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Chapter 32: Problem 21

Man berechne die Laurent-Reihenentwicklung der Funktion $$ f(z)=\frac{1}{(z-1)(z-2)} $$ (a) f眉r \(|z|<1\), (b) f眉r \(1<|z|<2\) und (c) f眉r \(|z|>2\).

Short Answer

Expert verified
Based on the solution provided, the Laurent series expansion of the function $$f(z)=\frac{1}{(z-1)(z-2)}$$ in each of the three regions can be written as follows: 1. For \(|z|<1\), the Laurent series expansion is $$ f(z)=\sum_{n=0}^{\infty} z^n - \sum_{n=0}^{\infty} (z-1)^n $$ 2. For \(1<|z|<2\), the Laurent series expansion is $$ f(z)=\sum_{n=0}^{\infty} \frac{2(1-z)^n}{z-2} - \sum_{n=0}^{\infty} (z-1)^n $$ 3. For \(|z|>2\), the Laurent series expansion is $$ f(z) = \frac{1}{z}\Big(\sum_{n=0}^{\infty} \frac{1}{z^n} - \sum_{n=0}^{\infty} 2^n\cdot\frac{1}{z^n}\Big) $$

Step by step solution

01

Partial fraction decomposition

Firstly, we need to decompose the given function into partial fractions. We can write: $$ f(z)=\frac{1}{(z-1)(z-2)} = \frac{A}{z-1}+\frac{B}{z-2} $$ To determine A and B, we multiply both sides by the denominator \((z-1)(z-2)\), we get: $$ 1= A(z-2)+B(z-1) $$ Now, let's set \(z=1\) and \(z=2\) to calculate \(A\) and \(B\), respectively: 1. If \(z=1\), \(1= -A\), so \(A=-1\). 2. If \(z=2\), \(1= B\), so \(B=1\). Thus, the function becomes: $$ f(z) = - \frac{1}{z-1} + \frac{1}{z-2} $$
02

Laurent series expansion for \(|z|

In this case, we use the geometric series expansion: $$ \frac{1}{1-w}=\sum_{n=0}^{\infty} w^n $$ For \(f(z)\) in the region \(|z|<1\), we can write: $$ f(z) = -\frac{1}{z-1} + \frac{1}{z-2} = \frac{1}{1-z} - \frac{1}{2-z} $$ Now, let \(w=z\) for the first term and \(w = z-1\) for the second term to use the geometric series expansion: $$ f(z)=\sum_{n=0}^{\infty} z^n - \sum_{n=0}^{\infty} (z-1)^n $$
03

Laurent series expansion for \(1

In this case, we simplify \(f(z)\): $$ f(z) = \frac{1}{z-2}\Big(\frac{2}{z-1}\Big) - \frac{1}{z-1} $$ Let \(w=1-z\) for the first term and \(w = z-1\) for the second term to use the geometric series expansion: $$ f(z)=\sum_{n=0}^{\infty} \frac{2(1-z)^n}{z-2} - \sum_{n=0}^{\infty} (z-1)^n $$
04

Laurent series expansion for \(|z|>2\)

For \(f(z)\) in the region \(|z|>2\), we can write: $$ f(z) = -\frac{1}{z-1} + \frac{1}{z-2} = \frac{1}{z}\Big( -\frac{1}{1-\frac{1}{z}} + \frac{1}{2-\frac{2}{z}} \Big) $$ Let \(w=\frac{1}{z}\) for both terms to use the geometric series expansion: $$ f(z) = \frac{1}{z}\Big(\sum_{n=0}^{\infty} \frac{1}{z^n} - \sum_{n=0}^{\infty} 2^n\cdot\frac{1}{z^n}\Big) $$

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