Chapter 9: Chapter 9

Problem 2

Expand \(f(z)=\frac{3 z-1}{z^{2}-2 z-3}\) in Laurent series valid for (i) \(1<|z|<3\) (ii) \(|z|>3\) (iii) \(|z|<1\).

Short Answer

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The Laurent series of \(f(z)\) are \(-\frac{3}{z} + \frac{7}{z^2}\) for \(|z|>3\), \(\frac{1}{z} + \frac{1}{z^2}\) for \(1<|z|<3\), and \(\sum_{n=0}^{\infty} (-z)^n + \sum_{n=0}^{\infty} (\frac{2}{3}z)^n\) for \(|z|<1\).

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Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Factorize the Denominator

First, factorize the denominator of the fraction. The quadratic \(z^{2}-2z-3\) can be factored into \((z-3)(z+1)\). So the function can be rewritten as \(f(z) = \frac{3z - 1}{(z-3)(z+1)}\).

Step 2: Partial Fraction Decomposition

Perform a partial fraction decomposition for the above expression, you get \(f(z) = \frac{1}{z+1} - \frac{2}{z-3}\). This is the function we will need to expand.

Step 3: Laurent Expansion for |z|>3

For \(|z|>3\), expand the function as a power series in \(1/z\). The first factor becomes \(\frac{1}{z(-1 + 1/z)}\) and the second becomes \(-\frac{2}{z(1 - 3/z)}\). Expanding both expressions you get \(\frac{-1}{z} + \frac{1}{z^2} - \frac{2}{z} + \frac{6}{z^2}\), so the Laurent series for \(|z|>3\) is \(-\frac{3}{z} + \frac{7}{z^2}\).

Step 4: Laurent Expansion for 1

For \(1<|z|<3\), express the function as a power series in \((z-3)\) and \(z\). The corresponding expression for the first factor would be \(-\frac1{z}(\frac{1}{1 + \frac{1}{z}})\) and for the second factor \(2\frac{1}{1 - \frac{3}{z}}\). Expanding both we get \(-\frac{1}{z} + \frac{1}{z^2} + \frac{2}{z}\), so the Laurent series for \(1<|z|<3\) is \( \frac{1}{z} + \frac{1}{z^2}\).

Step 5: Laurent Expansion for |z|

For \(|z|<1\), express the function as a power series in \(z\). Express the first factor as \(\frac{1}{1 + z}\) and the second one as \(-\frac{2}{3-z}\). The power series expansions will be \(\sum_{n=0}^{\infty} (-z)^n\) and \(\sum_{n=0}^{\infty} (\frac{2}{3}z)^n\). So the Laurent series for \(|z|<1\) are these two sums.

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Chapter 9: Chapter 9

Expand \(f(z)=\frac{3 z-1}{z^{2}-2 z-3}\) in Laurent series valid for (i) \(1<|z|<3\) (ii) \(|z|>3\) (iii) \(|z|<1\).

Short Answer

Step by step solution

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Step 1: Factorize the Denominator

Step 2: Partial Fraction Decomposition

Step 3: Laurent Expansion for |z|>3

Step 4: Laurent Expansion for 1

Step 5: Laurent Expansion for |z|

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More chapters from the book ‘Complex Variables With Applications’

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 10

Chapter 11

Chapter 12

Chapter 13

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