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.) $$ \sin \frac{1}{z}, z=0 $$

Short answer

The Laurent series for \sin \frac{1}{z} \ about \ z=0 \ is \ \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \cdots \. The residue at \ z = 0 \ is 1.

Step-by-step solution

- Rewrite the Function

Rewrite the original function in terms of a familiar series. The function given is \( \frac{1}{z} \). Let's consider the sine function series expansion: \[ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \] Now substitute \( x = \frac{1}{z} \) into this series.

- Substitute and Simplify

Using \( x = \frac{1}{z} \), we get \: \ \[ \sin(\frac{1}{z}) = \sum_{n=0}^{\infty} \frac{(-1)^n (\frac{1}{z})^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\finite} \frac{(-1)^n}{(2n+1)!} z^{-(2n+1)} \]

- Identify the Laurent Series and Convergence

The expression \sin(\frac{1}{z}) \ can be written in its Laurent series form as: \[ \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \cdots \] The series converges for \ |z| > 0 \ due to the nature of the terms containing negative powers of \ z \.

- Find the Residue

The residue of \ \sin \frac{1}{z} \ at \ z = 0 \ is the coefficient of the \ \frac{1}{z} \ term in the Laurent series. From the series: \ \[ \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \cdots, \] the coefficient of \ \frac{1}{z} \ is 1.

Key concepts

series expansion

Understanding Series Expansion:
Series expansion is a way to represent a function as a sum of terms of a sequence or series. It's particularly useful in approximating functions that aren’t easy to handle.
In the context of complex analysis, series expansions include Taylor and Laurent series.
While a Taylor series represents a function as a sum of its derivatives at a single point, a Laurent series can include terms with negative powers. This makes it suitable for handling functions with singularities.
For the function \(\frac{1}{z}\), we use the sine series: \[ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\] Substituting \(\frac{1}{z}\) into this series expands the function into a Laurent series.

residue theorem

The Residue Theorem Simplified:
The residue theorem is a powerful tool in complex analysis used to evaluate complex integrals. Instead of finding antiderivatives, we focus on singularities (poles) of a function inside a contour. The theorem states: For a meromorphic function \(f(z)\) and a closed contour \(\text{C}\), the integral: \[ \int_C f(z) dz = 2\pi i \sum \text{Residues of f(z)} \] The residues are the coefficients of \(\frac{1}{z - z_0}\) in the Laurent series of \(\text{C}\). In our function \(\sin \frac{1}{z}\text{,}\) at \(z = 0\), the residue equals the coefficient of \(\frac{1}{z}\)\text{, which is 1 in this case.}

complex analysis

Diving into Complex Analysis:
Complex analysis studies functions of complex variables. It extends real analysis concepts to complex numbers.
Core concepts include complex differentiation, integration, and series expansions. Laurent series and the residue theorem are integral parts.
Laurent series allow functions to be expressed with positive and negative power terms, offering insights into singular behavior.
The residue theorem leverages residues for integrals, simplifying computations involving complex functions.

mathematical functions

Grasping Mathematical Functions:
Mathematical functions map inputs to outputs. In complex analysis, functions deal with complex numbers.
For instance, \(\sin(\frac{1}{z})\) maps complex inputs \(z\) to output values based on the sine function.
Functions can be represented via series expansions (Taylor or Laurent series). Through these expansions, we analyze their behavior and singularities.
Understanding functions in terms of series aids in solving integrals, finding residues, and exploring deeper properties within complex analysis.