Chapter 14: Problem 24
Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer. $$\frac{1-\cos 2 z}{z^{3}} \text { at } z=0$$
Short Answer
Expert verified
The residue at \(z = 0\) is \(2\).
Step by step solution
01
Identify the Laurant series
To find the residue at a particular point, identify the Laurent series for the given function around the point. For the function \(\frac{1-\text{cos} \, 2z}{z^3}\) at \(z = 0\), we start with the Taylor series expansion of \(\text{cos} \, 2z\): \[ \text{cos} \, 2z = 1 - \frac{(2z)^2}{2!} + \frac{(2z)^4}{4!} - \frac{(2z)^6}{6!} + \cdots = 1 - 2z^2 + \frac{(2z)^4}{24} - \cdots \]
02
Substitute Taylor series into the function
Substitute the Taylor series of \(\text{cos} \, 2z\) into the original function: \[ \frac{1 - \left(1 - 2z^2 + \frac{(2z)^4}{24} - \cdots \right)}{z^3} = \frac{1 - 1 + 2z^2 - \frac{(2z)^4}{24} + \cdots}{z^3} = \frac{2z^2 - \frac{(2z)^4}{24} + \cdots}{z^3} \]
03
Simplify the fraction
Simplify the fraction: \[ \frac{2z^2}{z^3} - \frac{(2z)^4}{24z^3} + \cdots = \frac{2}{z} - \frac{2^4z}{24} + \cdots = \frac{2}{z} - \frac{16z}{24} + \cdots \]
04
Identify the residue
The residue is the coefficient of the \(\frac{1}{z}\) term in the Laurent series. In this case, the coefficient is \(2\). So, the residue at \(z = 0\) is \(\text{Res}(f, 0) = 2\).
05
Verify using a computer algebra system
To ensure the result is correct, use a computer algebra system like Mathematica, Maple, or an online tool to compute the residue. Confirm that the residue at \(z = 0\) is indeed \(2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laurent series
The Laurent series is a way to represent complex functions. It extends the idea of the Taylor series by including terms with negative powers of the variable. Laurent series are particularly useful when dealing with functions that have singularities (points where they are not defined or have infinite values).
In simple terms, while a Taylor series represents a function as an infinite sum of its derivatives at a single point, a Laurent series does that and more. It also captures the behavior of the function near singularities. For example, the given function, \[ \frac{1-\cos 2 z}{z^3} \], can be tricky near \( z = 0 \). Using a Laurent series helps break down and understand this behavior.
When you work with Laurent series, you typically aim to identify the coefficient of the term with \( \frac{1}{z} \). This coefficient is known as the residue of the function at that point. For instance, in our function, the residue at \( z = 0 \) turns out to be 2. Remember, finding the correct residue can be crucial as it often simplifies complex calculations within the realm of complex analysis.
In simple terms, while a Taylor series represents a function as an infinite sum of its derivatives at a single point, a Laurent series does that and more. It also captures the behavior of the function near singularities. For example, the given function, \[ \frac{1-\cos 2 z}{z^3} \], can be tricky near \( z = 0 \). Using a Laurent series helps break down and understand this behavior.
When you work with Laurent series, you typically aim to identify the coefficient of the term with \( \frac{1}{z} \). This coefficient is known as the residue of the function at that point. For instance, in our function, the residue at \( z = 0 \) turns out to be 2. Remember, finding the correct residue can be crucial as it often simplifies complex calculations within the realm of complex analysis.
Taylor series
The Taylor series is a representation of a function as an infinite sum of terms. Each term is calculated from the derivatives of the function at a single point. In essence, it allows you to express complex functions as simpler polynomial series. This is immensely useful in both real and complex analysis.
For instance, consider the exponential function \( e^z \). Its Taylor series around \( z = 0 \) is given by: \[ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \]
Utilizing the Taylor series in our problem, we expand \( \cos 2z \) near \( z = 0 \): \[ \cos 2z = 1 - 2z^2 + \frac{(2z)^4}{4!} - \cdots \]
This helps to transform our initial function \( \frac{1-\cos 2 z}{z^3} \) into a form that's easier to analyze. Substituting this series back into the function simplifies the problem considerably. By expressing the function as a series, you can more easily identify and extract terms that help determine the residue.
For instance, consider the exponential function \( e^z \). Its Taylor series around \( z = 0 \) is given by: \[ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \]
Utilizing the Taylor series in our problem, we expand \( \cos 2z \) near \( z = 0 \): \[ \cos 2z = 1 - 2z^2 + \frac{(2z)^4}{4!} - \cdots \]
This helps to transform our initial function \( \frac{1-\cos 2 z}{z^3} \) into a form that's easier to analyze. Substituting this series back into the function simplifies the problem considerably. By expressing the function as a series, you can more easily identify and extract terms that help determine the residue.
Complex functions
Complex functions are mappings from complex numbers to complex numbers. They can often exhibit rich and intriguing behaviors not seen in real functions. Some essential concepts linked to complex functions include:
For example, the function \( \frac{1-\cos 2 z}{z^3} \) has a singularity at \( z = 0 \). Using tools like the Laurent and Taylor series allows us to understand and work with these singularities effectively.
By breaking down the function into parts, series expansions offer a deeper understanding of how complex functions operate around critical points. This insight makes it easier to solve complex analysis problems, such as finding residues.
- Analyticity: A function is analytic if it is differentiable at every point in its domain. Analytic functions can be represented by a Taylor series.
- Singularities: Points where the function does not behave nicely (e.g., becomes infinite or is not defined). Understanding these points using Laurent series is crucial in complex analysis.
For example, the function \( \frac{1-\cos 2 z}{z^3} \) has a singularity at \( z = 0 \). Using tools like the Laurent and Taylor series allows us to understand and work with these singularities effectively.
By breaking down the function into parts, series expansions offer a deeper understanding of how complex functions operate around critical points. This insight makes it easier to solve complex analysis problems, such as finding residues.