Complex Analysis: Evaluating an Integral over C. where C is the contour |z| = 1. The 2019 Stack Overflow Developer Survey Results Are InEvaluating a Complex Trigonometric IntegralComplex contour integral with residue theoryEvaluating series by contour integration, the residue theorem, and cotangentComplex Contour Integration - Complex AnalysisComplex Analysis- Finding Laurent SeriesIntegral along closed contourComplex Laurent Series and Contour IntegralComplex Analysis - Contour IntegralEvaluating contour integral along the boundary of the fundamental domain of $SL_2(mathbbZ)$ near polesContour Integral of Complex Power Function
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Complex Analysis: Evaluating an Integral over C. where C is the contour |z| = 1.
The 2019 Stack Overflow Developer Survey Results Are InEvaluating a Complex Trigonometric IntegralComplex contour integral with residue theoryEvaluating series by contour integration, the residue theorem, and cotangentComplex Contour Integration - Complex AnalysisComplex Analysis- Finding Laurent SeriesIntegral along closed contourComplex Laurent Series and Contour IntegralComplex Analysis - Contour IntegralEvaluating contour integral along the boundary of the fundamental domain of $SL_2(mathbbZ)$ near polesContour Integral of Complex Power Function
$begingroup$
I've got a homework question that I believe requires me to use Laurent series/method of residues.
The question itself is:
Evaluate $int_C frac1z^2(z^2-16)dz$ where C is the contour $|z| = 1$.
I'm confused by this question because it doesn't say anything about the orientation of C.
I know that the function is not analytic at z = 0, 4, -4. However beyond that, i'm ashamed to say I don't even know how to approach this.
So far, I've tried breaking it into:
$int_C frac1z^2 frac1z+4i frac1z-4idz$. But I don't really know how to get the Taylor/Laurent series for these three pieces.
Am I approaching this in the right way?
Can somebody help me move forward?
complex-analysis
asked Apr 6 at 22:19
QhefQhef
111
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I've got a homework question that I believe requires me to use Laurent series/method of residues.
The question itself is:
Evaluate $int_C frac1z^2(z^2-16)dz$ where C is the contour $|z| = 1$.
I'm confused by this question because it doesn't say anything about the orientation of C.
I know that the function is not analytic at z = 0, 4, -4. However beyond that, i'm ashamed to say I don't even know how to approach this.
So far, I've tried breaking it into:
$int_C frac1z^2 frac1z+4i frac1z-4idz$. But I don't really know how to get the Taylor/Laurent series for these three pieces.
Am I approaching this in the right way?
Can somebody help me move forward?
complex-analysis
asked Apr 6 at 22:19
QhefQhef
111
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
The contour "|z|= 1" is the circle, of radius 1, centered at the origin. Both 4i and -4i lie outside that contour so, immediately, those two parts contribute nothing to the integral. You have reduced the problem to integrating $frac1z^2$ around that circle. One way to do that is to observe that, on that circle,$z= e^itheta$ with $theta$ going from 0 to $2pi$. $dz= ie^ithetadtheta$ so the integral becomes $int_0^2pi e^-2itheta ie^ithetadtheta= iint_0^2pi e^-ithetadtheta$.
$endgroup$
– user247327
Apr 6 at 22:35
$begingroup$
There is a very easy way to get the Taylor series for $f(z) = frac1z^2-16$; think about geometric series. Then you just have to divide it by $z^2$.
$endgroup$
– Nate Eldredge
Apr 6 at 23:27
$begingroup$
Note that $frac1z^2-16$ factors as $frac1(z+4)(z-4)$, not $4i$. Not that it makes a lot of difference for this problem.
$endgroup$
– Nate Eldredge
Apr 6 at 23:28
$begingroup$
The question probably should have specified an orientation for $C$. However you are going to find that the value of the integral is 0 so in fact it doesn't matter which way you orient it.
$endgroup$
– Nate Eldredge
Apr 6 at 23:30
add a comment |
$begingroup$
I've got a homework question that I believe requires me to use Laurent series/method of residues.
The question itself is:
Evaluate $int_C frac1z^2(z^2-16)dz$ where C is the contour $|z| = 1$.
I'm confused by this question because it doesn't say anything about the orientation of C.
I know that the function is not analytic at z = 0, 4, -4. However beyond that, i'm ashamed to say I don't even know how to approach this.
So far, I've tried breaking it into:
$int_C frac1z^2 frac1z+4i frac1z-4idz$. But I don't really know how to get the Taylor/Laurent series for these three pieces.
Am I approaching this in the right way?
Can somebody help me move forward?
complex-analysis
asked Apr 6 at 22:19
QhefQhef
111
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I've got a homework question that I believe requires me to use Laurent series/method of residues.
The question itself is:
Evaluate $int_C frac1z^2(z^2-16)dz$ where C is the contour $|z| = 1$.
I'm confused by this question because it doesn't say anything about the orientation of C.
I know that the function is not analytic at z = 0, 4, -4. However beyond that, i'm ashamed to say I don't even know how to approach this.
So far, I've tried breaking it into:
$int_C frac1z^2 frac1z+4i frac1z-4idz$. But I don't really know how to get the Taylor/Laurent series for these three pieces.
Am I approaching this in the right way?
Can somebody help me move forward?
complex-analysis
complex-analysis
asked Apr 6 at 22:19
QhefQhef
111
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 22:19
QhefQhef
111
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 22:19
QhefQhef
111
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 22:19
QhefQhef
111
asked Apr 6 at 22:19
QhefQhef
111
111
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Qhef is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
The contour "|z|= 1" is the circle, of radius 1, centered at the origin. Both 4i and -4i lie outside that contour so, immediately, those two parts contribute nothing to the integral. You have reduced the problem to integrating $frac1z^2$ around that circle. One way to do that is to observe that, on that circle,$z= e^itheta$ with $theta$ going from 0 to $2pi$. $dz= ie^ithetadtheta$ so the integral becomes $int_0^2pi e^-2itheta ie^ithetadtheta= iint_0^2pi e^-ithetadtheta$.
$endgroup$
– user247327
Apr 6 at 22:35
$begingroup$
There is a very easy way to get the Taylor series for $f(z) = frac1z^2-16$; think about geometric series. Then you just have to divide it by $z^2$.
$endgroup$
– Nate Eldredge
Apr 6 at 23:27
$begingroup$
Note that $frac1z^2-16$ factors as $frac1(z+4)(z-4)$, not $4i$. Not that it makes a lot of difference for this problem.
$endgroup$
– Nate Eldredge
Apr 6 at 23:28
$begingroup$
The question probably should have specified an orientation for $C$. However you are going to find that the value of the integral is 0 so in fact it doesn't matter which way you orient it.
$endgroup$
– Nate Eldredge
Apr 6 at 23:30
add a comment |
1
$begingroup$
The contour "|z|= 1" is the circle, of radius 1, centered at the origin. Both 4i and -4i lie outside that contour so, immediately, those two parts contribute nothing to the integral. You have reduced the problem to integrating $frac1z^2$ around that circle. One way to do that is to observe that, on that circle,$z= e^itheta$ with $theta$ going from 0 to $2pi$. $dz= ie^ithetadtheta$ so the integral becomes $int_0^2pi e^-2itheta ie^ithetadtheta= iint_0^2pi e^-ithetadtheta$.
$endgroup$
– user247327
Apr 6 at 22:35
$begingroup$
There is a very easy way to get the Taylor series for $f(z) = frac1z^2-16$; think about geometric series. Then you just have to divide it by $z^2$.
$endgroup$
– Nate Eldredge
Apr 6 at 23:27
$begingroup$
Note that $frac1z^2-16$ factors as $frac1(z+4)(z-4)$, not $4i$. Not that it makes a lot of difference for this problem.
$endgroup$
– Nate Eldredge
Apr 6 at 23:28
$begingroup$
The question probably should have specified an orientation for $C$. However you are going to find that the value of the integral is 0 so in fact it doesn't matter which way you orient it.
$endgroup$
– Nate Eldredge
Apr 6 at 23:30
1
1
$begingroup$
The contour "|z|= 1" is the circle, of radius 1, centered at the origin. Both 4i and -4i lie outside that contour so, immediately, those two parts contribute nothing to the integral. You have reduced the problem to integrating $frac1z^2$ around that circle. One way to do that is to observe that, on that circle,$z= e^itheta$ with $theta$ going from 0 to $2pi$. $dz= ie^ithetadtheta$ so the integral becomes $int_0^2pi e^-2itheta ie^ithetadtheta= iint_0^2pi e^-ithetadtheta$.
$endgroup$
– user247327
Apr 6 at 22:35
$begingroup$
The contour "|z|= 1" is the circle, of radius 1, centered at the origin. Both 4i and -4i lie outside that contour so, immediately, those two parts contribute nothing to the integral. You have reduced the problem to integrating $frac1z^2$ around that circle. One way to do that is to observe that, on that circle,$z= e^itheta$ with $theta$ going from 0 to $2pi$. $dz= ie^ithetadtheta$ so the integral becomes $int_0^2pi e^-2itheta ie^ithetadtheta= iint_0^2pi e^-ithetadtheta$.
$endgroup$
– user247327
Apr 6 at 22:35
$begingroup$
There is a very easy way to get the Taylor series for $f(z) = frac1z^2-16$; think about geometric series. Then you just have to divide it by $z^2$.
$endgroup$
– Nate Eldredge
Apr 6 at 23:27
$begingroup$
There is a very easy way to get the Taylor series for $f(z) = frac1z^2-16$; think about geometric series. Then you just have to divide it by $z^2$.
$endgroup$
– Nate Eldredge
Apr 6 at 23:27
$begingroup$
Note that $frac1z^2-16$ factors as $frac1(z+4)(z-4)$, not $4i$. Not that it makes a lot of difference for this problem.
$endgroup$
– Nate Eldredge
Apr 6 at 23:28
$begingroup$
Note that $frac1z^2-16$ factors as $frac1(z+4)(z-4)$, not $4i$. Not that it makes a lot of difference for this problem.
$endgroup$
– Nate Eldredge
Apr 6 at 23:28
$begingroup$
The question probably should have specified an orientation for $C$. However you are going to find that the value of the integral is 0 so in fact it doesn't matter which way you orient it.
$endgroup$
– Nate Eldredge
Apr 6 at 23:30
$begingroup$
The question probably should have specified an orientation for $C$. However you are going to find that the value of the integral is 0 so in fact it doesn't matter which way you orient it.
$endgroup$
– Nate Eldredge
Apr 6 at 23:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Orientation of $C$ does not matter here. The value of the integral is $0$ and you can prove it without any calculation! The only pole inside $C$ is $z=0$. Near $0$ the function $frac 1 z^2-16$ is analytic and when you divide the power series of this analytic function by $z^2$ the residue becomes $0$. [ Residue is the coefficient of $frac 1 z$. Note that the power series of this function has only even powers of $z$]. Hence the integral is $0$.
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
add a comment |
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$begingroup$
Orientation of $C$ does not matter here. The value of the integral is $0$ and you can prove it without any calculation! The only pole inside $C$ is $z=0$. Near $0$ the function $frac 1 z^2-16$ is analytic and when you divide the power series of this analytic function by $z^2$ the residue becomes $0$. [ Residue is the coefficient of $frac 1 z$. Note that the power series of this function has only even powers of $z$]. Hence the integral is $0$.
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
add a comment |
$begingroup$
Orientation of $C$ does not matter here. The value of the integral is $0$ and you can prove it without any calculation! The only pole inside $C$ is $z=0$. Near $0$ the function $frac 1 z^2-16$ is analytic and when you divide the power series of this analytic function by $z^2$ the residue becomes $0$. [ Residue is the coefficient of $frac 1 z$. Note that the power series of this function has only even powers of $z$]. Hence the integral is $0$.
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
add a comment |
$begingroup$
Orientation of $C$ does not matter here. The value of the integral is $0$ and you can prove it without any calculation! The only pole inside $C$ is $z=0$. Near $0$ the function $frac 1 z^2-16$ is analytic and when you divide the power series of this analytic function by $z^2$ the residue becomes $0$. [ Residue is the coefficient of $frac 1 z$. Note that the power series of this function has only even powers of $z$]. Hence the integral is $0$.
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
$endgroup$
Orientation of $C$ does not matter here. The value of the integral is $0$ and you can prove it without any calculation! The only pole inside $C$ is $z=0$. Near $0$ the function $frac 1 z^2-16$ is analytic and when you divide the power series of this analytic function by $z^2$ the residue becomes $0$. [ Residue is the coefficient of $frac 1 z$. Note that the power series of this function has only even powers of $z$]. Hence the integral is $0$.
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
answered Apr 6 at 23:23
Kavi Rama MurthyKavi Rama Murthy
73.9k53170
73.9k53170
add a comment |
add a comment |
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$begingroup$
The contour "|z|= 1" is the circle, of radius 1, centered at the origin. Both 4i and -4i lie outside that contour so, immediately, those two parts contribute nothing to the integral. You have reduced the problem to integrating $frac1z^2$ around that circle. One way to do that is to observe that, on that circle,$z= e^itheta$ with $theta$ going from 0 to $2pi$. $dz= ie^ithetadtheta$ so the integral becomes $int_0^2pi e^-2itheta ie^ithetadtheta= iint_0^2pi e^-ithetadtheta$.
$endgroup$
– user247327
Apr 6 at 22:35
$begingroup$
There is a very easy way to get the Taylor series for $f(z) = frac1z^2-16$; think about geometric series. Then you just have to divide it by $z^2$.
$endgroup$
– Nate Eldredge
Apr 6 at 23:27
$begingroup$
Note that $frac1z^2-16$ factors as $frac1(z+4)(z-4)$, not $4i$. Not that it makes a lot of difference for this problem.
$endgroup$
– Nate Eldredge
Apr 6 at 23:28
$begingroup$
The question probably should have specified an orientation for $C$. However you are going to find that the value of the integral is 0 so in fact it doesn't matter which way you orient it.
$endgroup$
– Nate Eldredge
Apr 6 at 23:30