Problem 14 - Advanced Engineering Mathematics

Question

In Problems 13-24, determine the order of the poles for the given function. $$ f(z)=5-\frac{6}{z^{2}} $$

Step-by-Step Solution

Answer

The function has a pole of order 2 at \(z = 0\).

Step 1: Understand the Function Components

The given function is \( f(z) = 5 - \frac{6}{z^2} \). This function consists of a constant term, \(5\), and a term with a negative power of \(z\), \(\frac{-6}{z^2}\). We need to determine if this creates a singularity, specifically a pole.

Step 2: Identify Singularity of the Function

A function has a pole where it is undefined but can be made finite by multiplying by a suitable power of \(z\). For \(f(z) = 5 - \frac{6}{z^2}\), observe the term \(\frac{-6}{z^2}\): it causes the function to be undefined at \(z = 0\) due to division by zero.

Step 3: Determine the Order of the Pole

The term \(\frac{6}{z^2}\) suggests a singularity at \(z=0\) due to \(z^{-2}\). The form of the term \(z^{-n}\) indicates a pole of order \(n\). Here, since we have \(z^{-2}\), it implies a pole of order \(2\).

Step 4: Conclusion: State the Order of the Pole

After examining the singular term \(\frac{6}{z^2}\), we conclude that the function \(f(z) = 5 - \frac{6}{z^2}\) has a pole of order 2 at \(z = 0\).

Key Concepts

Poles and Residues

In complex analysis, understanding poles and residues is essential for evaluating integrals and analyzing the behavior of complex functions. A pole is a specific kind of singularity where a function goes to infinity. To find where these poles occur, you examine points where the function is not well-defined, often due to a division by zero. In the given function, \( f(z) = 5 - \frac{6}{z^2} \), the term \( \frac{6}{z^2} \) suggests a potential pole by making the function undefined at \( z = 0 \). Residues are closely related to poles and offer numerical value insights associated with them. At a pole, the residue is the coefficient of \( \frac{1}{z-a} \) in the Laurent series expansion around the point \( a \). Although the exercise did not require calculating the residue, recognizing its importance is valuable when dealing with more advanced problems. Residues often play a crucial role in evaluating complex integrals using the residue theorem, especially when dealing with closed contour integrals.

Singularity Analysis

Singularity analysis is a method used to study the points where a function cannot be defined or extended as a function in its immediate vicinity. In our exercise, we analyze the singularity of \( f(z) = 5 - \frac{6}{z^2} \). The term \( \frac{6}{z^2} \) indicates a singularity at \( z=0 \), since dividing by zero is mathematically undefined.Understanding singularities involves:

• Identifying zeros in the denominator, indicating potential singular points.
• Determining whether these singularities are removable or if they are poles, essential singularities, or branch points.

A removable singularity can be "fixed" by redefining the function at that point, but poles, like the one in this function, are more persistent and tell more about the function's behavior near that point. In this exercise, the singularity at \( z=0 \) is non-removable and specifically a pole.

Order of a Pole

The order of a pole is a key concept and tells us how a function behaves as it approaches a singular point. To determine the pole's order, we're essentially assessing the number of times the variable \( z \) is raised to the power of negative one in the function's singular term.For \( f(z) = 5 - \frac{6}{z^2} \), the singular term \( \frac{6}{z^2} \) provides insight into the order of the pole:

• The power \( z^{-2} \) indicates a pole of order 2.
• A pole of order 2 means that near this point, the function's magnitude is dramatically increasing or decreasing as \( z \) approaches zero.

Understanding the order helps predict how a function behaves near its singularities, which is crucial in complex analysis for both theoretical studies and practical applications, such as contour integration or mapping properties of complex functions.