Does the specific contour matter when integrating over a closed loop
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I've heard different people say yes and no, so I want to ask it here on math stack, but often, as in with cauchys integral theorem on wiki for instance, the contour is specified as a circle in the complex plane. Could it however be any shape (that's a closed contour) in the complex plane, and if so why?
To me, it doesn't seem to matter, and here's my reasoning: define the arbitrary contour as a parametric function of time, and since it's loop (not sure if there's a more formal definition, but here just take it to mean a closed contour) after some given time (let's call it T) it winds up back were it started at $t=0$. The derivative of the curve with respect to time is another complex valued function of time, and since after T the tangent is the same, the derivative is the same at T, meaning the derivative of the curve with respect to time also forms a loop (given of course that the contour is differentiable). So, the opposite also applies. Integrating a loop yields another loop, and integrating the loop over the time it takes to make one revolution (ie a definite integral from 0 to T) is 0 since you haven't moved anywhere since it's a loop. Then, if $dz=z'dt$ integrating a complex function over a complex variable is the same as a complex function (since the $z'$ that comes out is multiplied and complex numbers are closed under multiplication) integratied over time (a real variable), which is zero unless they cancel. Assuming they don't, symbolically (and I is the antiderivative of Z):
$oint_z f(z)dz=int^T_0 f(z)z'dt=int^T_0 Zdt=I(T)-I(0)=0$
$frac{1}{z}$ is an example of what I mean by cancel, if you were to integrate it like this. Making a change of coordinates to polar reveals that regardless of $r(t)$ (the loop) it evaluates to $2pi i$.
Again, even if this yields the right results I'm not sure if the intuition's correct. Do people just just use circles as the contour because they're convenient? Thanks in advance, anything that may be helpful would be great!
complex-analysis complex-integration
This question has an open bounty worth +50
reputation from Benjamin Thoburn ending in 7 days.
The current answers do not contain enough detail.
I know my result is wrong but I don't know where I went wrong. Unsure why this method is invalid.
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I've heard different people say yes and no, so I want to ask it here on math stack, but often, as in with cauchys integral theorem on wiki for instance, the contour is specified as a circle in the complex plane. Could it however be any shape (that's a closed contour) in the complex plane, and if so why?
To me, it doesn't seem to matter, and here's my reasoning: define the arbitrary contour as a parametric function of time, and since it's loop (not sure if there's a more formal definition, but here just take it to mean a closed contour) after some given time (let's call it T) it winds up back were it started at $t=0$. The derivative of the curve with respect to time is another complex valued function of time, and since after T the tangent is the same, the derivative is the same at T, meaning the derivative of the curve with respect to time also forms a loop (given of course that the contour is differentiable). So, the opposite also applies. Integrating a loop yields another loop, and integrating the loop over the time it takes to make one revolution (ie a definite integral from 0 to T) is 0 since you haven't moved anywhere since it's a loop. Then, if $dz=z'dt$ integrating a complex function over a complex variable is the same as a complex function (since the $z'$ that comes out is multiplied and complex numbers are closed under multiplication) integratied over time (a real variable), which is zero unless they cancel. Assuming they don't, symbolically (and I is the antiderivative of Z):
$oint_z f(z)dz=int^T_0 f(z)z'dt=int^T_0 Zdt=I(T)-I(0)=0$
$frac{1}{z}$ is an example of what I mean by cancel, if you were to integrate it like this. Making a change of coordinates to polar reveals that regardless of $r(t)$ (the loop) it evaluates to $2pi i$.
Again, even if this yields the right results I'm not sure if the intuition's correct. Do people just just use circles as the contour because they're convenient? Thanks in advance, anything that may be helpful would be great!
complex-analysis complex-integration
This question has an open bounty worth +50
reputation from Benjamin Thoburn ending in 7 days.
The current answers do not contain enough detail.
I know my result is wrong but I don't know where I went wrong. Unsure why this method is invalid.
Why does putting a loop "in the loop" make the integral invalid?
– Benjamin Thoburn
2 hours ago
add a comment |
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I've heard different people say yes and no, so I want to ask it here on math stack, but often, as in with cauchys integral theorem on wiki for instance, the contour is specified as a circle in the complex plane. Could it however be any shape (that's a closed contour) in the complex plane, and if so why?
To me, it doesn't seem to matter, and here's my reasoning: define the arbitrary contour as a parametric function of time, and since it's loop (not sure if there's a more formal definition, but here just take it to mean a closed contour) after some given time (let's call it T) it winds up back were it started at $t=0$. The derivative of the curve with respect to time is another complex valued function of time, and since after T the tangent is the same, the derivative is the same at T, meaning the derivative of the curve with respect to time also forms a loop (given of course that the contour is differentiable). So, the opposite also applies. Integrating a loop yields another loop, and integrating the loop over the time it takes to make one revolution (ie a definite integral from 0 to T) is 0 since you haven't moved anywhere since it's a loop. Then, if $dz=z'dt$ integrating a complex function over a complex variable is the same as a complex function (since the $z'$ that comes out is multiplied and complex numbers are closed under multiplication) integratied over time (a real variable), which is zero unless they cancel. Assuming they don't, symbolically (and I is the antiderivative of Z):
$oint_z f(z)dz=int^T_0 f(z)z'dt=int^T_0 Zdt=I(T)-I(0)=0$
$frac{1}{z}$ is an example of what I mean by cancel, if you were to integrate it like this. Making a change of coordinates to polar reveals that regardless of $r(t)$ (the loop) it evaluates to $2pi i$.
Again, even if this yields the right results I'm not sure if the intuition's correct. Do people just just use circles as the contour because they're convenient? Thanks in advance, anything that may be helpful would be great!
complex-analysis complex-integration
I've heard different people say yes and no, so I want to ask it here on math stack, but often, as in with cauchys integral theorem on wiki for instance, the contour is specified as a circle in the complex plane. Could it however be any shape (that's a closed contour) in the complex plane, and if so why?
To me, it doesn't seem to matter, and here's my reasoning: define the arbitrary contour as a parametric function of time, and since it's loop (not sure if there's a more formal definition, but here just take it to mean a closed contour) after some given time (let's call it T) it winds up back were it started at $t=0$. The derivative of the curve with respect to time is another complex valued function of time, and since after T the tangent is the same, the derivative is the same at T, meaning the derivative of the curve with respect to time also forms a loop (given of course that the contour is differentiable). So, the opposite also applies. Integrating a loop yields another loop, and integrating the loop over the time it takes to make one revolution (ie a definite integral from 0 to T) is 0 since you haven't moved anywhere since it's a loop. Then, if $dz=z'dt$ integrating a complex function over a complex variable is the same as a complex function (since the $z'$ that comes out is multiplied and complex numbers are closed under multiplication) integratied over time (a real variable), which is zero unless they cancel. Assuming they don't, symbolically (and I is the antiderivative of Z):
$oint_z f(z)dz=int^T_0 f(z)z'dt=int^T_0 Zdt=I(T)-I(0)=0$
$frac{1}{z}$ is an example of what I mean by cancel, if you were to integrate it like this. Making a change of coordinates to polar reveals that regardless of $r(t)$ (the loop) it evaluates to $2pi i$.
Again, even if this yields the right results I'm not sure if the intuition's correct. Do people just just use circles as the contour because they're convenient? Thanks in advance, anything that may be helpful would be great!
complex-analysis complex-integration
complex-analysis complex-integration
edited Nov 22 at 15:09
José Carlos Santos
146k22117217
146k22117217
asked Nov 22 at 14:58
Benjamin Thoburn
11210
11210
This question has an open bounty worth +50
reputation from Benjamin Thoburn ending in 7 days.
The current answers do not contain enough detail.
I know my result is wrong but I don't know where I went wrong. Unsure why this method is invalid.
This question has an open bounty worth +50
reputation from Benjamin Thoburn ending in 7 days.
The current answers do not contain enough detail.
I know my result is wrong but I don't know where I went wrong. Unsure why this method is invalid.
Why does putting a loop "in the loop" make the integral invalid?
– Benjamin Thoburn
2 hours ago
add a comment |
Why does putting a loop "in the loop" make the integral invalid?
– Benjamin Thoburn
2 hours ago
Why does putting a loop "in the loop" make the integral invalid?
– Benjamin Thoburn
2 hours ago
Why does putting a loop "in the loop" make the integral invalid?
– Benjamin Thoburn
2 hours ago
add a comment |
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I am not sure that I understood your question, but it seems to me that you are claiming that if you integrate $frac1z$ along any loop around $0$, then you will always get $2pi i$. This is not true. If $gammacolon[0,2pi]longrightarrowmathbb C$ is defined by $gamma(t)=e^{-it}$, then$$int_gammafrac1z,mathrm dz=-2pi i.$$ So, in that sense at least, the choice of the loop matters.
On the other hand, if we assume that we are not talking about loops on general, but only about simple loops, that is, loops such their winding number with respect to $0$ is $1$, then, yes, you are right: the shape doesn't matter. It can be a square, a triangle, an ellipse, or whatever you want. Then $int_gammafrac1z,mathrm dz$ will always be $2pi i$.
Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $frac{1}{z}$
– Benjamin Thoburn
Nov 22 at 15:12
What do you mean by “shape”? Are you thinking about the possibility that the image of $gamma$ is, say, an ellipse or a square?
– José Carlos Santos
Nov 22 at 15:15
yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$.
– Benjamin Thoburn
Nov 22 at 15:40
I have added another paragraph to my answer. What do you think now?
– José Carlos Santos
Nov 22 at 15:45
And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero?
– Benjamin Thoburn
Nov 22 at 16:24
|
show 6 more comments
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I am not sure that I understood your question, but it seems to me that you are claiming that if you integrate $frac1z$ along any loop around $0$, then you will always get $2pi i$. This is not true. If $gammacolon[0,2pi]longrightarrowmathbb C$ is defined by $gamma(t)=e^{-it}$, then$$int_gammafrac1z,mathrm dz=-2pi i.$$ So, in that sense at least, the choice of the loop matters.
On the other hand, if we assume that we are not talking about loops on general, but only about simple loops, that is, loops such their winding number with respect to $0$ is $1$, then, yes, you are right: the shape doesn't matter. It can be a square, a triangle, an ellipse, or whatever you want. Then $int_gammafrac1z,mathrm dz$ will always be $2pi i$.
Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $frac{1}{z}$
– Benjamin Thoburn
Nov 22 at 15:12
What do you mean by “shape”? Are you thinking about the possibility that the image of $gamma$ is, say, an ellipse or a square?
– José Carlos Santos
Nov 22 at 15:15
yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$.
– Benjamin Thoburn
Nov 22 at 15:40
I have added another paragraph to my answer. What do you think now?
– José Carlos Santos
Nov 22 at 15:45
And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero?
– Benjamin Thoburn
Nov 22 at 16:24
|
show 6 more comments
up vote
1
down vote
I am not sure that I understood your question, but it seems to me that you are claiming that if you integrate $frac1z$ along any loop around $0$, then you will always get $2pi i$. This is not true. If $gammacolon[0,2pi]longrightarrowmathbb C$ is defined by $gamma(t)=e^{-it}$, then$$int_gammafrac1z,mathrm dz=-2pi i.$$ So, in that sense at least, the choice of the loop matters.
On the other hand, if we assume that we are not talking about loops on general, but only about simple loops, that is, loops such their winding number with respect to $0$ is $1$, then, yes, you are right: the shape doesn't matter. It can be a square, a triangle, an ellipse, or whatever you want. Then $int_gammafrac1z,mathrm dz$ will always be $2pi i$.
Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $frac{1}{z}$
– Benjamin Thoburn
Nov 22 at 15:12
What do you mean by “shape”? Are you thinking about the possibility that the image of $gamma$ is, say, an ellipse or a square?
– José Carlos Santos
Nov 22 at 15:15
yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$.
– Benjamin Thoburn
Nov 22 at 15:40
I have added another paragraph to my answer. What do you think now?
– José Carlos Santos
Nov 22 at 15:45
And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero?
– Benjamin Thoburn
Nov 22 at 16:24
|
show 6 more comments
up vote
1
down vote
up vote
1
down vote
I am not sure that I understood your question, but it seems to me that you are claiming that if you integrate $frac1z$ along any loop around $0$, then you will always get $2pi i$. This is not true. If $gammacolon[0,2pi]longrightarrowmathbb C$ is defined by $gamma(t)=e^{-it}$, then$$int_gammafrac1z,mathrm dz=-2pi i.$$ So, in that sense at least, the choice of the loop matters.
On the other hand, if we assume that we are not talking about loops on general, but only about simple loops, that is, loops such their winding number with respect to $0$ is $1$, then, yes, you are right: the shape doesn't matter. It can be a square, a triangle, an ellipse, or whatever you want. Then $int_gammafrac1z,mathrm dz$ will always be $2pi i$.
I am not sure that I understood your question, but it seems to me that you are claiming that if you integrate $frac1z$ along any loop around $0$, then you will always get $2pi i$. This is not true. If $gammacolon[0,2pi]longrightarrowmathbb C$ is defined by $gamma(t)=e^{-it}$, then$$int_gammafrac1z,mathrm dz=-2pi i.$$ So, in that sense at least, the choice of the loop matters.
On the other hand, if we assume that we are not talking about loops on general, but only about simple loops, that is, loops such their winding number with respect to $0$ is $1$, then, yes, you are right: the shape doesn't matter. It can be a square, a triangle, an ellipse, or whatever you want. Then $int_gammafrac1z,mathrm dz$ will always be $2pi i$.
edited Nov 22 at 15:44
answered Nov 22 at 15:09
José Carlos Santos
146k22117217
146k22117217
Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $frac{1}{z}$
– Benjamin Thoburn
Nov 22 at 15:12
What do you mean by “shape”? Are you thinking about the possibility that the image of $gamma$ is, say, an ellipse or a square?
– José Carlos Santos
Nov 22 at 15:15
yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$.
– Benjamin Thoburn
Nov 22 at 15:40
I have added another paragraph to my answer. What do you think now?
– José Carlos Santos
Nov 22 at 15:45
And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero?
– Benjamin Thoburn
Nov 22 at 16:24
|
show 6 more comments
Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $frac{1}{z}$
– Benjamin Thoburn
Nov 22 at 15:12
What do you mean by “shape”? Are you thinking about the possibility that the image of $gamma$ is, say, an ellipse or a square?
– José Carlos Santos
Nov 22 at 15:15
yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$.
– Benjamin Thoburn
Nov 22 at 15:40
I have added another paragraph to my answer. What do you think now?
– José Carlos Santos
Nov 22 at 15:45
And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero?
– Benjamin Thoburn
Nov 22 at 16:24
Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $frac{1}{z}$
– Benjamin Thoburn
Nov 22 at 15:12
Oh yeah, I forgot about integrating in another direction. But aside from that... does the shape of the path matter? What about any function, not just $frac{1}{z}$
– Benjamin Thoburn
Nov 22 at 15:12
What do you mean by “shape”? Are you thinking about the possibility that the image of $gamma$ is, say, an ellipse or a square?
– José Carlos Santos
Nov 22 at 15:15
What do you mean by “shape”? Are you thinking about the possibility that the image of $gamma$ is, say, an ellipse or a square?
– José Carlos Santos
Nov 22 at 15:15
yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$.
– Benjamin Thoburn
Nov 22 at 15:40
yes, ie the shape of the contour if you graph it. By a different shape that isn't a circle I mean a contour that doesn't look like $e^{it}$.
– Benjamin Thoburn
Nov 22 at 15:40
I have added another paragraph to my answer. What do you think now?
– José Carlos Santos
Nov 22 at 15:45
I have added another paragraph to my answer. What do you think now?
– José Carlos Santos
Nov 22 at 15:45
And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero?
– Benjamin Thoburn
Nov 22 at 16:24
And I guess it follows that this goes for any general f(z) without a negative one term in the Laurent series integrating to zero?
– Benjamin Thoburn
Nov 22 at 16:24
|
show 6 more comments
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Why does putting a loop "in the loop" make the integral invalid?
– Benjamin Thoburn
2 hours ago