Problem 22
Question
Expand \(f(z)=\frac{1}{z(1-z)^{2}}\) in a Laurent series valid for the given annular domain. $$ |z|>1 $$
Step-by-Step Solution
Verified Answer
The Laurent series for \( f(z) = \frac{1}{z(1-z)^2} \) around \(|z|>1\) is \( \sum_{n=0}^{\infty} (n+1) z^{n-1} \).
1Step 1: Recognize the Function Requirement
We need a Laurent series expansion for the function \( f(z) = \frac{1}{z(1-z)^2} \) valid in the annular region where \( |z| > 1 \). This involves rewriting the function in a form suitable for expansion in this region.
2Step 2: Rewrite the Denominator
The function \( f(z) = \frac{1}{z(1-z)^2} \) involves the term \((1-z)^2\). Expand \((1-z)^{-2}\) using the binomial series formula for \(|1/z|<1\) as \((1-z)^{-2} = \sum_{n=0}^{\infty} \binom{n+1}{1} z^n = \sum_{n=0}^{\infty} (n+1) z^n \) after rewriting for our domain.
3Step 3: Substitute Binomial Expansion into the Function
Substitute the expanded series from Step 2 into \( f(z) \): \[ f(z) = \frac{1}{z} \cdot \sum_{n=0}^{\infty} (n+1) z^n \]. Simplify to get \[ f(z) = \sum_{n=0}^{\infty} (n+1) z^{n-1} \], resulting in a Laurent series.
4Step 4: Reorganize Terms for the Region |z| > 1
Recognize that for \(|z|>1\), the series may need adjustment, but here it is sufficient as is because it represents a series consisting of terms of powers of \(z\) with no negative powers beyond \(z^{-1}\), matching the domain requirement.
Key Concepts
Complex AnalysisBinomial SeriesFunction Expansion
Complex Analysis
Complex analysis is a fascinating branch of mathematics that focuses on functions of complex numbers. With their roots in real numbers, complex numbers include an imaginary unit, defined as \(i\), with the property \(i^2 = -1\). This property makes complex analysis a rich area with unique techniques and results.
Unlike real analysis, complex analysis leverages the properties of complex differentiability, leading to the development of powerful tools such as the Laurent series. The Laurent series is particularly useful for representing complex functions in regions where they might have singularities.
In the context of the given exercise, we explore the expansion of a function in a Laurent series. A fundamental concept in complex analysis, the Laurent series allows representation of a function on an annular domain. This is crucial when working with functions that possess regions of convergence beyond simple disks, a common scenario when the domain is defined as \(|z|>1\).
Unlike real analysis, complex analysis leverages the properties of complex differentiability, leading to the development of powerful tools such as the Laurent series. The Laurent series is particularly useful for representing complex functions in regions where they might have singularities.
In the context of the given exercise, we explore the expansion of a function in a Laurent series. A fundamental concept in complex analysis, the Laurent series allows representation of a function on an annular domain. This is crucial when working with functions that possess regions of convergence beyond simple disks, a common scenario when the domain is defined as \(|z|>1\).
- It extends the idea of a Taylor series, including terms with negative powers of \(z\).
- The inclusion of negative exponentials allows representation around singular points.
Binomial Series
The binomial series plays a key role in the expansion of functions, especially when dealing with powers that aren't integers. It extends the ordinary binomial theorem, which you might recognize from algebra, into the realm of infinite series.
The binomial series is expressed as:
\[(1 + x)^{u} = \sum_{n=0}^{ ext{∞}} \binom{u}{n} x^n\]
where \(u\) is any real number and \( \binom{u}{n} \) represents the generalized binomial coefficient. This projectively allows us to deal with fractional and negative powers.
In our solution, we used the binomial series to expand \((1-z)^{-2}\) for the function \(f(z)=\frac{1}{z(1-z)^2}\). This transformation is valid when \(|z| > 1\), hence allowing us to express the function in terms of a power series suitable for the Laurent series.
The binomial series is expressed as:
\[(1 + x)^{u} = \sum_{n=0}^{ ext{∞}} \binom{u}{n} x^n\]
where \(u\) is any real number and \( \binom{u}{n} \) represents the generalized binomial coefficient. This projectively allows us to deal with fractional and negative powers.
In our solution, we used the binomial series to expand \((1-z)^{-2}\) for the function \(f(z)=\frac{1}{z(1-z)^2}\). This transformation is valid when \(|z| > 1\), hence allowing us to express the function in terms of a power series suitable for the Laurent series.
- Enables expansion of expressions with real exponents or negative integers.
- Adapts traditional binomial coefficients for more complex calculations.
Function Expansion
Function expansion is an important tool in mathematics, allowing functions to be expressed as series. This technique is especially useful in complex analysis, where functions may be expanded using either Taylor or Laurent series, depending on the nature of the function and the region of interest.
In our given problem, when we looked at \(f(z) = \frac{1}{z(1-z)^2}\), we found success by employing the binomial expansion followed by organizing the function into a Laurent series valid in the region \(|z| > 1\). This was achieved by:
Laurent series are special since they allow us to include terms with negative powers of \(z\). This makes them suited for domains that include annular regions around singularities, which occurs in this case of \(|z| > 1\). Thus, through function expansion, we can get a clearer understanding and representation of complex functions across various domains.
In our given problem, when we looked at \(f(z) = \frac{1}{z(1-z)^2}\), we found success by employing the binomial expansion followed by organizing the function into a Laurent series valid in the region \(|z| > 1\). This was achieved by:
- First using a binomial series to handle the expression \((1-z)^{-2}\).
- Then rewriting the function in terms of a series expansion.
Laurent series are special since they allow us to include terms with negative powers of \(z\). This makes them suited for domains that include annular regions around singularities, which occurs in this case of \(|z| > 1\). Thus, through function expansion, we can get a clearer understanding and representation of complex functions across various domains.
Other exercises in this chapter
Problem 22
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