Expansion Around a Singularity Examples

Laurent Expansion

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction

1. If a function is analytic everywhere on and in a circle, except forthe center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center

, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0

, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either

, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.

3. This presentation shows how an isolated singularity (a placewhere f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity

(a placewhere f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it)

can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion

, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Introduction1. If a function is analytic everywhere on and in a circle, except for

the center, then we cannot apply the Cauchy-Goursat Theoremor Cauchy’s Integral Formula to integrals of the function aroundthe circle.

2. But, looking back at Cauchy’s Integral Formula, the integrand in

the formula wasf (z)

z− z0, which is not analytic at z0 either, and we

still were able to get a useful result.3. This presentation shows how an isolated singularity (a place

where f is not analytic, but so that f is analytic near it) can behandled.

4. We will obtain a power series type expansion, but it will involvenegative as well as positive integer exponents.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem.

Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion.

Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R.

Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented.

Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z)

=1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ

− 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Theorem. Laurent expansion. Let 0 < r < R, let z0 ∈ C and let thefunction f be analytic in a region that contains the annulus A given byr ≤ |z− z0| ≤ R. Let CR be the circle of radius R around z0 and let Cr

be the circle of radius r around z0, both positively oriented. Then forany z ∈ A we have

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

=:∞

∑n=−∞

an(z− z0)n

where an =1

2πi

∮C

f (ξ )(ξ − z0)n+1 dξ for all integers n, bn = a−n for all

positive integers n and C is any circle around z0 whose radius is in[r,R].

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

q

z0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0

w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

q

z ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz

��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz �

�

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ

− 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

qz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

qz ��

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

12πi

∫CR

f (ξ )ξ − z

dξ

=1

2πi

∫CR

f (ξ )ξ − z0− (z− z0)

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

11− z−z0

ξ−z0

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

[N

∑n=0

(z− z0

ξ − z0

)n

+∞

∑n=N+1

(z− z0

ξ − z0

)n]

dξ

=N

∑n=0

(z− z0)n 1

2πi

∫CR

f (ξ )(ξ − z0)n+1 dξ

+1

2πi

∫CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)1

2πi

∫CR

f (ξ )ξ − z

dξ

=1

2πi

∫CR

f (ξ )ξ − z0− (z− z0)

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

11− z−z0

ξ−z0

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

[N

∑n=0

(z− z0

ξ − z0

)n

+∞

∑n=N+1

(z− z0

ξ − z0

)n]

dξ

=N

∑n=0

(z− z0)n 1

2πi

∫CR

f (ξ )(ξ − z0)n+1 dξ

+1

2πi

∫CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)1

2πi

∫CR

f (ξ )ξ − z

dξ

=1

2πi

∫CR

f (ξ )ξ − z0− (z− z0)

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

11− z−z0

ξ−z0

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

[N

∑n=0

(z− z0

ξ − z0

)n

+∞

∑n=N+1

(z− z0

ξ − z0

)n]

dξ

=N

∑n=0

(z− z0)n 1

2πi

∫CR

f (ξ )(ξ − z0)n+1 dξ

+1

2πi

∫CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)1

2πi

∫CR

f (ξ )ξ − z

dξ

=1

2πi

∫CR

f (ξ )ξ − z0− (z− z0)

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

11− z−z0

ξ−z0

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

[N

∑n=0

(z− z0

ξ − z0

)n

+∞

∑n=N+1

(z− z0

ξ − z0

)n]

dξ

=N

∑n=0

(z− z0)n 1

2πi

∫CR

f (ξ )(ξ − z0)n+1 dξ

+1

2πi

∫CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)1

2πi

∫CR

f (ξ )ξ − z

dξ

=1

2πi

∫CR

f (ξ )ξ − z0− (z− z0)

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

11− z−z0

ξ−z0

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

[N

∑n=0

(z− z0

ξ − z0

)n

+∞

∑n=N+1

(z− z0

ξ − z0

)n]

dξ

=N

∑n=0

(z− z0)n 1

2πi

∫CR

f (ξ )(ξ − z0)n+1 dξ

+1

2πi

∫CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)1

2πi

∫CR

f (ξ )ξ − z

dξ

=1

2πi

∫CR

f (ξ )ξ − z0− (z− z0)

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

11− z−z0

ξ−z0

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

[N

∑n=0

(z− z0

ξ − z0

)n

+∞

∑n=N+1

(z− z0

ξ − z0

)n]

dξ

=N

∑n=0

(z− z0)n 1

2πi

∫CR

f (ξ )(ξ − z0)n+1 dξ

+1

2πi

∫CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)1

2πi

∫CR

f (ξ )ξ − z

dξ

=1

2πi

∫CR

f (ξ )ξ − z0− (z− z0)

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

11− z−z0

ξ−z0

dξ

=1

2πi

∫CR

f (ξ )ξ − z0

[N

∑n=0

(z− z0

ξ − z0

)n

+∞

∑n=N+1

(z− z0

ξ − z0

)n]

dξ

=N

∑n=0

(z− z0)n 1

2πi

∫CR

f (ξ )(ξ − z0)n+1 dξ

+1

2πi

∫CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

Because z is fixed and |z− z0|< R = |ξ − z0| for all ξ on CR,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣≤

∫CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|< R

= |ξ − z0| for all ξ on CR,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣≤

∫CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|< R = |ξ − z0| for all ξ on CR

,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣≤

∫CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|< R = |ξ − z0| for all ξ on CR,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣≤

∫CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|< R = |ξ − z0| for all ξ on CR,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣

≤∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|< R = |ξ − z0| for all ξ on CR,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣≤

∫CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|< R = |ξ − z0| for all ξ on CR,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣≤

∫CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

→ 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|< R = |ξ − z0| for all ξ on CR,

we have that there is a q < 1 so that∣∣∣∣ z− z0

ξ − z0

∣∣∣∣< q for all ξ on CR.

Therefore∣∣∣∣∣∫

CR

f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

dξ

∣∣∣∣∣≤

∫CR

∣∣∣∣∣ f (ξ )ξ − z0

(z− z0

ξ − z0

)N+1 11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

CR

∣∣∣∣∣ f (ξ )ξ − z0

11− z−z0

ξ−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

− 12πi

∫Cr

f (ξ )ξ − z

dξ

=1

2πi

∫Cr

f (ξ )z− z0− (ξ − z0)

dξ

=1

2πi

∫Cr

f (ξ )z− z0

1

1− ξ−z0z−z0

dξ

=1

2πi

∫Cr

f (ξ )z− z0

[N

∑n=0

(ξ − z0

z− z0

)n

+∞

∑n=N+1

(ξ − z0

z− z0

)n]

dξ

=N+1

∑n=1

(z− z0)−n 1

2πi

∫Cr

f (ξ )(ξ − z0)−n+1 dξ

+1

2πi

∫Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

− 12πi

∫Cr

f (ξ )ξ − z

dξ

=1

2πi

∫Cr

f (ξ )z− z0− (ξ − z0)

dξ

=1

2πi

∫Cr

f (ξ )z− z0

1

1− ξ−z0z−z0

dξ

=1

2πi

∫Cr

f (ξ )z− z0

[N

∑n=0

(ξ − z0

z− z0

)n

+∞

∑n=N+1

(ξ − z0

z− z0

)n]

dξ

=N+1

∑n=1

(z− z0)−n 1

2πi

∫Cr

f (ξ )(ξ − z0)−n+1 dξ

+1

2πi

∫Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

− 12πi

∫Cr

f (ξ )ξ − z

dξ

=1

2πi

∫Cr

f (ξ )z− z0− (ξ − z0)

dξ

=1

2πi

∫Cr

f (ξ )z− z0

1

1− ξ−z0z−z0

dξ

=1

2πi

∫Cr

f (ξ )z− z0

[N

∑n=0

(ξ − z0

z− z0

)n

+∞

∑n=N+1

(ξ − z0

z− z0

)n]

dξ

=N+1

∑n=1

(z− z0)−n 1

2πi

∫Cr

f (ξ )(ξ − z0)−n+1 dξ

+1

2πi

∫Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

− 12πi

∫Cr

f (ξ )ξ − z

dξ

=1

2πi

∫Cr

f (ξ )z− z0− (ξ − z0)

dξ

=1

2πi

∫Cr

f (ξ )z− z0

1

1− ξ−z0z−z0

dξ

=1

2πi

∫Cr

f (ξ )z− z0

[N

∑n=0

(ξ − z0

z− z0

)n

+∞

∑n=N+1

(ξ − z0

z− z0

)n]

dξ

=N+1

∑n=1

(z− z0)−n 1

2πi

∫Cr

f (ξ )(ξ − z0)−n+1 dξ

+1

2πi

∫Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

− 12πi

∫Cr

f (ξ )ξ − z

dξ

=1

2πi

∫Cr

f (ξ )z− z0− (ξ − z0)

dξ

=1

2πi

∫Cr

f (ξ )z− z0

1

1− ξ−z0z−z0

dξ

=1

2πi

∫Cr

f (ξ )z− z0

[N

∑n=0

(ξ − z0

z− z0

)n

+∞

∑n=N+1

(ξ − z0

z− z0

)n]

dξ

=N+1

∑n=1

(z− z0)−n 1

2πi

∫Cr

f (ξ )(ξ − z0)−n+1 dξ

+1

2πi

∫Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

− 12πi

∫Cr

f (ξ )ξ − z

dξ

=1

2πi

∫Cr

f (ξ )z− z0− (ξ − z0)

dξ

=1

2πi

∫Cr

f (ξ )z− z0

1

1− ξ−z0z−z0

dξ

=1

2πi

∫Cr

f (ξ )z− z0

[N

∑n=0

(ξ − z0

z− z0

)n

+∞

∑n=N+1

(ξ − z0

z− z0

)n]

dξ

=N+1

∑n=1

(z− z0)−n 1

2πi

∫Cr

f (ξ )(ξ − z0)−n+1 dξ

+1

2πi

∫Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof (cont.)

− 12πi

∫Cr

f (ξ )ξ − z

dξ

=1

2πi

∫Cr

f (ξ )z− z0− (ξ − z0)

dξ

=1

2πi

∫Cr

f (ξ )z− z0

1

1− ξ−z0z−z0

dξ

=1

2πi

∫Cr

f (ξ )z− z0

[N

∑n=0

(ξ − z0

z− z0

)n

+∞

∑n=N+1

(ξ − z0

z− z0

)n]

dξ

=N+1

∑n=1

(z− z0)−n 1

2πi

∫Cr

f (ξ )(ξ − z0)−n+1 dξ

+1

2πi

∫Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

Because z is fixed and |z− z0|> r = |ξ − z0| for all ξ on Cr,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣≤

∫Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|> r

= |ξ − z0| for all ξ on Cr,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣≤

∫Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|> r = |ξ − z0| for all ξ on Cr

,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣≤

∫Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|> r = |ξ − z0| for all ξ on Cr,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣≤

∫Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|> r = |ξ − z0| for all ξ on Cr,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣

≤∫

Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|> r = |ξ − z0| for all ξ on Cr,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣≤

∫Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|> r = |ξ − z0| for all ξ on Cr,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣≤

∫Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

→ 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Because z is fixed and |z− z0|> r = |ξ − z0| for all ξ on Cr,

we have that there is a q < 1 so that∣∣∣∣ξ − z0

z− z0

∣∣∣∣< q for all ξ on Cr.

Therefore∣∣∣∣∣∫

Cr

f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

dξ

∣∣∣∣∣≤

∫Cr

∣∣∣∣∣ f (ξ )z− z0

(ξ − z0

z− z0

)N+1 1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ |

≤ qN+1∫

Cr

∣∣∣∣∣ f (ξ )z− z0

1

1− ξ−z0z−z0

∣∣∣∣∣ d|ξ | → 0 (N→ ∞)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z)

=1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ

− 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ

+∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed

(for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain)

, except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. Therefore

f (z) =1

2πi

∮CR

f (ξ )ξ − z

dξ − 12πi

∮Cr

f (ξ )ξ − z

dξ

=∞

∑n=0

(z−z0)n 1

2πi

∫CR

f (ξ )(ξ−z0)n+1 dξ+

∞

∑n=1

(z−z0)−n 1

2πi

∫Cr

f (ξ )(ξ−z0)−n+1 dξ

=∞

∑n=0

an(z− z0)n +

∞

∑n=1

bn(z− z0)−n

and the coefficients an and bn are as claimed (for an explicit formulafor the bn, chasing the negative signs is a bit of a pain), except that theclaim is that the integral can be taken over any circle around z0 withradius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof.

For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R

, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

p

z0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0

w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

�

�

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Proof. For the replacement of the circles Cr and CR with any circle of

radius between r and R, note that the integral off (ξ )

(ξ − z0)k over Cr is

the same as over CR ...

pz0w 9

o�1

Cr (backwards)

�

�

U

-

�

K

CR

��

... and similarly, the integral is the same over any circle around z0with radius between r and R.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example.

Find the Laurent expansion of f (z) = sin(

1z

)around

z = 0.

sin(z) =∞

∑n=0

(−1)n

(2n+1)!z2n+1

sin(

1z

)=

∞

∑n=0

(−1)n

(2n+1)!

(1z

)2n+1

=∞

∑n=0

(−1)n

(2n+1)!z−(2n+1)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) = sin(

1z

)around

z = 0.

sin(z) =∞

∑n=0

(−1)n

(2n+1)!z2n+1

sin(

1z

)=

∞

∑n=0

(−1)n

(2n+1)!

(1z

)2n+1

=∞

∑n=0

(−1)n

(2n+1)!z−(2n+1)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) = sin(

1z

)around

z = 0.

sin(z)

=∞

∑n=0

(−1)n

(2n+1)!z2n+1

sin(

1z

)=

∞

∑n=0

(−1)n

(2n+1)!

(1z

)2n+1

=∞

∑n=0

(−1)n

(2n+1)!z−(2n+1)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) = sin(

1z

)around

z = 0.

sin(z) =∞

∑n=0

(−1)n

(2n+1)!z2n+1

sin(

1z

)=

∞

∑n=0

(−1)n

(2n+1)!

(1z

)2n+1

=∞

∑n=0

(−1)n

(2n+1)!z−(2n+1)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) = sin(

1z

)around

z = 0.

sin(z) =∞

∑n=0

(−1)n

(2n+1)!z2n+1

sin(

1z

)

=∞

∑n=0

(−1)n

(2n+1)!

(1z

)2n+1

=∞

∑n=0

(−1)n

(2n+1)!z−(2n+1)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) = sin(

1z

)around

z = 0.

sin(z) =∞

∑n=0

(−1)n

(2n+1)!z2n+1

sin(

1z

)=

∞

∑n=0

(−1)n

(2n+1)!

(1z

)2n+1

=∞

∑n=0

(−1)n

(2n+1)!z−(2n+1)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) = sin(

1z

)around

z = 0.

sin(z) =∞

∑n=0

(−1)n

(2n+1)!z2n+1

sin(

1z

)=

∞

∑n=0

(−1)n

(2n+1)!

(1z

)2n+1

=∞

∑n=0

(−1)n

(2n+1)!z−(2n+1)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example.

Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z)

=z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)

= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn

=−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 0 < |z|< 1 we have

f (z) =z+1z−1

= (z+1)(− 1

1− z

)= (z+1)

(−

∞

∑n=0

zn

)

= −∞

∑n=0

zn+1−∞

∑n=0

zn =−∞

∑m=1

zm−∞

∑n=0

zn

= −1−∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z)

=z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion

Expansion Around a Singularity Examples

Example. Find the Laurent expansion of f (z) =z+1z−1

for 0 < |z|< 1

and for 1 < |z|< ∞.

For 1 < |z|< ∞ note that 0 <

∣∣∣∣1z∣∣∣∣< 1.

f (z) =z+1z−1

·1z1z

=1+ 1

z

1− 1z

=

(1+

1z

)1

1− 1z

=

(1+

1z

)∞

∑n=0

(1z

)n

=∞

∑n=0

1zn +

∞

∑n=0

1zn+1

= 1+∞

∑n=1

2zn

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Laurent Expansion