m=2 \). The function \( f(z) = \frac{\cot \pi z}{z^2} \) leads to \( Res(f,0) = \lim_{z \to 0} \frac{d}{dz}((z-0)^2 \frac{\cot \pi z}{z^2}) = \lim_{z \to 0} \frac{d}{dz}(\cot \pi z) = -\pi \).

Step 4: Apply Cauchy's Residue Theorem

Cauchy's Residue Theorem states that for a function \( f(z) \) analytic inside and on some simple closed contour \( C \) except for isolated singularities, \( \oint_C f(z) \, dz = 2\pi i \sum \text{Res}(f, \text{inside } C) \). Here, we only have the pole at \( z=0 \), so the integral is \( 2\pi i(-\pi) = -2\pi^2 i \).

Key Concepts

Laurent Series

The Laurent series is an important tool in complex analysis. It extends the idea of a Taylor series to include negative powers and is used to describe functions with singularities. The series takes the form:

• For a function defined in a punctured neighborhood surrounding a singularity, it can be expressed as: \[ f(z) = \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n \]This allows us to capture the pole behaviour of functions where Taylor series cannot.
• This is essential when dealing with poles in contour integrations, as it helps to identify how a function behaves around its singular points

In the given exercise, we express \( \cot \pi z \) using its Laurent series to understand its behaviour around the pole at \( z = 0 \). This expansion reveals how the function acts, especially when combined with the pole at \( z^2 \) in the integrand \( \frac{\cot \pi z}{z^2} \). This series highlights how each term contributes to the complex nature of the function around its singularity.

Complex Contour Integration

Complex contour integration is a method used to evaluate complex integrals over specific paths, or contours, in the complex plane. This method is a central part of complex analysis and helps solve integrals that would be difficult to tackle using real integration techniques.

• A contour is a closed curve or path in the complex plane. In our exercise, the contour \( C \) is defined as a circle with radius \( \frac{1}{2} \), centered at the origin.
• When evaluating integrals using complex contour integration, singularities (points where the function does not behave well) inside the contour affect the integral’s value significantly.

The goal of contour integration is to use the information about these singularities—captured through concepts like the Laurent series or residue calculation—to compute the integral around the contour efficiently. In many cases, such as with our exercise, Cauchy's Residue Theorem is employed to streamline this process by focusing on the residues at singularities within the contour.

Complex Poles and Residues

Complex poles and residues are fundamental concepts in dealing with integrals of complex functions. Understanding their role is crucial when using powerful tools like Cauchy's Residue Theorem.

• A pole is a type of singularity where the function's limit goes to infinity. The order of the pole signifies how rapidly this infinity is approached. In the given function \( \frac{\cot \pi z}{z^2} \), the pole at \( z = 0 \) is of order 2.
• The residue at a pole is, intuitively, the coefficient of the \( \frac{1}{z} \) term in the Laurent series expansion of a function about that pole.

To find the residue at \( z=0 \) in our exercise, we incorporate the derivative of \( \cot \pi z \) evaluated at \( z=0 \) due to the pole's order. Through calculating derivatives and using limits, it is found to be \( -\pi \). This residue is crucial as it directly contributes to the integral's value when applying Cauchy's Residue Theorem, simplifying what would otherwise be a complex calculation into a straightforward evaluation using the pole's residues.