Question

Find the Laurent series for the following functions about the indicated points; hence find the residue of the function at the point. (Be sure you have the Laurent series which converges near the point.) $$\frac{1+\cos z}{(z-\pi)^{2}}, z=\pi$$

Short answer

The Laurent series is \( \frac{1}{2} - \frac{(z-\text{π})^2}{24} + \frac{(z-\text{π})^4}{720} - \text{...} \) and the residue is \(0\).

Step-by-step solution

Rewrite the function around the specified point

Given function is \(\frac{1+\text{cos} z}{(z-\text{π})^{2}}\). Substitute \(w=z-\text{π}\) to re-center the function around \(z=\text{π}\). This transforms the function to \(\frac{1+\text{cos} (w+\text{π})}{w^2}\). Using the trigonometric identity \(\text{cos}(a + b) = \text{cos} a \text{cos} b - \text{sin} a \text{sin} b\), we have \(\text{cos} (w + \text{π}) = -\text{cos} w\). The function becomes: \( \frac{1 - \text{cos} w}{w^2} \).

Expand \(1 - \text{cos} w\) using its Taylor Series

The Taylor series for \(\text{cos} w\) around \(w = 0\) is \(\text{cos} w = 1 - \frac{w^2}{2!} + \frac{w^4}{4!} - \frac{w^6}{6!} + \text{...}\). Therefore, \(1 - \text{cos} w = 1 - \bigg(1 - \frac{w^2}{2!} + \frac{w^4}{4!} - \frac{w^6}{6!} + \text{...}\bigg) = \frac{w^2}{2!} - \frac{w^4}{4!} + \frac{w^6}{6!} - \text{...}\).

Substitute the Taylor series into the function

The function \( \frac{1 - \text{cos} w}{w^2} \) thus becomes \( \frac{\frac{w^2}{2!} - \frac{w^4}{4!} + \frac{w^6}{6!} - \text{...}}{w^2} = \frac{1}{2!} - \frac{w^2}{4!} + \frac{w^4}{6!} - \text{...} \)

Identify the Laurent series

The Laurent series for the function \( \frac{1 + \text{cos} z}{(z - \text{π})^2} \) around \( z = \text{π} \) can be written as \( \frac{1}{2} - \frac{(z-\text{π})^2}{24} + \frac{(z-\text{π})^4}{720} - \text{...} \).

Find the residue

The residue of the function at \( z = \text{π} \) is the coefficient of \( \frac{1}{w} \) term in its Laurent series expansion. In this series, there is no \( \frac{1}{w} \) or \( \frac{1}{(z-\text{π})} \) term. Therefore, the residue is \(0\).

Key concepts

residue theorem

The residue theorem is an essential tool in complex analysis for evaluating contour integrals of complex functions. It states that for a function with isolated singularities inside a closed contour, the integral around the contour is equal to \(2\text{π}i\) times the sum of the residues at these singularities. There's a quick formula to remember: \( \text{∮} f(z) \text{dz} = 2\text{π}i \text{Σ Res}(f, z_k) \).

To use the residue theorem, you must:

• Find the singular points of the complex function.
• Calculate the residue at each singularity.
• Sum the residues and multiply by \(2\text{π}i\).

The residue of a function at a singularity gives crucial information about the behavior of the function near that point. In the provided exercise, we determined that the residue of the function \( \frac{1+\text{cos} z}{(z-\text{π})^{2}} \) at \(z=\text{π}\) is \(0\). This indicates that there is no pole of order 1 at this point. The residue theorem simplifies integration, especially for complex functions with multiple singular points.

complex functions

Complex functions are functions that take complex numbers as inputs and give complex numbers as outputs. They play a vital role in fields like engineering, physics, and applied mathematics.

Key aspects include:

• The function \(f(z)\), where \(z\) is a complex variable.
• Analyticity, meaning the function is differentiable at every point in its domain.
• Poles and singularities, which are points where the function does not behave normally (e.g., it goes to infinity).

In our exercise, the function in question \( \frac{1+\text{cos} z}{(z-\text{π})^{2}} \) has a pole of order 2 at \(z=\text{π}\), indicated by the squared term in the denominator. Understanding complex functions includes being comfortable with different forms of series like Taylor and Laurent series, which help in expanding and analyzing functions around specific points.

Taylor series expansion

The Taylor series is a way to represent a function as an infinite sum of terms calculated from the function's derivatives at a single point. For a function \(f(z)\) analytic at \(z=a\), its Taylor series is written as: \( f(z) = f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + ...\).

This series converges to the function within a certain radius. Key points include:

• Taylor series are used for functions analytic at the expansion point.
• They provide useful approximations for functions around that point.

In our exercise, we used the Taylor series expansion of \( \text{cos} w \) to write \( 1 - \text{cos} w \) as \( \frac{w^2}{2!} - \frac{w^4}{4!} + \frac{w^6}{6!} - \text{...} \). This expansion made it possible to formulate the Laurent series of the given function. The Taylor series is foundational in complex analysis, providing a powerful tool for manipulating and understanding complex functions around specific points.