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.) $$ \frac{1}{z^{2}-5 z+6}, z=2 $$

Short answer

Residue at \( z = 2 \) is \( -1 \).

Step-by-step solution

Factor the Denominator

First, factorize the denominator of the function. Given function is \( \frac{1}{z^2 - 5z + 6} \). The denominator can be factored as: \( (z - 2)(z - 3) \). Therefore, the function becomes \( \frac{1}{(z - 2)(z - 3)} \).

Express the Function in Partial Fractions

Express \( \frac{1}{(z - 2)(z - 3)} \) as a sum of partial fractions: \( \frac{A}{z - 2} + \frac{B}{z - 3} \). Solve for A and B by clearing the fractions: \( 1 = A(z - 3) + B(z - 2) \).

Determine the Constants A and B

Set \( z = 2 \): \( 1 = A(2-3) + B(2-2) \Rightarrow 1 = -A \Rightarrow A = -1 \). Set \( z = 3 \): \( 1 = A(3-3) + B(3-2) \Rightarrow 1 = B \Rightarrow B = 1 \). Therefore, \( \frac{1}{(z - 2)(z - 3)} = \frac{-1}{z - 2} + \frac{1}{z - 3} \).

Expand the Partial Fractions

For \( \frac{-1}{z - 2} \), it is already in the required form. For \( \frac{1}{z - 3} \), write it in terms of \( z - 2 \): \( \frac{1}{z - 3} = \frac{1}{(z - 2) - 1} = \frac{1}{-(1 - (z - 2))} \Rightarrow - \frac{1}{1 - (z - 2)} \). Use the geometric series \(\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n \) to expand it: \( -\frac{1}{1 - (z - 2)} = - \sum_{n=0}^{\infty} (z-2)^n \).

Write the Laurent Series

Combining both terms, the Laurent series of \( \frac{1}{z-2} + -\sum_{n=0}^{\infty} (z-2)^n \), convergence near \( z = 2 \), is: \( -\frac{1}{z-2} - \sum_{n=0}^{\infty}(z-2)^n \).

Identify the Residue

The residue of the function at \( z = 2 \) is the coefficient of \( \frac{1}{z-2} \) term in the Laurent series. From our series, it is \( -1 \).

Key concepts

complex analysis

Complex analysis deals with functions of a complex variable. These functions are defined on the complex plane, which includes real and imaginary parts. One important concept in complex analysis is the Laurent series. A Laurent series is like a Taylor series but allows for terms with negative powers. This makes it useful for representing functions with singularities (points where functions go to infinity or are otherwise undefined). Laurent series can be used to analyze and understand these singularities and to find residues of functions at specific points.

residue theorem

The residue theorem is a powerful tool in complex analysis. It relates complex line integrals to the residues of the function's singularities. A residue at a point is a specific coefficient in the Laurent series expansion around that point. The residue theorem states that if you have a function with isolated singularities surrounded by a closed contour, the integral of the function around that contour is equal to \(2πi\) times the sum of the residues inside the contour. This makes it easier to evaluate complex integrals and has applications in various fields, like theoretical physics and engineering.

partial fractions

Partial fractions are a method used to break down rational functions into simpler fractions. This is helpful in integration and in finding the Laurent series. To use partial fractions, you factor the denominator of the rational function and express it as a sum of simpler fractions. For example, given the function \(\frac{1}{z^2 - 5z + 6}\), we first factorize the denominator to get \((z - 2)(z - 3)\). Then, we express the function as a sum of partial fractions like \(\frac{A}{z - 2} + \frac{B}{z - 3}\) and solve for constants \(A\) and \(B\).

geometric series

A geometric series is a series with a constant ratio between successive terms. The general form is \(\frac{1}{1 - w} = \text{} _{n=0}^{∞} w^n\). This concept is useful in complex analysis, particularly in expanding functions into Laurent series. For instance, if we have \(\frac{1}{z - 3}\), we can rework it into a form suitable for geometric series expansion, resulting in \(- \text{}^{∞}_{n=0} (z-2)^n\). This approach simplifies dealing with functions around a point, like in the example function \(\frac{1}{z^2 - 5z + 6}\) near \(z = 2\).