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.) $$\frac{\sin z}{z^{4}}, z=0$$

Short answer

Laurent series: \ \[ z^{-3} - \frac{1}{6}z^{-1} + \frac{1}{120}z + \frac{1}{5040}z^3 + \frac{1}{362880}z^5 + \text{ ...} \] \ Residue: \ \( -\frac{1}{6} \)

Step-by-step solution

- Express the given function in simpler terms

Write the function in a way that separates the known series expansions. The given function is \ \( \frac{\text{sin} z}{z^4} \).

- Recall the Taylor series expansion for sine

The Taylor series expansion for \ \( \text{sin} z \) around \ \( z=0 \) is: \ \[ \text{sin} z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \frac{z^9}{9!} - \frac{z^{11}}{11!} + \text{ ... } \]

- Divide by \ \( z^4 \)

Divide each term in the Taylor series for \ \( \text{sin} z \) by \ \( z^4 \): \ \[ \frac{\text{sin} z}{z^4} = \frac{z}{z^4} - \frac{z^3}{3!z^4} + \frac{z^5}{5!z^4} - \frac{z^7}{7!z^4} + \frac{z^9}{9!z^4} - \frac{z^{11}}{11!z^4} + \text{ ... } \]

- Simplify the expression

Simplify each term in the series: \ \[ \frac{\text{sin} z}{z^4} = z^{-3} - \frac{1}{6}z^{-1} + \frac{1}{120}z + \frac{1}{5040}z^3 + \frac{1}{362880}z^5 + \text{ ...} \]

- Identify the Laurent series and the residue

Identify the Laurent series for \ \( \frac{\text{sin} z}{z^4} \) near \ \( z=0 \): \ \[ \frac{\text{sin} z}{z^4} = z^{-3} - \frac{1}{6}z^{-1} + \frac{1}{120}z + \frac{1}{5040}z^3 + \frac{1}{362880}z^5 + \text{ ...} \] \ The residue is the coefficient of the \ \( z^{-1} \) term, which is \ \( -\frac{1}{6} \).

Key concepts

complex functions

Complex functions are functions that map complex numbers to complex numbers. These functions can be expressed in the form f(z) where z is a complex number. In our exercise, the function given is \ \( \frac{\sin z}{z^4} \) \, which is an example of a complex function.

Understanding complex functions involves:

• Domain: The set of complex numbers where the function is defined.
• Range: The set of complex numbers that result from applying the function to the domain.

Complex functions are vital in fields like engineering, physics, and applied mathematics. They are also fundamental when discussing series like Taylor and Laurent series in complex analysis.

Taylor series

The Taylor series is a way to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is typically written as:

\ \[ f(z) = f(a) + f'(a)(z-a) + \frac{f''(a)}{2!}(z-a)^2 + \frac{f'''(a)}{3!}(z-a)^3 + \cdots \] \
In our exercise, the Taylor series of \ \( \sin z \) \ around \ \( z=0 \) \ is crucial. The series expansion of \ \( \sin z \) \ is:

\ \[ \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \frac{z^9}{9!} - \frac{z^{11}}{11!} + \cdots \]
Each term is derived from evaluating the function and its higher-order derivatives at \ \( 0 \). Notice how alternating signs and increasing factorials appear in the series. This is typical of many Taylor series.

Taylor series help break down complex functions into simpler infinite sums, making them pivotal in studying and solving problems involving complex functions.

residue calculus

Residue calculus involves finding residues of functions, which are useful in evaluating complex integrals. The residue of a function at a point is essentially the coefficient of the \ \( z^{-1} \) term in a Laurent series expansion around that point.

In our exercise, we found that the Laurent series for \ \( \frac{\sin z}{z^4} \) is:

\ \[ \frac{\sin z}{z^4} = z^{-3} - \frac{1}{6}z^{-1} + \frac{1}{120}z + \frac{1}{5040}z^3 + \frac{1}{362880}z^5 + \cdots \]

Here, the residue at \ \( z=0 \) is \ \( -\frac{1}{6} \), the coefficient of the \ \( z^{-1} \) term.

Residue calculus is extensively used in complex analysis, particularly for evaluating contour integrals using the residue theorem. The theorem states that the integral of a function around a closed contour is \ \( 2\pi i \) times the sum of residues within the contour. Thus, understanding residues helps in solving complex integrals, making problems like this exercise far more approachable.

series expansion

Series expansion involves expressing functions as sums of sequences, often infinite, of terms. It's a powerful method for simplifying complex functions and solving equations in various fields of mathematics and physics.

The Laurent series is a type of series expansion used specifically for functions with singularities (points where they go to infinity or become undefined). Unlike the Taylor series, which works well for functions analytic at a point, the Laurent series also includes negative powers of \ \( z \).

In our exercise, we derive the Laurent series by dividing the Taylor series of \ \(\sin z\) by \ \(z^4\):

\ \[ \frac{\sin z}{z^4} = \frac{z}{z^4} - \frac{z^3}{3!z^4} + \frac{z^5}{5!z^4} - \frac{z^7}{7!z^4} + \cdots \]

Simplifying this, we get:

\ \[ \frac{\sin z}{z^4} = z^{-3} - \frac{1}{6}z^{-1} + \frac{1}{120}z + \frac{1}{5040}z^3 + \cdots \]

Understanding series expansion is key to dealing with complex functions and their behaviors around singularities, as demonstrated in resolving the given exercise.