Problem 22 - Advanced Engineering Mathematics

Question

Expand \(f(z)=\frac{1}{z(1-z)^{2}}\) in a Laurent series valid for the indicated annular domain. \(|z|>1\)

Step-by-Step Solution

Answer

The Laurent series for \(|z|>1\) is \( \sum_{n=0}^{\infty} (n+1) z^{-(n+1)} \).

Step 1: Rewriting the Function

To expand the given function in a Laurent series, let's first rewrite the function in a form that is easier to expand. The given function is \( f(z) = \frac{1}{z(1-z)^2} \). We aim to express it in a form suitable for expansion around \(|z|>1\). By factoring out the \(z\), the function becomes \( \frac{1}{z} \cdot \frac{1}{(1-z)^2} \).

Step 2: Expressing in Binomial Series

The term \( \frac{1}{(1-z)^2} \) can be expressed using the binomial series expansion formula for \(|z|>1\). Recall that \( \frac{1}{(1-w)^n} = \sum_{k=0}^{\infty} \binom{n+k-1}{k} w^k \) for \(|w| < 1\). Thus, we consider \(w = -z\), which gives us \( \frac{1}{1-z} = \sum_{k=0}^{\infty} z^k \) and subsequently \( \frac{1}{(1-z)^2} = \sum_{k=0}^{\infty} (k+1) z^k \) for \(|z|>1\).

Step 3: Substituting the Series

Substitute the series back into the original expression. Therefore, \( f(z) = \frac{1}{z} \sum_{k=0}^{\infty} (k+1) z^k = \sum_{k=0}^{\infty} (k+1) z^{k-1} \).

Step 4: Adjusting for \\( |z| > 1 \\\)

For \( |z| > 1 \), each term \( z^{k-1} \) should transform. To do that, let's express \( z^{k-1} \) in the inverse power form to ensure a valid Laurent series for \(|z|>1\), specifically transforming \( z^k \) to \( \frac{1}{z^{-k}} \). Hence, \( f(z) = \sum_{k=0}^{\infty} (k+1) z^{k-1} = \sum_{k=1}^{\infty} k z^{-k} \).

Step 5: Final Series Representation for \\( |z| > 1 \\\)

Thus, we rearranged the terms to obtain a Laurent series expansion valid for \(|z|>1\): \( f(z) = \sum_{n=0}^{\infty} (n+1) z^{-(n+1)} \). Therefore, \( f(z) = z^{-1} + 2z^{-2} + 3z^{-3} + \cdots \).

Key Concepts

Complex Analysis

Complex analysis is a branch of mathematics that studies functions of complex variables. It is a powerful tool that provides deep insights, especially when dealing with problems in the field of engineering and physics.
One core concept of complex analysis is the function of a complex variable, denoted commonly as \( f(z) \), where \( z \) is a complex number. Complex functions behave differently compared to real-valued functions because they involve both real and imaginary components.
In complex analysis, series expansions like the Laurent series or Taylor series help analyze and approximate functions. These expansions can reveal the function behavior near singularities, like poles, or along specific regions such as annular domains.

Series Expansion

Series expansion in complex analysis is a way to express a complex function as an infinite sum of simpler terms. Two common series types are the Taylor series and the Laurent series.
A Laurent series generalizes the Taylor series by allowing for terms with negative exponents. This makes it particularly useful for functions with singularities, where just positive powers wouldn't suffice.

• The Laurent series is written as: \( \sum_{n=-\infty}^{\infty} c_n z^n \)
• The coefficients \( c_n \) are derived from the function being expanded.

For instance, given the function \( f(z) = \frac{1}{z(1-z)^2} \), expanding it into a Laurent series involves representing it around specific domains like \(|z| > 1\), ensuring convergence and tractability in those regions.

Annular Domain

An annular domain in complex analysis is a ring-shaped region in the complex plane defined by two concentric circles with different radii. It is often represented as \( r_1 < |z - z_0| < r_2 \).
Annular domains are particularly useful for functions that behave differently in different regions of the complex plane. They allow for functions to be analytically continued or expanded in series like the Laurent series.

• In the exercise, the annular domain of \(|z| > 1\) indicates that we are considering a region outside the unit circle centered at the origin.
• This domain affects how the Laurent series is formed because the series must converge outside the boundary specified by \(|z| = 1\).

Understanding an annular domain's role is crucial for properly expanding a function and ensuring that the series complies with the function's behavior over that specific region.