Problem 23 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The function has a pole of order 1 at \( z=0 \) and a pole of order 1 at \( z=1 \).

Step 1: Identify Denominator Zeroes

The denominator of the function is given by \( z^2 - z = 0 \). To find the zeroes, solve the equation: \( z(z - 1) = 0 \). This gives us zeroes at \( z = 0 \) and \( z = 1 \). These are the potential poles of the function.

Step 2: Check Numerator at Poles

The numerator is \( \sin z \). First, substitute \( z = 0 \) into \( \sin z \). We know \( \sin(0) = 0 \), so \( z = 0 \) is not a simple pole; it is actually a potential higher order pole since both numerator and denominator are zero at \( z = 0 \).

Step 3: Determine Order at Pole 0

Around \( z = 0 \), \( \sin z \) can be expanded using its Taylor series: \( \sin z = z - \frac{z^3}{3!} + \cdots \). This series shows that \( \sin z \) has a zero of order 1 at \( z = 0 \). Therefore, replace \( \sin z \) with \( z imes g(z) \), where \( g(z) \) is a series that is non-zero at \( z = 0 \). The denominator has zero of order 2 at \( z = 0 \), so the pole at \( z = 0 \) is of order \( 2 - 1 = 1 \).

Step 4: Determine Order at Pole 1

At \( z = 1 \), \( \sin(1) eq 0 \), meaning the numerator does not vanish. The denominator presents a zero of order 1 at \( z = 1 \). Thus, there is a simple pole of order 1 at \( z = 1 \).

Key Concepts

Poles of a Function

In complex analysis, poles of a function are pivotal in understanding its behavior around certain points. A pole is essentially a point where a function goes off to infinity. If you think about a graph, it is where the curve "shoots" upwards or downwards infinitely. Poles assist in characterizing singularities, places where functions are not well-defined in a particular sense.

To determine poles, you'll need to identify zeroes in the denominator of a given function. These are spots where the function could become undefined, thus potentially rendering them as poles. In the given function, \( f(z) = \frac{\sin z}{z^{2}-z} \), finding poles begins with solving \( z^{2} - z = 0 \) leading to \( z = 0 \) and \( z = 1 \). Each of these zeroes provides a point of interest where poles might exist.

Depending on the order of zeroes found in the denominator and the behavior of the numerator, you can determine the order of these poles as we will see more into when discussing simple poles.

Taylor Series in Complex Analysis

Taylor series are a powerful tool in complex analysis for approximating functions around a specific point. This series expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point.

To dive deeper: if you have a function \( f(z) \), its Taylor series around \( z = 0 \) can be written as:

• \( f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \cdots \)

This representation is vital in analyzing functions at points where they are not straightforward.

In the context of our exercise, we used the Taylor series for \( \sin z \) at \( z = 0 \), which is \( \sin z = z - \frac{z^3}{3!} + \cdots \). This expansion indicates a zero of order 1 at \( z = 0 \), showing how such series accounts for the behavior of functions.

Taylor series in complex analysis becomes more helpful when it shows how both the numerator and denominator behave, particularly at potential poles to assess their order.

Simple Pole Determination

Determining simple poles involves assessing both the numerator and denominator of a function at specific points of interest—usually the zeroes of the denominator. A simple pole is essentially a zero of the denominator that is not canceled by the numerator.

Consider the function \( f(z) = \frac{\sin z}{z^{2}-z} \) as an example:

• At \( z = 0 \), \( \sin z = 0 \) results in both numerator and denominator being zero. This cancels out one order of zero from the denominator, resulting in a pole of order \( 2 - 1 = 1 \).
• At \( z = 1 \), \( \sin(1) eq 0 \), so only the denominator has a zero. Thus, \( z = 1 \) is a simple pole, as there is an order 1 zero only in the denominator.

It’s crucial to note that simple poles are the points where the function sharply turns undefined, contributing greatly to understanding complex functions' singular structures.

Recognizing simple poles and understanding their implications provide essential insight into how analytic properties evolve in complex analysis.