Problem 26 - A First Course in Complex Analysis with Applications

Question

Determine the order of the poles for the given function. $$ f(z)=\frac{\cos z-\cos 2 z}{z^{6}} $$

Step-by-Step Solution

Answer

The order of the pole at \(z = 0\) is 4.

Step 1: Identify the Poles

A pole occurs where the denominator goes to zero but the numerator does not. Here, the denominator is \(z^6\), which has a zero at \(z=0\). Therefore, \(z=0\) is a candidate for a pole.

Step 2: Find the Order of the Zero in the Numerator at \(z=0\)

The function \(\cos z\) has a Taylor series expansion about \(z=0\) given by \(1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \ldots\), and \(\cos 2z\) is \(1 - \frac{(2z)^2}{2!} + \frac{(2z)^4}{4!} - \ldots\). Thus, \(\cos z - \cos 2z\) starts as \(-\frac{z^2}{2} + 2z^2 = \frac{3z^2}{2} - \frac{z^4}{12} + \ldots\).This shows that \(\cos z - \cos 2z\) has a zero of order 2 at \(z=0\).

Step 3: Determine the Order of the Pole

The order of the pole is the order of the zero in the denominator minus the order of the zero in the numerator. Here, the denominator has a zero of order 6 at \(z=0\), and the numerator has a zero of order 2 at \(z=0\). The order of the pole at \(z=0\) is therefore \(6 - 2 = 4\).

Key Concepts

Poles and Zeros

In complex analysis, poles and zeros are critical concepts when studying the behavior of complex functions. A **zero** of a function refers to a point where the function's value becomes zero. Conversely, a **pole** is a type of singularity where a complex function goes to infinity.

• **Zeros**: Indicate points where the numerator of a function equals zero.
• **Poles**: Occur at points where the denominator equals zero and the numerator doesn't.

For the function in the original exercise, \( f(z) = \frac{\cos z - \cos 2z}{z^6}\)the denominator \(z^6\) equals zero when \(z = 0\). Thus, \(z = 0\) is a candidate for a pole, provided the numerator does not contribute a zero of equal or higher order at that point.

Taylor Series

The **Taylor series** is an incredibly useful tool in mathematics for expressing functions as infinite sums of terms calculated from the values of their derivatives at a single point. When analyzing \(\cos z - \cos 2z\)the Taylor series expansion helps identify zeros and evaluate the function's behavior around point \(z=0\).

• **Taylor expansion of \(\cos z\)** starts as: \(1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \ldots\)
• **Taylor expansion of \(\cos 2z\)** starts as: \(1 - \frac{(2z)^2}{2!} + \frac{(2z)^4}{4!} - \ldots\)

Calculating \(\cos z - \cos 2z\) shows the function's leading term is \(\frac{3z^2}{2}\), indicating a zero of order 2 at \(z=0\). Understanding Taylor series allows us to see the difference between the starting terms of the series, essential for understanding the zero's order in the numerator.

Order of a Pole

The **order of a pole** provides insight into how singular a function is at a given point. Determining the order involves comparing the zeros present in both the numerator and denominator. For a function \(\frac{P(z)}{Q(z)}\), if point \(z = a\) is where both \(P(z)\) and \(Q(z)\) equal zero, the order of the pole at \(z = a\) is the difference in the orders of zero of \(Q(z)\) minus \(P(z)\).

• **Numerator zero** order at \(z=0\) was found to be 2 from \(\cos z - \cos 2z\).
• **Denominator zero** order for \(z^6\) is 6.

Therefore, the order of the pole for the original exercise at \(z=0\) is \(6 - 2 = 4\).
This concept provides a clear method to identify and classify the nature of poles, allowing deeper insight into function behavior.