Problem 16 - Advanced Engineering Mathematics

Question

In Problems 13-24, determine the order of the poles for the given function. $$ f(z)=\frac{z-1}{(z+1)\left(z^{3}+1\right)} $$

Step-by-Step Solution

Answer

The pole at \(z = -1\) is of order 2; the other poles from \(z^2-z+1 = 0\) are of order 1.

Step 1: Identify the factors of the denominator

First, factor the denominator of the function. The denominator is \((z+1)(z^3+1)\). The term \(z^3+1\) can be factored using the sum of cubes formula: \(z^3+1 = (z+1)(z^2-z+1)\). Therefore, the denominator becomes \((z+1)^2(z^2-z+1)\).

Step 2: Identify the poles of the function

A pole is a value of \(z\) for which the function becomes undefined due to division by zero in the denominator. Set the denominator equal to zero: \((z+1)^2(z^2-z+1) = 0\). Solving \((z+1)^2 = 0\), we get a pole at \(z = -1\) with multiplicity 2. Solving \(z^2-z+1 = 0\) using the quadratic formula yields two complex poles, but these are not repeated, thus each has multiplicity 1.

Step 3: Determine the order of each pole

The order of a pole is equivalent to its multiplicity. For \(z = -1\), the pole is of order 2, because \((z+1)^2\) implies it is repeated twice. The poles from \(z^2-z+1 = 0\) are of order 1, because they appear once without repetition.

Key Concepts

Poles in Complex Functions

In complex analysis, poles represent points in the complex plane where a function takes an infinite value. Understanding poles is crucial because they describe the behavior of a complex function near these points.
To find poles:

• Look for values of the variable at which the denominator of a rational function equals zero.
• The function becomes undefined at these values, indicating the presence of a pole.

For example, in the function \( f(z)= \frac{z-1}{(z+1)*(z^3+1)}\), a pole exists where \( (z+1)(z^3+1) = 0 \).
Poles are critical when analyzing the properties of a complex function, particularly in relation to residues and contour integration, where they play a significant role in evaluating complex integrals.

Order of Poles

The order of a pole indicates how many times a particular value causes the function to be undefined. It is synonymous with the concept of multiplicity from algebra. The larger the order, the stronger the asymptotic behavior of the function near that pole.
For example:

• If a function \( f(z) \) has a pole at \( z=a \) and can be expressed as \( rac{g(z)}{(z-a)^n} \), where \( g(z) \) is analytic and \( g(a) eq 0 \), it has a pole of order \( n \).
• In the function example given, \( z = -1 \) is a pole with an order of 2 because \( (z+1)^2 \) appears in the factorization of the denominator.
• Other poles found using the quadratic formula appear with order 1, meaning they do not repeat.

Analyzing the order of poles helps to comprehend the intensity and nature of the singularity exhibited by the function there.

Factoring Polynomials

Factoring polynomials is an essential skill in determining the behavior of complex functions. By breaking down a complex expression into simpler polynomials, it is easier to identify zeros and poles.

• The process involves expressing a polynomial as a product of its root factors or smaller polynomials.
• This enables the understanding of the solution set of a polynomial equation.

In the given exercise, the denominator \(z^3+1\) was factored into \((z+1)(z^2-z+1)\), illustrating how factoring helps to identify solutions like \(z = -1\), which influence the presence of poles. Practice factoring regularly to build a strong foundation in handling polynomial expressions efficiently.

Sum of Cubes Formula

The sum of cubes formula is a technique used in algebra to factor certain expressions involving cubes.
It is represented as:\[ a^3 + b^3 = (a+b)(a^2-ab+b^2) \]This formula is particularly helpful when examining complex functions that include cubic components, such as \(z^3+1\).
Using this formula:

• Factoring \(z^3+1\) becomes simpler, resulting in \((z+1)(z^2-z+1)\).
• This breakdown assists in identifying the structure and roots of polynomial equations.

Beyond just factoring, understanding this formula is advantageous in showing the relationship between cube components and simplifying complex expressions.