Mixing
Contour integral vanishes
$begingroup$
I need some help regarding why the contour integral $$int_{gamma} frac{f(w)}{w-frac{1}{bar{z}}} mathrm dw$$ is equal to zero, where $gamma$ is the unit circle and $f$ is holomorphic on the unit disk and continuous on its closure.
I think I have to use Cauchy's Theorem, but can't understand why the integrand in this case is holomorphic and why, for example, the integrand in Cauchy's Formula (i.e. $frac{f(w)}{w-z}$) being very similar, is not.
Thank you all for any clues you can give me
complex-analysis cauchy-integral-formula
edited Jan 30 at 16:07
mrtaurho
6,09271641
asked Jan 30 at 16:04
xan32xan32
33
$endgroup$
add a comment |
$begingroup$
I need some help regarding why the contour integral $$int_{gamma} frac{f(w)}{w-frac{1}{bar{z}}} mathrm dw$$ is equal to zero, where $gamma$ is the unit circle and $f$ is holomorphic on the unit disk and continuous on its closure.
I think I have to use Cauchy's Theorem, but can't understand why the integrand in this case is holomorphic and why, for example, the integrand in Cauchy's Formula (i.e. $frac{f(w)}{w-z}$) being very similar, is not.
Thank you all for any clues you can give me
complex-analysis cauchy-integral-formula
edited Jan 30 at 16:07
mrtaurho
6,09271641
asked Jan 30 at 16:04
xan32xan32
33
$endgroup$
$begingroup$
Where is $z$ located relative to $gamma$?
$endgroup$
– md2perpe
Jan 30 at 16:58
$begingroup$
z in on the open unit disk, so that $frac{1}{bar{z}}$ is outside the unit circle $gamma$. The thing is that I don't know how's that related to the function being holomorphic, since $w$ is in the unit circle in both of the cases I mentioned in my question
$endgroup$
– xan32
Jan 31 at 0:03
add a comment |
$begingroup$
I need some help regarding why the contour integral $$int_{gamma} frac{f(w)}{w-frac{1}{bar{z}}} mathrm dw$$ is equal to zero, where $gamma$ is the unit circle and $f$ is holomorphic on the unit disk and continuous on its closure.
I think I have to use Cauchy's Theorem, but can't understand why the integrand in this case is holomorphic and why, for example, the integrand in Cauchy's Formula (i.e. $frac{f(w)}{w-z}$) being very similar, is not.
Thank you all for any clues you can give me
complex-analysis cauchy-integral-formula
edited Jan 30 at 16:07
mrtaurho
6,09271641
asked Jan 30 at 16:04
xan32xan32
33
$endgroup$
I need some help regarding why the contour integral $$int_{gamma} frac{f(w)}{w-frac{1}{bar{z}}} mathrm dw$$ is equal to zero, where $gamma$ is the unit circle and $f$ is holomorphic on the unit disk and continuous on its closure.
I think I have to use Cauchy's Theorem, but can't understand why the integrand in this case is holomorphic and why, for example, the integrand in Cauchy's Formula (i.e. $frac{f(w)}{w-z}$) being very similar, is not.
Thank you all for any clues you can give me
complex-analysis cauchy-integral-formula
complex-analysis cauchy-integral-formula
edited Jan 30 at 16:07
mrtaurho
6,09271641
asked Jan 30 at 16:04
xan32xan32
33
edited Jan 30 at 16:07
mrtaurho
6,09271641
asked Jan 30 at 16:04
xan32xan32
33
edited Jan 30 at 16:07
mrtaurho
6,09271641
edited Jan 30 at 16:07
mrtaurho
6,09271641
edited Jan 30 at 16:07
mrtaurho
6,09271641
6,09271641
asked Jan 30 at 16:04
xan32xan32
33
asked Jan 30 at 16:04
xan32xan32
33
asked Jan 30 at 16:04
xan32xan32
33
33
$begingroup$
Where is $z$ located relative to $gamma$?
$endgroup$
– md2perpe
Jan 30 at 16:58
$begingroup$
z in on the open unit disk, so that $frac{1}{bar{z}}$ is outside the unit circle $gamma$. The thing is that I don't know how's that related to the function being holomorphic, since $w$ is in the unit circle in both of the cases I mentioned in my question
$endgroup$
– xan32
Jan 31 at 0:03
add a comment |
$begingroup$
Where is $z$ located relative to $gamma$?
$endgroup$
– md2perpe
Jan 30 at 16:58
$begingroup$
z in on the open unit disk, so that $frac{1}{bar{z}}$ is outside the unit circle $gamma$. The thing is that I don't know how's that related to the function being holomorphic, since $w$ is in the unit circle in both of the cases I mentioned in my question
$endgroup$
– xan32
Jan 31 at 0:03
$begingroup$
Where is $z$ located relative to $gamma$?
$endgroup$
– md2perpe
Jan 30 at 16:58
$begingroup$
Where is $z$ located relative to $gamma$?
$endgroup$
– md2perpe
Jan 30 at 16:58
$begingroup$
z in on the open unit disk, so that $frac{1}{bar{z}}$ is outside the unit circle $gamma$. The thing is that I don't know how's that related to the function being holomorphic, since $w$ is in the unit circle in both of the cases I mentioned in my question
$endgroup$
– xan32
Jan 31 at 0:03
$begingroup$
z in on the open unit disk, so that $frac{1}{bar{z}}$ is outside the unit circle $gamma$. The thing is that I don't know how's that related to the function being holomorphic, since $w$ is in the unit circle in both of the cases I mentioned in my question
$endgroup$
– xan32
Jan 31 at 0:03
add a comment |
1 Answer
1
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$begingroup$
When $zeta$ is inside $gamma$ then $1/(w-zeta)$ is not holomorphic inside $gamma$ since it has a pole at $w=zeta$. When $zeta$ is outside $gamma$ then $1/(w-zeta)$ has no pole inside $gamma$ and is therefore holomorphic there. Thus, if $z$ is inside $gamma$ then $1/(w-z)$ is not holomorphic, but $1/(w-1/bar z)$ is. Therefore the integral in your question vanishes.
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When $zeta$ is inside $gamma$ then $1/(w-zeta)$ is not holomorphic inside $gamma$ since it has a pole at $w=zeta$. When $zeta$ is outside $gamma$ then $1/(w-zeta)$ has no pole inside $gamma$ and is therefore holomorphic there. Thus, if $z$ is inside $gamma$ then $1/(w-z)$ is not holomorphic, but $1/(w-1/bar z)$ is. Therefore the integral in your question vanishes.
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
$endgroup$
add a comment |
$begingroup$
When $zeta$ is inside $gamma$ then $1/(w-zeta)$ is not holomorphic inside $gamma$ since it has a pole at $w=zeta$. When $zeta$ is outside $gamma$ then $1/(w-zeta)$ has no pole inside $gamma$ and is therefore holomorphic there. Thus, if $z$ is inside $gamma$ then $1/(w-z)$ is not holomorphic, but $1/(w-1/bar z)$ is. Therefore the integral in your question vanishes.
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
$endgroup$
add a comment |
$begingroup$
When $zeta$ is inside $gamma$ then $1/(w-zeta)$ is not holomorphic inside $gamma$ since it has a pole at $w=zeta$. When $zeta$ is outside $gamma$ then $1/(w-zeta)$ has no pole inside $gamma$ and is therefore holomorphic there. Thus, if $z$ is inside $gamma$ then $1/(w-z)$ is not holomorphic, but $1/(w-1/bar z)$ is. Therefore the integral in your question vanishes.
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
$endgroup$
When $zeta$ is inside $gamma$ then $1/(w-zeta)$ is not holomorphic inside $gamma$ since it has a pole at $w=zeta$. When $zeta$ is outside $gamma$ then $1/(w-zeta)$ has no pole inside $gamma$ and is therefore holomorphic there. Thus, if $z$ is inside $gamma$ then $1/(w-z)$ is not holomorphic, but $1/(w-1/bar z)$ is. Therefore the integral in your question vanishes.
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
answered Jan 31 at 7:04
md2perpemd2perpe
8,32411028
8,32411028
add a comment |
add a comment |
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$begingroup$
Where is $z$ located relative to $gamma$?
$endgroup$
– md2perpe
Jan 30 at 16:58
$begingroup$
z in on the open unit disk, so that $frac{1}{bar{z}}$ is outside the unit circle $gamma$. The thing is that I don't know how's that related to the function being holomorphic, since $w$ is in the unit circle in both of the cases I mentioned in my question
$endgroup$
– xan32
Jan 31 at 0:03