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

Laurent series: \(\text{sin}( \frac{1}{z}) = \sum_{n=0}^{\text{∞}} (-1)^n \frac{1}{(2n+1)!z^{2n+1}} \); Residue: 1.

Step-by-step solution

- Express the Function in Terms of Known Series

Given the function \(\frac{1}{z}\), we need to find its Laurent series around \(z=0\). We start with the known Taylor series expansion for \(\frac{1}{z}\): \(\frac{1}{z} = z^{-1}\).

- Substitute into the Sine Series

Replace \(z\) in \(\frac{1}{z}\) into the Taylor series of \(\text{sin}(z)\). The Taylor series for \( \text{sin}(z)\) is: \(\text{sin}(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \frac{z^9}{9!} + \text{ ... } \).

- Create the Laurent Series

Substituting \(z^{-1}\) in place of \(z\) in the Taylor series for \(\text{sin}(z)\), the Laurent series for \(\text{sin}( \frac{1}{z} )\) around \(z = 0\): \(\text{sin}( \frac{1}{z}) = \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \frac{1}{5040z^7} + ... \) \( = \sum_{n=0}^{\text{∞}} (-1)^n \frac{1}{(2n+1)!z^{2n+1}}\)

- Identify the Residue

The residue of a function at a point is the coefficient of the \( \frac{1}{z} \) term in the Laurent series. Looking at the series: \(\text{sin}( \frac{1}{z} ) = \frac{1}{z} - \frac{1}{6z^3} + \frac{1}{120z^5} - \frac{1}{5040z^7} + ...\), The coefficient of \(\frac{1}{z}\) is 1.

Key concepts

Complex Analysis

Complex analysis is a powerful branch of mathematics that deals with functions of a complex variable. These functions are defined on the complex plane and exhibit richer behavior compared to real functions.
In complex analysis, we study various properties and representations of complex functions, such as their analyticity, which means being expressible by a convergent power series around any point within a certain region.
One of the central tools in this field is the Laurent series, which allows us to represent more complex functions by series that may include powers of negative exponents.

Residue Theorem

The Residue Theorem is a key result in complex analysis that simplifies the evaluation of contour integrals. It tells us that the integral of a function around a closed contour is related to the sum of residues of the function within the contour.
A residue at a pole is essentially the coefficient of the \frac{1}{z-z_0}\ term in the Laurent series expansion of a function around a pole \(z_0\).
This connection makes the Laurent series particularly useful because it breaks a function into simpler additive parts, allowing for easier identification and isolation of singularities.

Taylor Series

The Taylor series is a way to represent functions as infinite sums of their derivatives at a point. For a function \(f(z)\) that is analytic (i.e., infinitely differentiable) at a point \(z=a\), the Taylor series expansion around \(z=a\) is given by: \[f(z) = \frac{f(a)}{0!} + \frac{f'(a)(z-a)}{1!} + \frac{f''(a)(z-a)^2}{2!} + \frac{f^{(n)}(a)(z-a)^n}{n!} + \text{ ... } \] In the Step By Step Solution, \sin(z)\ is expanded using a Taylor series, but with \z\ replaced by \frac{1}{z}\ to form the Laurent series. This transition allows the expression to include negative powers of \z\, capturing more complex behavior near singularities like \z=0\.