Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated...
Transcript of Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated...
![Page 1: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/1.jpg)
Expansion Around a Singularity Examples
Laurent Expansion
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Laurent Expansion
![Page 2: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/2.jpg)
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
![Page 3: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/3.jpg)
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
![Page 4: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/4.jpg)
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
![Page 5: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/5.jpg)
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
![Page 6: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/6.jpg)
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
![Page 7: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/7.jpg)
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
![Page 8: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/8.jpg)
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
![Page 9: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/9.jpg)
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
![Page 10: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/10.jpg)
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
![Page 11: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/11.jpg)
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
![Page 12: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/12.jpg)
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
![Page 13: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/13.jpg)
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
![Page 14: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/14.jpg)
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
![Page 15: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/15.jpg)
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
![Page 16: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/16.jpg)
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
![Page 17: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/17.jpg)
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
![Page 18: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/18.jpg)
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
![Page 19: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/19.jpg)
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
![Page 20: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/20.jpg)
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
![Page 21: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/21.jpg)
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
![Page 22: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/22.jpg)
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
![Page 23: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/23.jpg)
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
![Page 24: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/24.jpg)
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
![Page 25: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/25.jpg)
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
![Page 26: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/26.jpg)
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
![Page 27: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/27.jpg)
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
![Page 28: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/28.jpg)
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
![Page 29: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/29.jpg)
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
![Page 30: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/30.jpg)
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
![Page 31: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/31.jpg)
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
![Page 32: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/32.jpg)
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
![Page 33: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/33.jpg)
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
![Page 34: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/34.jpg)
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
![Page 35: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/35.jpg)
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
![Page 36: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/36.jpg)
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
![Page 37: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/37.jpg)
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
![Page 38: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/38.jpg)
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
![Page 39: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/39.jpg)
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
![Page 40: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/40.jpg)
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
![Page 41: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/41.jpg)
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
![Page 42: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/42.jpg)
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
![Page 43: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/43.jpg)
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
![Page 44: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/44.jpg)
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
![Page 45: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/45.jpg)
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
![Page 46: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/46.jpg)
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
![Page 47: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/47.jpg)
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
![Page 48: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/48.jpg)
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
![Page 49: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/49.jpg)
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
![Page 50: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/50.jpg)
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
![Page 51: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/51.jpg)
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
![Page 52: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/52.jpg)
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
![Page 53: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/53.jpg)
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
![Page 54: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/54.jpg)
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
![Page 55: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/55.jpg)
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
![Page 56: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/56.jpg)
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
![Page 57: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/57.jpg)
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
![Page 58: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/58.jpg)
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
![Page 59: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/59.jpg)
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
![Page 60: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/60.jpg)
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
![Page 61: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/61.jpg)
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
![Page 62: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/62.jpg)
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
![Page 63: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/63.jpg)
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
![Page 64: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/64.jpg)
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
![Page 65: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/65.jpg)
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
![Page 66: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/66.jpg)
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
![Page 67: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/67.jpg)
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
![Page 68: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/68.jpg)
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
![Page 69: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/69.jpg)
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
![Page 70: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/70.jpg)
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
![Page 71: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/71.jpg)
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
![Page 72: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/72.jpg)
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
![Page 73: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/73.jpg)
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
![Page 74: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/74.jpg)
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
![Page 75: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/75.jpg)
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
![Page 76: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/76.jpg)
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
![Page 77: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/77.jpg)
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
![Page 78: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/78.jpg)
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
![Page 79: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/79.jpg)
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
![Page 80: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/80.jpg)
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
![Page 81: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/81.jpg)
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
![Page 82: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/82.jpg)
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
![Page 83: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/83.jpg)
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
![Page 84: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/84.jpg)
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
![Page 85: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/85.jpg)
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
![Page 86: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/86.jpg)
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
![Page 87: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/87.jpg)
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
![Page 88: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/88.jpg)
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
![Page 89: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/89.jpg)
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
![Page 90: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/90.jpg)
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
![Page 91: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/91.jpg)
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
![Page 92: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/92.jpg)
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
![Page 93: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/93.jpg)
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
![Page 94: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/94.jpg)
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
![Page 95: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/95.jpg)
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
![Page 96: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/96.jpg)
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
![Page 97: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/97.jpg)
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
![Page 98: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/98.jpg)
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
![Page 99: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/99.jpg)
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
![Page 100: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/100.jpg)
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
![Page 101: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/101.jpg)
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
![Page 102: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/102.jpg)
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
![Page 103: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/103.jpg)
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
![Page 104: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/104.jpg)
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
![Page 105: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/105.jpg)
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
![Page 106: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/106.jpg)
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
![Page 107: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/107.jpg)
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
![Page 108: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/108.jpg)
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
![Page 109: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/109.jpg)
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
![Page 110: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/110.jpg)
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
![Page 111: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/111.jpg)
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
![Page 112: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/112.jpg)
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
![Page 113: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/113.jpg)
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
![Page 114: Laurent Expansion - University of Southern Mississippi3.This presentation shows how an isolated singularity (a place where f is not analytic, but so that f is analytic near it) can](https://reader036.fdocuments.us/reader036/viewer/2022071214/6042ab92d4d3a467b92fd488/html5/thumbnails/114.jpg)
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