A quick question about complex integrals and Cauchy's integral formula
$begingroup$
I'll spare the specifics for brevity's sake, but in essence the problem I'm posed is finding
$$int_C frac{(z-1)^3 cdot e^z cdot cos(z)}{z}dz$$
along two different closed loops $C$. Each is a rectangle, oriented clockwise. One of them encloses the discontinuity of this function (i.e. $z=0$) and another doesn't.
Post-Script (December 2018): I recognize that the "discontinuity" mentioned is in reality a singularity.
I'm mostly just wanting to double-check my approach to this since it's explicitly specified that "this shouldn't take much computation," and want to double-check I'm on the right path.
My thoughts on the matter:
For $C$ being the closed loop not enclosing the discontinuity, the integral would be $0$ per Cauchy's integral theorem.
For $C$ being the closed loop that encloses the discontinuity, the integral would be $2pi i f(0)$, from Cauchy's integral formula, where $f(z)$ is the numerator of the integrand (sans the $dz$ of course), and "$0$" coming from being the point of discontinuity.
I have a rough intuition for why this might be - it's fairly heuristic and informal though - so I just wanted to make sure I was on the right track.
Thanks in advance.
complex-analysis complex-integration
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
$endgroup$
|
show 2 more comments
$begingroup$
I'll spare the specifics for brevity's sake, but in essence the problem I'm posed is finding
$$int_C frac{(z-1)^3 cdot e^z cdot cos(z)}{z}dz$$
along two different closed loops $C$. Each is a rectangle, oriented clockwise. One of them encloses the discontinuity of this function (i.e. $z=0$) and another doesn't.
Post-Script (December 2018): I recognize that the "discontinuity" mentioned is in reality a singularity.
I'm mostly just wanting to double-check my approach to this since it's explicitly specified that "this shouldn't take much computation," and want to double-check I'm on the right path.
My thoughts on the matter:
For $C$ being the closed loop not enclosing the discontinuity, the integral would be $0$ per Cauchy's integral theorem.
For $C$ being the closed loop that encloses the discontinuity, the integral would be $2pi i f(0)$, from Cauchy's integral formula, where $f(z)$ is the numerator of the integrand (sans the $dz$ of course), and "$0$" coming from being the point of discontinuity.
I have a rough intuition for why this might be - it's fairly heuristic and informal though - so I just wanted to make sure I was on the right track.
Thanks in advance.
complex-analysis complex-integration
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
$endgroup$
1
$begingroup$
You are right and the value of the integral in the second case is $-1$.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 7:50
$begingroup$
You have already been told that you are right. I just want to add that the function that you are integrating has no point of discontinuity. The point $0$ is a singularity.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:00
1
$begingroup$
Actually @KaviRamaMurthy I did have a second thought: the orientation of the curve. Since it's clockwise, and the formulation of Cauchy's formula states counterclockwise, would it not be the negative of that? (Another minor nitpick but I think it should be noted $f(0)=-1)$ as opposed to the integral which is $2pi i f(0)$. Not trying to be a jerk though.)
$endgroup$
– Eevee Trainer
Oct 12 '18 at 8:01
2
$begingroup$
@EeveeTrainer The correct answer is $2pi i$. I was a bit careless in my earlier comment.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 8:15
1
$begingroup$
@EeveeTrainer A function can only be continuous (or discontinuous) at points of its domain. An isolated singularity is a point outside the domain such that all points nearby belong to it.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:17
|
show 2 more comments
$begingroup$
I'll spare the specifics for brevity's sake, but in essence the problem I'm posed is finding
$$int_C frac{(z-1)^3 cdot e^z cdot cos(z)}{z}dz$$
along two different closed loops $C$. Each is a rectangle, oriented clockwise. One of them encloses the discontinuity of this function (i.e. $z=0$) and another doesn't.
Post-Script (December 2018): I recognize that the "discontinuity" mentioned is in reality a singularity.
I'm mostly just wanting to double-check my approach to this since it's explicitly specified that "this shouldn't take much computation," and want to double-check I'm on the right path.
My thoughts on the matter:
For $C$ being the closed loop not enclosing the discontinuity, the integral would be $0$ per Cauchy's integral theorem.
For $C$ being the closed loop that encloses the discontinuity, the integral would be $2pi i f(0)$, from Cauchy's integral formula, where $f(z)$ is the numerator of the integrand (sans the $dz$ of course), and "$0$" coming from being the point of discontinuity.
I have a rough intuition for why this might be - it's fairly heuristic and informal though - so I just wanted to make sure I was on the right track.
Thanks in advance.
complex-analysis complex-integration
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
$endgroup$
I'll spare the specifics for brevity's sake, but in essence the problem I'm posed is finding
$$int_C frac{(z-1)^3 cdot e^z cdot cos(z)}{z}dz$$
along two different closed loops $C$. Each is a rectangle, oriented clockwise. One of them encloses the discontinuity of this function (i.e. $z=0$) and another doesn't.
Post-Script (December 2018): I recognize that the "discontinuity" mentioned is in reality a singularity.
I'm mostly just wanting to double-check my approach to this since it's explicitly specified that "this shouldn't take much computation," and want to double-check I'm on the right path.
My thoughts on the matter:
For $C$ being the closed loop not enclosing the discontinuity, the integral would be $0$ per Cauchy's integral theorem.
For $C$ being the closed loop that encloses the discontinuity, the integral would be $2pi i f(0)$, from Cauchy's integral formula, where $f(z)$ is the numerator of the integrand (sans the $dz$ of course), and "$0$" coming from being the point of discontinuity.
I have a rough intuition for why this might be - it's fairly heuristic and informal though - so I just wanted to make sure I was on the right track.
Thanks in advance.
complex-analysis complex-integration
complex-analysis complex-integration
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
edited Dec 30 '18 at 8:43
Eevee Trainer
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
asked Oct 12 '18 at 7:48
Eevee TrainerEevee Trainer
8,38921439
8,38921439
1
$begingroup$
You are right and the value of the integral in the second case is $-1$.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 7:50
$begingroup$
You have already been told that you are right. I just want to add that the function that you are integrating has no point of discontinuity. The point $0$ is a singularity.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:00
1
$begingroup$
Actually @KaviRamaMurthy I did have a second thought: the orientation of the curve. Since it's clockwise, and the formulation of Cauchy's formula states counterclockwise, would it not be the negative of that? (Another minor nitpick but I think it should be noted $f(0)=-1)$ as opposed to the integral which is $2pi i f(0)$. Not trying to be a jerk though.)
$endgroup$
– Eevee Trainer
Oct 12 '18 at 8:01
2
$begingroup$
@EeveeTrainer The correct answer is $2pi i$. I was a bit careless in my earlier comment.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 8:15
1
$begingroup$
@EeveeTrainer A function can only be continuous (or discontinuous) at points of its domain. An isolated singularity is a point outside the domain such that all points nearby belong to it.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:17
|
show 2 more comments
1
$begingroup$
You are right and the value of the integral in the second case is $-1$.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 7:50
$begingroup$
You have already been told that you are right. I just want to add that the function that you are integrating has no point of discontinuity. The point $0$ is a singularity.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:00
1
$begingroup$
Actually @KaviRamaMurthy I did have a second thought: the orientation of the curve. Since it's clockwise, and the formulation of Cauchy's formula states counterclockwise, would it not be the negative of that? (Another minor nitpick but I think it should be noted $f(0)=-1)$ as opposed to the integral which is $2pi i f(0)$. Not trying to be a jerk though.)
$endgroup$
– Eevee Trainer
Oct 12 '18 at 8:01
2
$begingroup$
@EeveeTrainer The correct answer is $2pi i$. I was a bit careless in my earlier comment.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 8:15
1
$begingroup$
@EeveeTrainer A function can only be continuous (or discontinuous) at points of its domain. An isolated singularity is a point outside the domain such that all points nearby belong to it.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:17
1
1
$begingroup$
You are right and the value of the integral in the second case is $-1$.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 7:50
$begingroup$
You are right and the value of the integral in the second case is $-1$.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 7:50
$begingroup$
You have already been told that you are right. I just want to add that the function that you are integrating has no point of discontinuity. The point $0$ is a singularity.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:00
$begingroup$
You have already been told that you are right. I just want to add that the function that you are integrating has no point of discontinuity. The point $0$ is a singularity.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:00
1
1
$begingroup$
Actually @KaviRamaMurthy I did have a second thought: the orientation of the curve. Since it's clockwise, and the formulation of Cauchy's formula states counterclockwise, would it not be the negative of that? (Another minor nitpick but I think it should be noted $f(0)=-1)$ as opposed to the integral which is $2pi i f(0)$. Not trying to be a jerk though.)
$endgroup$
– Eevee Trainer
Oct 12 '18 at 8:01
$begingroup$
Actually @KaviRamaMurthy I did have a second thought: the orientation of the curve. Since it's clockwise, and the formulation of Cauchy's formula states counterclockwise, would it not be the negative of that? (Another minor nitpick but I think it should be noted $f(0)=-1)$ as opposed to the integral which is $2pi i f(0)$. Not trying to be a jerk though.)
$endgroup$
– Eevee Trainer
Oct 12 '18 at 8:01
2
2
$begingroup$
@EeveeTrainer The correct answer is $2pi i$. I was a bit careless in my earlier comment.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 8:15
$begingroup$
@EeveeTrainer The correct answer is $2pi i$. I was a bit careless in my earlier comment.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 8:15
1
1
$begingroup$
@EeveeTrainer A function can only be continuous (or discontinuous) at points of its domain. An isolated singularity is a point outside the domain such that all points nearby belong to it.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:17
$begingroup$
@EeveeTrainer A function can only be continuous (or discontinuous) at points of its domain. An isolated singularity is a point outside the domain such that all points nearby belong to it.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:17
|
show 2 more comments
1 Answer
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$begingroup$
Full disclosure: I was going back through some of the questions I asked over the past semester and noticed this one was left unanswered, at least in the "official answer" sense, i.e. no one gave me an answer to accept. Since this was basically an "am I right" sort of question, I'm just going to make this a community wiki answer and accept it for the sake of closing the question. (I chose community wiki because it was other people who verified my solution and I don't have much to add to the solution. Plus if others have discussion to offer, they should do so.)
The thoughts posted in the OP are indeed correct, and follow directly from Cauchy's integral theorem/formula. The first contour yields an integral evaluating to $0$, the second to $2pi i$.
Thanks to José Carlos Santos and Kavi Rama Murthy for their input on the post.
$endgroup$
add a comment |
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$begingroup$
Full disclosure: I was going back through some of the questions I asked over the past semester and noticed this one was left unanswered, at least in the "official answer" sense, i.e. no one gave me an answer to accept. Since this was basically an "am I right" sort of question, I'm just going to make this a community wiki answer and accept it for the sake of closing the question. (I chose community wiki because it was other people who verified my solution and I don't have much to add to the solution. Plus if others have discussion to offer, they should do so.)
The thoughts posted in the OP are indeed correct, and follow directly from Cauchy's integral theorem/formula. The first contour yields an integral evaluating to $0$, the second to $2pi i$.
Thanks to José Carlos Santos and Kavi Rama Murthy for their input on the post.
$endgroup$
add a comment |
$begingroup$
Full disclosure: I was going back through some of the questions I asked over the past semester and noticed this one was left unanswered, at least in the "official answer" sense, i.e. no one gave me an answer to accept. Since this was basically an "am I right" sort of question, I'm just going to make this a community wiki answer and accept it for the sake of closing the question. (I chose community wiki because it was other people who verified my solution and I don't have much to add to the solution. Plus if others have discussion to offer, they should do so.)
The thoughts posted in the OP are indeed correct, and follow directly from Cauchy's integral theorem/formula. The first contour yields an integral evaluating to $0$, the second to $2pi i$.
Thanks to José Carlos Santos and Kavi Rama Murthy for their input on the post.
$endgroup$
add a comment |
$begingroup$
Full disclosure: I was going back through some of the questions I asked over the past semester and noticed this one was left unanswered, at least in the "official answer" sense, i.e. no one gave me an answer to accept. Since this was basically an "am I right" sort of question, I'm just going to make this a community wiki answer and accept it for the sake of closing the question. (I chose community wiki because it was other people who verified my solution and I don't have much to add to the solution. Plus if others have discussion to offer, they should do so.)
The thoughts posted in the OP are indeed correct, and follow directly from Cauchy's integral theorem/formula. The first contour yields an integral evaluating to $0$, the second to $2pi i$.
Thanks to José Carlos Santos and Kavi Rama Murthy for their input on the post.
$endgroup$
Full disclosure: I was going back through some of the questions I asked over the past semester and noticed this one was left unanswered, at least in the "official answer" sense, i.e. no one gave me an answer to accept. Since this was basically an "am I right" sort of question, I'm just going to make this a community wiki answer and accept it for the sake of closing the question. (I chose community wiki because it was other people who verified my solution and I don't have much to add to the solution. Plus if others have discussion to offer, they should do so.)
The thoughts posted in the OP are indeed correct, and follow directly from Cauchy's integral theorem/formula. The first contour yields an integral evaluating to $0$, the second to $2pi i$.
Thanks to José Carlos Santos and Kavi Rama Murthy for their input on the post.
answered Dec 30 '18 at 8:34
community wiki
Eevee Trainer
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$begingroup$
You are right and the value of the integral in the second case is $-1$.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 7:50
$begingroup$
You have already been told that you are right. I just want to add that the function that you are integrating has no point of discontinuity. The point $0$ is a singularity.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:00
1
$begingroup$
Actually @KaviRamaMurthy I did have a second thought: the orientation of the curve. Since it's clockwise, and the formulation of Cauchy's formula states counterclockwise, would it not be the negative of that? (Another minor nitpick but I think it should be noted $f(0)=-1)$ as opposed to the integral which is $2pi i f(0)$. Not trying to be a jerk though.)
$endgroup$
– Eevee Trainer
Oct 12 '18 at 8:01
2
$begingroup$
@EeveeTrainer The correct answer is $2pi i$. I was a bit careless in my earlier comment.
$endgroup$
– Kavi Rama Murthy
Oct 12 '18 at 8:15
1
$begingroup$
@EeveeTrainer A function can only be continuous (or discontinuous) at points of its domain. An isolated singularity is a point outside the domain such that all points nearby belong to it.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 8:17