Integral by Residue Theorem
I'm working through a Complex Analysis text and am working through the Residue chapter. I am not sure if I am approaching this question correctly.
$$ int_gamma frac{1}{(z-1)^2(z^2+1)}$$
Such that $gamma$ is a circle of radius of 2 centered at 0. Traversed counter clockwise.
My approach was as follows:
I know that there are poles at $z=1,i,-i$ where $1$ is of order $2$.
First I found the residues at the poles as follows:
$$operatorname{Res} _{z=1} :lim_{zrightarrow 1} frac{d}{dz} frac{(z-1)^2}{(z-1)^2(z^2+1)}$$
$$lim_{zrightarrow 1} frac{d}{dz} frac{1}{(z^2+1)}$$
$$lim_{zrightarrow 1} -frac{2z}{(z^2+1)^2}$$
$$ = -1/2$$
$$operatorname{Res} _{z=i} :lim_{zrightarrow i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(i-1)^2}$$
$$operatorname{Res} _{z=-i} :lim_{zrightarrow -i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(-i-1)^2}$$
Therefore the integral is $2pi i$*sum of the Residues.
$$= 2pi i(frac{1}{(i-1)^2} - frac{1}{2} + frac{1}{(-i-1)^2})$$
Thank you for any guidance. The repeated root is really where I'm wondering if it makes a difference.
complex-analysis complex-numbers residue-calculus cauchy-integral-formula
asked Nov 29 '18 at 0:43
SafderSafder
15110
|
show 1 more comment
I'm working through a Complex Analysis text and am working through the Residue chapter. I am not sure if I am approaching this question correctly.
$$ int_gamma frac{1}{(z-1)^2(z^2+1)}$$
Such that $gamma$ is a circle of radius of 2 centered at 0. Traversed counter clockwise.
My approach was as follows:
I know that there are poles at $z=1,i,-i$ where $1$ is of order $2$.
First I found the residues at the poles as follows:
$$operatorname{Res} _{z=1} :lim_{zrightarrow 1} frac{d}{dz} frac{(z-1)^2}{(z-1)^2(z^2+1)}$$
$$lim_{zrightarrow 1} frac{d}{dz} frac{1}{(z^2+1)}$$
$$lim_{zrightarrow 1} -frac{2z}{(z^2+1)^2}$$
$$ = -1/2$$
$$operatorname{Res} _{z=i} :lim_{zrightarrow i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(i-1)^2}$$
$$operatorname{Res} _{z=-i} :lim_{zrightarrow -i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(-i-1)^2}$$
Therefore the integral is $2pi i$*sum of the Residues.
$$= 2pi i(frac{1}{(i-1)^2} - frac{1}{2} + frac{1}{(-i-1)^2})$$
Thank you for any guidance. The repeated root is really where I'm wondering if it makes a difference.
complex-analysis complex-numbers residue-calculus cauchy-integral-formula
asked Nov 29 '18 at 0:43
SafderSafder
15110
There is also a pole at $z=-i$
– Seth
Nov 29 '18 at 0:48
You forgot the pole $-i$.
– Bernard
Nov 29 '18 at 0:49
I'm pretty sure the only difference when you have a double pole is that if you are using the differentiation formula for residues you have to differentiate twice
– Seth
Nov 29 '18 at 0:56
I think you have accidentally asked this question twice
– Seth
Nov 29 '18 at 0:57
Woops! I think my internet must've done that. Okay I see where the pole comes from. I checked back with the formula. Wouldn't it be differentiate it once? I'll give it a try and edit the post with my answer.
– Safder
Nov 29 '18 at 1:21
|
show 1 more comment
I'm working through a Complex Analysis text and am working through the Residue chapter. I am not sure if I am approaching this question correctly.
$$ int_gamma frac{1}{(z-1)^2(z^2+1)}$$
Such that $gamma$ is a circle of radius of 2 centered at 0. Traversed counter clockwise.
My approach was as follows:
I know that there are poles at $z=1,i,-i$ where $1$ is of order $2$.
First I found the residues at the poles as follows:
$$operatorname{Res} _{z=1} :lim_{zrightarrow 1} frac{d}{dz} frac{(z-1)^2}{(z-1)^2(z^2+1)}$$
$$lim_{zrightarrow 1} frac{d}{dz} frac{1}{(z^2+1)}$$
$$lim_{zrightarrow 1} -frac{2z}{(z^2+1)^2}$$
$$ = -1/2$$
$$operatorname{Res} _{z=i} :lim_{zrightarrow i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(i-1)^2}$$
$$operatorname{Res} _{z=-i} :lim_{zrightarrow -i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(-i-1)^2}$$
Therefore the integral is $2pi i$*sum of the Residues.
$$= 2pi i(frac{1}{(i-1)^2} - frac{1}{2} + frac{1}{(-i-1)^2})$$
Thank you for any guidance. The repeated root is really where I'm wondering if it makes a difference.
complex-analysis complex-numbers residue-calculus cauchy-integral-formula
asked Nov 29 '18 at 0:43
SafderSafder
15110
I'm working through a Complex Analysis text and am working through the Residue chapter. I am not sure if I am approaching this question correctly.
$$ int_gamma frac{1}{(z-1)^2(z^2+1)}$$
Such that $gamma$ is a circle of radius of 2 centered at 0. Traversed counter clockwise.
My approach was as follows:
I know that there are poles at $z=1,i,-i$ where $1$ is of order $2$.
First I found the residues at the poles as follows:
$$operatorname{Res} _{z=1} :lim_{zrightarrow 1} frac{d}{dz} frac{(z-1)^2}{(z-1)^2(z^2+1)}$$
$$lim_{zrightarrow 1} frac{d}{dz} frac{1}{(z^2+1)}$$
$$lim_{zrightarrow 1} -frac{2z}{(z^2+1)^2}$$
$$ = -1/2$$
$$operatorname{Res} _{z=i} :lim_{zrightarrow i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(i-1)^2}$$
$$operatorname{Res} _{z=-i} :lim_{zrightarrow -i} frac{1}{(z-1)^2}$$
$$ = frac{1}{(-i-1)^2}$$
Therefore the integral is $2pi i$*sum of the Residues.
$$= 2pi i(frac{1}{(i-1)^2} - frac{1}{2} + frac{1}{(-i-1)^2})$$
Thank you for any guidance. The repeated root is really where I'm wondering if it makes a difference.
complex-analysis complex-numbers residue-calculus cauchy-integral-formula
complex-analysis complex-numbers residue-calculus cauchy-integral-formula
asked Nov 29 '18 at 0:43
SafderSafder
15110
asked Nov 29 '18 at 0:43
SafderSafder
15110
edited Nov 29 '18 at 1:27
Safder
asked Nov 29 '18 at 0:43
SafderSafder
15110
asked Nov 29 '18 at 0:43
SafderSafder
15110
asked Nov 29 '18 at 0:43
SafderSafder
15110
15110
There is also a pole at $z=-i$
– Seth
Nov 29 '18 at 0:48
You forgot the pole $-i$.
– Bernard
Nov 29 '18 at 0:49
I'm pretty sure the only difference when you have a double pole is that if you are using the differentiation formula for residues you have to differentiate twice
– Seth
Nov 29 '18 at 0:56
I think you have accidentally asked this question twice
– Seth
Nov 29 '18 at 0:57
Woops! I think my internet must've done that. Okay I see where the pole comes from. I checked back with the formula. Wouldn't it be differentiate it once? I'll give it a try and edit the post with my answer.
– Safder
Nov 29 '18 at 1:21
|
show 1 more comment
There is also a pole at $z=-i$
– Seth
Nov 29 '18 at 0:48
You forgot the pole $-i$.
– Bernard
Nov 29 '18 at 0:49
I'm pretty sure the only difference when you have a double pole is that if you are using the differentiation formula for residues you have to differentiate twice
– Seth
Nov 29 '18 at 0:56
I think you have accidentally asked this question twice
– Seth
Nov 29 '18 at 0:57
Woops! I think my internet must've done that. Okay I see where the pole comes from. I checked back with the formula. Wouldn't it be differentiate it once? I'll give it a try and edit the post with my answer.
– Safder
Nov 29 '18 at 1:21
There is also a pole at $z=-i$
– Seth
Nov 29 '18 at 0:48
There is also a pole at $z=-i$
– Seth
Nov 29 '18 at 0:48
You forgot the pole $-i$.
– Bernard
Nov 29 '18 at 0:49
You forgot the pole $-i$.
– Bernard
Nov 29 '18 at 0:49
I'm pretty sure the only difference when you have a double pole is that if you are using the differentiation formula for residues you have to differentiate twice
– Seth
Nov 29 '18 at 0:56
I'm pretty sure the only difference when you have a double pole is that if you are using the differentiation formula for residues you have to differentiate twice
– Seth
Nov 29 '18 at 0:56
I think you have accidentally asked this question twice
– Seth
Nov 29 '18 at 0:57
I think you have accidentally asked this question twice
– Seth
Nov 29 '18 at 0:57
Woops! I think my internet must've done that. Okay I see where the pole comes from. I checked back with the formula. Wouldn't it be differentiate it once? I'll give it a try and edit the post with my answer.
– Safder
Nov 29 '18 at 1:21
Woops! I think my internet must've done that. Okay I see where the pole comes from. I checked back with the formula. Wouldn't it be differentiate it once? I'll give it a try and edit the post with my answer.
– Safder
Nov 29 '18 at 1:21
|
show 1 more comment
1 Answer
1
active
oldest
votes
The residue at $1$ is computed correctly. The residues at $pm i$ are a bit off. E.g.
$$ operatorname{Res}_i(f) =lim_{z to i} (z - i)f(z) = lim_{z to i} frac{z - i}{(z - 1)^2(z^2 + 1)} = lim_{z to i}frac{1}{(z-1)^2(z + i)} $$
This should simplify to $frac14$. The residue at $-i$ is also $frac14$ so the sum of residues is
$$ underbrace{-frac12}_{z = 0} + underbrace{frac14}_{z = i} + underbrace{frac14}_{z = -i} = 0. $$
Alternatively, you can look at the question as
$$ int_gamma f = -int_{- gamma} f $$
where now we are integrating clockwise. Going clockwise around 0 is the same as going counterclockwise around $infty$. Since all the poles in $mathbb{C}$ are contained in ${|z| < 2}$, on ${|z| > 2}$ we only have the pole at infinity. See Residue at infinity (Wikipedia). So
$$ -int_{- gamma} f = -2pi ioperatorname{Res}_{infty} (f) = 2pi ioperatorname{Res}_{0} left( frac{1}{z^2} fleft( frac1z right) right). $$
and
$$ operatorname{Res}_{0} left[ frac{1}{z^2} frac{1}{left( frac1z - 1 right)^2left( frac1{z^2} + 1 right)} right] = operatorname{Res}_{0} left[ frac{z^2}{left( 1 - z right)^2left( 1 + z^2 right)} right] = 0. $$
As you can see, using the residue at infinity saves you some computation. It's not going to make every computation easier, but it is still worth being aware of.
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you
– Safder
Nov 29 '18 at 4:43
@Safder Yes, the residue of $f$ at $infty$ is the same as the residue of $-frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information.
– Trevor Gunn
Nov 29 '18 at 5:09
add a comment |
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The residue at $1$ is computed correctly. The residues at $pm i$ are a bit off. E.g.
$$ operatorname{Res}_i(f) =lim_{z to i} (z - i)f(z) = lim_{z to i} frac{z - i}{(z - 1)^2(z^2 + 1)} = lim_{z to i}frac{1}{(z-1)^2(z + i)} $$
This should simplify to $frac14$. The residue at $-i$ is also $frac14$ so the sum of residues is
$$ underbrace{-frac12}_{z = 0} + underbrace{frac14}_{z = i} + underbrace{frac14}_{z = -i} = 0. $$
Alternatively, you can look at the question as
$$ int_gamma f = -int_{- gamma} f $$
where now we are integrating clockwise. Going clockwise around 0 is the same as going counterclockwise around $infty$. Since all the poles in $mathbb{C}$ are contained in ${|z| < 2}$, on ${|z| > 2}$ we only have the pole at infinity. See Residue at infinity (Wikipedia). So
$$ -int_{- gamma} f = -2pi ioperatorname{Res}_{infty} (f) = 2pi ioperatorname{Res}_{0} left( frac{1}{z^2} fleft( frac1z right) right). $$
and
$$ operatorname{Res}_{0} left[ frac{1}{z^2} frac{1}{left( frac1z - 1 right)^2left( frac1{z^2} + 1 right)} right] = operatorname{Res}_{0} left[ frac{z^2}{left( 1 - z right)^2left( 1 + z^2 right)} right] = 0. $$
As you can see, using the residue at infinity saves you some computation. It's not going to make every computation easier, but it is still worth being aware of.
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you
– Safder
Nov 29 '18 at 4:43
@Safder Yes, the residue of $f$ at $infty$ is the same as the residue of $-frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information.
– Trevor Gunn
Nov 29 '18 at 5:09
add a comment |
The residue at $1$ is computed correctly. The residues at $pm i$ are a bit off. E.g.
$$ operatorname{Res}_i(f) =lim_{z to i} (z - i)f(z) = lim_{z to i} frac{z - i}{(z - 1)^2(z^2 + 1)} = lim_{z to i}frac{1}{(z-1)^2(z + i)} $$
This should simplify to $frac14$. The residue at $-i$ is also $frac14$ so the sum of residues is
$$ underbrace{-frac12}_{z = 0} + underbrace{frac14}_{z = i} + underbrace{frac14}_{z = -i} = 0. $$
Alternatively, you can look at the question as
$$ int_gamma f = -int_{- gamma} f $$
where now we are integrating clockwise. Going clockwise around 0 is the same as going counterclockwise around $infty$. Since all the poles in $mathbb{C}$ are contained in ${|z| < 2}$, on ${|z| > 2}$ we only have the pole at infinity. See Residue at infinity (Wikipedia). So
$$ -int_{- gamma} f = -2pi ioperatorname{Res}_{infty} (f) = 2pi ioperatorname{Res}_{0} left( frac{1}{z^2} fleft( frac1z right) right). $$
and
$$ operatorname{Res}_{0} left[ frac{1}{z^2} frac{1}{left( frac1z - 1 right)^2left( frac1{z^2} + 1 right)} right] = operatorname{Res}_{0} left[ frac{z^2}{left( 1 - z right)^2left( 1 + z^2 right)} right] = 0. $$
As you can see, using the residue at infinity saves you some computation. It's not going to make every computation easier, but it is still worth being aware of.
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you
– Safder
Nov 29 '18 at 4:43
@Safder Yes, the residue of $f$ at $infty$ is the same as the residue of $-frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information.
– Trevor Gunn
Nov 29 '18 at 5:09
add a comment |
The residue at $1$ is computed correctly. The residues at $pm i$ are a bit off. E.g.
$$ operatorname{Res}_i(f) =lim_{z to i} (z - i)f(z) = lim_{z to i} frac{z - i}{(z - 1)^2(z^2 + 1)} = lim_{z to i}frac{1}{(z-1)^2(z + i)} $$
This should simplify to $frac14$. The residue at $-i$ is also $frac14$ so the sum of residues is
$$ underbrace{-frac12}_{z = 0} + underbrace{frac14}_{z = i} + underbrace{frac14}_{z = -i} = 0. $$
Alternatively, you can look at the question as
$$ int_gamma f = -int_{- gamma} f $$
where now we are integrating clockwise. Going clockwise around 0 is the same as going counterclockwise around $infty$. Since all the poles in $mathbb{C}$ are contained in ${|z| < 2}$, on ${|z| > 2}$ we only have the pole at infinity. See Residue at infinity (Wikipedia). So
$$ -int_{- gamma} f = -2pi ioperatorname{Res}_{infty} (f) = 2pi ioperatorname{Res}_{0} left( frac{1}{z^2} fleft( frac1z right) right). $$
and
$$ operatorname{Res}_{0} left[ frac{1}{z^2} frac{1}{left( frac1z - 1 right)^2left( frac1{z^2} + 1 right)} right] = operatorname{Res}_{0} left[ frac{z^2}{left( 1 - z right)^2left( 1 + z^2 right)} right] = 0. $$
As you can see, using the residue at infinity saves you some computation. It's not going to make every computation easier, but it is still worth being aware of.
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
The residue at $1$ is computed correctly. The residues at $pm i$ are a bit off. E.g.
$$ operatorname{Res}_i(f) =lim_{z to i} (z - i)f(z) = lim_{z to i} frac{z - i}{(z - 1)^2(z^2 + 1)} = lim_{z to i}frac{1}{(z-1)^2(z + i)} $$
This should simplify to $frac14$. The residue at $-i$ is also $frac14$ so the sum of residues is
$$ underbrace{-frac12}_{z = 0} + underbrace{frac14}_{z = i} + underbrace{frac14}_{z = -i} = 0. $$
Alternatively, you can look at the question as
$$ int_gamma f = -int_{- gamma} f $$
where now we are integrating clockwise. Going clockwise around 0 is the same as going counterclockwise around $infty$. Since all the poles in $mathbb{C}$ are contained in ${|z| < 2}$, on ${|z| > 2}$ we only have the pole at infinity. See Residue at infinity (Wikipedia). So
$$ -int_{- gamma} f = -2pi ioperatorname{Res}_{infty} (f) = 2pi ioperatorname{Res}_{0} left( frac{1}{z^2} fleft( frac1z right) right). $$
and
$$ operatorname{Res}_{0} left[ frac{1}{z^2} frac{1}{left( frac1z - 1 right)^2left( frac1{z^2} + 1 right)} right] = operatorname{Res}_{0} left[ frac{z^2}{left( 1 - z right)^2left( 1 + z^2 right)} right] = 0. $$
As you can see, using the residue at infinity saves you some computation. It's not going to make every computation easier, but it is still worth being aware of.
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
edited Nov 29 '18 at 5:01
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
answered Nov 29 '18 at 3:03
Trevor GunnTrevor Gunn
14.2k32046
14.2k32046
What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you
– Safder
Nov 29 '18 at 4:43
@Safder Yes, the residue of $f$ at $infty$ is the same as the residue of $-frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information.
– Trevor Gunn
Nov 29 '18 at 5:09
add a comment |
What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you
– Safder
Nov 29 '18 at 4:43
@Safder Yes, the residue of $f$ at $infty$ is the same as the residue of $-frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information.
– Trevor Gunn
Nov 29 '18 at 5:09
What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you
– Safder
Nov 29 '18 at 4:43
What is $Res_0$? Is it just the residue? Also, what are the conditions for when I can use residue at infinity. Thank you
– Safder
Nov 29 '18 at 4:43
@Safder Yes, the residue of $f$ at $infty$ is the same as the residue of $-frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information.
– Trevor Gunn
Nov 29 '18 at 5:09
@Safder Yes, the residue of $f$ at $infty$ is the same as the residue of $-frac{1}{z^2}f(1/z)$ at $0$. You can use this whenever you have a function that is holomorphic for large values of $z$. I.e. for $|z| > R$ for some $R$. See the Wikipedia entry. Also see math.stackexchange.com/q/571510 or math.usm.edu/schroeder/slides/comp_var/18_residue_theorem.pdf or whatever other source you can dig up if you want more information.
– Trevor Gunn
Nov 29 '18 at 5:09
add a comment |
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There is also a pole at $z=-i$
– Seth
Nov 29 '18 at 0:48
You forgot the pole $-i$.
– Bernard
Nov 29 '18 at 0:49
I'm pretty sure the only difference when you have a double pole is that if you are using the differentiation formula for residues you have to differentiate twice
– Seth
Nov 29 '18 at 0:56
I think you have accidentally asked this question twice
– Seth
Nov 29 '18 at 0:57
Woops! I think my internet must've done that. Okay I see where the pole comes from. I checked back with the formula. Wouldn't it be differentiate it once? I'll give it a try and edit the post with my answer.
– Safder
Nov 29 '18 at 1:21