Conjugate of contour integrals
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Let $varphi : { z in C | |z| = 1} rightarrow C$ be a continuous function and let $gamma$ be the circular contour centered at 0 of radius 1 positively oriented. Prove that
$$overline{int_{gamma}! varphi (z) mathrm{d}z} = - int_gamma! overline{varphi (z)} frac{mathrm{d}z}{z^2}$$
I think you might be able to use Cauchy's residue theorem but I'm not too sure, any help is greatly appreciated
complex-analysis contour-integration
edited Nov 25 at 17:44
Bernard
117k637109
asked Nov 25 at 17:31
ProfessorPyg
32
add a comment |
up vote
0
down vote
favorite
Let $varphi : { z in C | |z| = 1} rightarrow C$ be a continuous function and let $gamma$ be the circular contour centered at 0 of radius 1 positively oriented. Prove that
$$overline{int_{gamma}! varphi (z) mathrm{d}z} = - int_gamma! overline{varphi (z)} frac{mathrm{d}z}{z^2}$$
I think you might be able to use Cauchy's residue theorem but I'm not too sure, any help is greatly appreciated
complex-analysis contour-integration
edited Nov 25 at 17:44
Bernard
117k637109
asked Nov 25 at 17:31
ProfessorPyg
32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $varphi : { z in C | |z| = 1} rightarrow C$ be a continuous function and let $gamma$ be the circular contour centered at 0 of radius 1 positively oriented. Prove that
$$overline{int_{gamma}! varphi (z) mathrm{d}z} = - int_gamma! overline{varphi (z)} frac{mathrm{d}z}{z^2}$$
I think you might be able to use Cauchy's residue theorem but I'm not too sure, any help is greatly appreciated
complex-analysis contour-integration
edited Nov 25 at 17:44
Bernard
117k637109
asked Nov 25 at 17:31
ProfessorPyg
32
Let $varphi : { z in C | |z| = 1} rightarrow C$ be a continuous function and let $gamma$ be the circular contour centered at 0 of radius 1 positively oriented. Prove that
$$overline{int_{gamma}! varphi (z) mathrm{d}z} = - int_gamma! overline{varphi (z)} frac{mathrm{d}z}{z^2}$$
I think you might be able to use Cauchy's residue theorem but I'm not too sure, any help is greatly appreciated
complex-analysis contour-integration
complex-analysis contour-integration
edited Nov 25 at 17:44
Bernard
117k637109
asked Nov 25 at 17:31
ProfessorPyg
32
edited Nov 25 at 17:44
Bernard
117k637109
asked Nov 25 at 17:31
ProfessorPyg
32
edited Nov 25 at 17:44
Bernard
117k637109
edited Nov 25 at 17:44
Bernard
117k637109
edited Nov 25 at 17:44
Bernard
117k637109
117k637109
asked Nov 25 at 17:31
ProfessorPyg
32
asked Nov 25 at 17:31
ProfessorPyg
32
asked Nov 25 at 17:31
ProfessorPyg
32
32
add a comment |
add a comment |
2 Answers
2
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You don't have to bother about the $phi$. (Note that $phi$ is only continuous.) The magic is with the $dz$. On $partial D_1$ one has $bar z={1over z}$. It follows that
$$dbar z= dleft({1over z}right)=-{1over z^2}>dz .$$
answered Nov 26 at 9:29
Christian Blatter
171k7111325
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Hint: take the obvious parametrization $z = e^{it}, tin [0,2pi]$ and do
$$
overline{int_{gamma}varphi(z),dz} =
overline{int_0^{2pi}varphi(e^{it})ie^{it},dt} = cdots
$$
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
I know how I would do this if i knew $varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks
– ProfessorPyg
Nov 25 at 22:56
@ProfessorPyg, the exact value of $varphi$ is irrelevant. Edited, can you continue or do the same for the RHS?
– Martín-Blas Pérez Pinilla
Nov 26 at 9:22
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You don't have to bother about the $phi$. (Note that $phi$ is only continuous.) The magic is with the $dz$. On $partial D_1$ one has $bar z={1over z}$. It follows that
$$dbar z= dleft({1over z}right)=-{1over z^2}>dz .$$
answered Nov 26 at 9:29
Christian Blatter
171k7111325
add a comment |
up vote
0
down vote
accepted
You don't have to bother about the $phi$. (Note that $phi$ is only continuous.) The magic is with the $dz$. On $partial D_1$ one has $bar z={1over z}$. It follows that
$$dbar z= dleft({1over z}right)=-{1over z^2}>dz .$$
answered Nov 26 at 9:29
Christian Blatter
171k7111325
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You don't have to bother about the $phi$. (Note that $phi$ is only continuous.) The magic is with the $dz$. On $partial D_1$ one has $bar z={1over z}$. It follows that
$$dbar z= dleft({1over z}right)=-{1over z^2}>dz .$$
answered Nov 26 at 9:29
Christian Blatter
171k7111325
You don't have to bother about the $phi$. (Note that $phi$ is only continuous.) The magic is with the $dz$. On $partial D_1$ one has $bar z={1over z}$. It follows that
$$dbar z= dleft({1over z}right)=-{1over z^2}>dz .$$
answered Nov 26 at 9:29
Christian Blatter
171k7111325
answered Nov 26 at 9:29
Christian Blatter
171k7111325
answered Nov 26 at 9:29
Christian Blatter
171k7111325
answered Nov 26 at 9:29
Christian Blatter
171k7111325
171k7111325
add a comment |
add a comment |
up vote
0
down vote
Hint: take the obvious parametrization $z = e^{it}, tin [0,2pi]$ and do
$$
overline{int_{gamma}varphi(z),dz} =
overline{int_0^{2pi}varphi(e^{it})ie^{it},dt} = cdots
$$
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
I know how I would do this if i knew $varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks
– ProfessorPyg
Nov 25 at 22:56
@ProfessorPyg, the exact value of $varphi$ is irrelevant. Edited, can you continue or do the same for the RHS?
– Martín-Blas Pérez Pinilla
Nov 26 at 9:22
add a comment |
up vote
0
down vote
Hint: take the obvious parametrization $z = e^{it}, tin [0,2pi]$ and do
$$
overline{int_{gamma}varphi(z),dz} =
overline{int_0^{2pi}varphi(e^{it})ie^{it},dt} = cdots
$$
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
I know how I would do this if i knew $varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks
– ProfessorPyg
Nov 25 at 22:56
@ProfessorPyg, the exact value of $varphi$ is irrelevant. Edited, can you continue or do the same for the RHS?
– Martín-Blas Pérez Pinilla
Nov 26 at 9:22
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: take the obvious parametrization $z = e^{it}, tin [0,2pi]$ and do
$$
overline{int_{gamma}varphi(z),dz} =
overline{int_0^{2pi}varphi(e^{it})ie^{it},dt} = cdots
$$
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
Hint: take the obvious parametrization $z = e^{it}, tin [0,2pi]$ and do
$$
overline{int_{gamma}varphi(z),dz} =
overline{int_0^{2pi}varphi(e^{it})ie^{it},dt} = cdots
$$
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
edited Nov 26 at 9:19
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
answered Nov 25 at 18:31
Martín-Blas Pérez Pinilla
33.9k42771
33.9k42771
I know how I would do this if i knew $varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks
– ProfessorPyg
Nov 25 at 22:56
@ProfessorPyg, the exact value of $varphi$ is irrelevant. Edited, can you continue or do the same for the RHS?
– Martín-Blas Pérez Pinilla
Nov 26 at 9:22
add a comment |
I know how I would do this if i knew $varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks
– ProfessorPyg
Nov 25 at 22:56
@ProfessorPyg, the exact value of $varphi$ is irrelevant. Edited, can you continue or do the same for the RHS?
– Martín-Blas Pérez Pinilla
Nov 26 at 9:22
I know how I would do this if i knew $varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks
– ProfessorPyg
Nov 25 at 22:56
I know how I would do this if i knew $varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks
– ProfessorPyg
Nov 25 at 22:56
@ProfessorPyg, the exact value of $varphi$ is irrelevant. Edited, can you continue or do the same for the RHS?
– Martín-Blas Pérez Pinilla
Nov 26 at 9:22
@ProfessorPyg, the exact value of $varphi$ is irrelevant. Edited, can you continue or do the same for the RHS?
– Martín-Blas Pérez Pinilla
Nov 26 at 9:22
add a comment |
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