A tricky contour integral.
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Let $f$ be holomorphic on a domain $D$ containing $overline{D}(0,1)$, and $gamma$ be the positively oriented circular contour centred at 0 of radius 1. Prove that
$$
frac{1}{2 pi i} int_{gamma} frac{overline{f(z)}}{(z-z_0)} dz=
begin{cases} overline{f(0)} &mbox{if} ,, |z_0|<1, \
overline{f(0)}-overline{f(1/ overline{z_0})} &mbox{if},, |z_0|>1.
end{cases}
$$
In the previous part of the question I was asked to prove that for a continuous function $phi:A={zinmathbb{C}:|z|=1}rightarrowmathbb{C}$ and $gamma$ as above,
$$overline{int_{gamma}phi(z),, dz} = -int_{gamma} overline{phi(z)},,frac{dz}{z^2}$$
which I've already done. So it's reasonable to think that the above would help solve the integral (perhaps along with Cauchy's Integral Formula) for the different cases of $z_0$ but I don't know how to get to a point where I can use it.
I've seen some suggestions of a variable change from $z$ to $1/z$ as $overline{f(bar{z})}$ is in fact holomorphic and so continuous on A since for $|z|=1 ,,bar{z}=1/z$ by the identity $bar{z}z=|z|^2$ but I don't know how to implement the change of variables on a contour integral because we haven't been taught how to do so yet which leads me to think that maybe it isn't what they want us to do.
Any guidance would be appreciated. Thank you in advance.
complex-analysis contour-integration
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up vote
1
down vote
favorite
Let $f$ be holomorphic on a domain $D$ containing $overline{D}(0,1)$, and $gamma$ be the positively oriented circular contour centred at 0 of radius 1. Prove that
$$
frac{1}{2 pi i} int_{gamma} frac{overline{f(z)}}{(z-z_0)} dz=
begin{cases} overline{f(0)} &mbox{if} ,, |z_0|<1, \
overline{f(0)}-overline{f(1/ overline{z_0})} &mbox{if},, |z_0|>1.
end{cases}
$$
In the previous part of the question I was asked to prove that for a continuous function $phi:A={zinmathbb{C}:|z|=1}rightarrowmathbb{C}$ and $gamma$ as above,
$$overline{int_{gamma}phi(z),, dz} = -int_{gamma} overline{phi(z)},,frac{dz}{z^2}$$
which I've already done. So it's reasonable to think that the above would help solve the integral (perhaps along with Cauchy's Integral Formula) for the different cases of $z_0$ but I don't know how to get to a point where I can use it.
I've seen some suggestions of a variable change from $z$ to $1/z$ as $overline{f(bar{z})}$ is in fact holomorphic and so continuous on A since for $|z|=1 ,,bar{z}=1/z$ by the identity $bar{z}z=|z|^2$ but I don't know how to implement the change of variables on a contour integral because we haven't been taught how to do so yet which leads me to think that maybe it isn't what they want us to do.
Any guidance would be appreciated. Thank you in advance.
complex-analysis contour-integration
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f$ be holomorphic on a domain $D$ containing $overline{D}(0,1)$, and $gamma$ be the positively oriented circular contour centred at 0 of radius 1. Prove that
$$
frac{1}{2 pi i} int_{gamma} frac{overline{f(z)}}{(z-z_0)} dz=
begin{cases} overline{f(0)} &mbox{if} ,, |z_0|<1, \
overline{f(0)}-overline{f(1/ overline{z_0})} &mbox{if},, |z_0|>1.
end{cases}
$$
In the previous part of the question I was asked to prove that for a continuous function $phi:A={zinmathbb{C}:|z|=1}rightarrowmathbb{C}$ and $gamma$ as above,
$$overline{int_{gamma}phi(z),, dz} = -int_{gamma} overline{phi(z)},,frac{dz}{z^2}$$
which I've already done. So it's reasonable to think that the above would help solve the integral (perhaps along with Cauchy's Integral Formula) for the different cases of $z_0$ but I don't know how to get to a point where I can use it.
I've seen some suggestions of a variable change from $z$ to $1/z$ as $overline{f(bar{z})}$ is in fact holomorphic and so continuous on A since for $|z|=1 ,,bar{z}=1/z$ by the identity $bar{z}z=|z|^2$ but I don't know how to implement the change of variables on a contour integral because we haven't been taught how to do so yet which leads me to think that maybe it isn't what they want us to do.
Any guidance would be appreciated. Thank you in advance.
complex-analysis contour-integration
Let $f$ be holomorphic on a domain $D$ containing $overline{D}(0,1)$, and $gamma$ be the positively oriented circular contour centred at 0 of radius 1. Prove that
$$
frac{1}{2 pi i} int_{gamma} frac{overline{f(z)}}{(z-z_0)} dz=
begin{cases} overline{f(0)} &mbox{if} ,, |z_0|<1, \
overline{f(0)}-overline{f(1/ overline{z_0})} &mbox{if},, |z_0|>1.
end{cases}
$$
In the previous part of the question I was asked to prove that for a continuous function $phi:A={zinmathbb{C}:|z|=1}rightarrowmathbb{C}$ and $gamma$ as above,
$$overline{int_{gamma}phi(z),, dz} = -int_{gamma} overline{phi(z)},,frac{dz}{z^2}$$
which I've already done. So it's reasonable to think that the above would help solve the integral (perhaps along with Cauchy's Integral Formula) for the different cases of $z_0$ but I don't know how to get to a point where I can use it.
I've seen some suggestions of a variable change from $z$ to $1/z$ as $overline{f(bar{z})}$ is in fact holomorphic and so continuous on A since for $|z|=1 ,,bar{z}=1/z$ by the identity $bar{z}z=|z|^2$ but I don't know how to implement the change of variables on a contour integral because we haven't been taught how to do so yet which leads me to think that maybe it isn't what they want us to do.
Any guidance would be appreciated. Thank you in advance.
complex-analysis contour-integration
complex-analysis contour-integration
edited Nov 18 at 12:04
asked Nov 18 at 10:07
MathWolf
569
569
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1 Answer
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2
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One of the quickest way to show this is to use power series. If $|zeta|<1$, we may write
$frac{1}{z-zeta} = sum_{j=0}^infty frac{zeta^j}{z^{j+1}}$, and if $|zeta|>1$, then we have $frac{1}{z-zeta} = -sum_{j=0}^infty frac{z^j}{zeta^{j+1}}.$ Note that $overline{z} = z^{-1}$ on the unit circle and
$$frac{1}{2pi i}int_gamma z^k dz = 1_{{k=-1}}.$$
Let $f(z) = sum_{i=0}^infty a_i z^i$. In case $|zeta|<1$,
$$
frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i-j-1}zeta^jright) dz = overline{a_0} = overline{f(0)}.
$$
Otherwise,
$$
-frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i+j}zeta^{-j-1}right) dz = -sum_{j=0}^infty overline{a_{j+1}}zeta^{-j-1} = -overline{f(frac{1}{overline{zeta}})} + f(0),
$$as desired.
We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time.
– MathWolf
Nov 18 at 14:05
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
One of the quickest way to show this is to use power series. If $|zeta|<1$, we may write
$frac{1}{z-zeta} = sum_{j=0}^infty frac{zeta^j}{z^{j+1}}$, and if $|zeta|>1$, then we have $frac{1}{z-zeta} = -sum_{j=0}^infty frac{z^j}{zeta^{j+1}}.$ Note that $overline{z} = z^{-1}$ on the unit circle and
$$frac{1}{2pi i}int_gamma z^k dz = 1_{{k=-1}}.$$
Let $f(z) = sum_{i=0}^infty a_i z^i$. In case $|zeta|<1$,
$$
frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i-j-1}zeta^jright) dz = overline{a_0} = overline{f(0)}.
$$
Otherwise,
$$
-frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i+j}zeta^{-j-1}right) dz = -sum_{j=0}^infty overline{a_{j+1}}zeta^{-j-1} = -overline{f(frac{1}{overline{zeta}})} + f(0),
$$as desired.
We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time.
– MathWolf
Nov 18 at 14:05
add a comment |
up vote
2
down vote
One of the quickest way to show this is to use power series. If $|zeta|<1$, we may write
$frac{1}{z-zeta} = sum_{j=0}^infty frac{zeta^j}{z^{j+1}}$, and if $|zeta|>1$, then we have $frac{1}{z-zeta} = -sum_{j=0}^infty frac{z^j}{zeta^{j+1}}.$ Note that $overline{z} = z^{-1}$ on the unit circle and
$$frac{1}{2pi i}int_gamma z^k dz = 1_{{k=-1}}.$$
Let $f(z) = sum_{i=0}^infty a_i z^i$. In case $|zeta|<1$,
$$
frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i-j-1}zeta^jright) dz = overline{a_0} = overline{f(0)}.
$$
Otherwise,
$$
-frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i+j}zeta^{-j-1}right) dz = -sum_{j=0}^infty overline{a_{j+1}}zeta^{-j-1} = -overline{f(frac{1}{overline{zeta}})} + f(0),
$$as desired.
We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time.
– MathWolf
Nov 18 at 14:05
add a comment |
up vote
2
down vote
up vote
2
down vote
One of the quickest way to show this is to use power series. If $|zeta|<1$, we may write
$frac{1}{z-zeta} = sum_{j=0}^infty frac{zeta^j}{z^{j+1}}$, and if $|zeta|>1$, then we have $frac{1}{z-zeta} = -sum_{j=0}^infty frac{z^j}{zeta^{j+1}}.$ Note that $overline{z} = z^{-1}$ on the unit circle and
$$frac{1}{2pi i}int_gamma z^k dz = 1_{{k=-1}}.$$
Let $f(z) = sum_{i=0}^infty a_i z^i$. In case $|zeta|<1$,
$$
frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i-j-1}zeta^jright) dz = overline{a_0} = overline{f(0)}.
$$
Otherwise,
$$
-frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i+j}zeta^{-j-1}right) dz = -sum_{j=0}^infty overline{a_{j+1}}zeta^{-j-1} = -overline{f(frac{1}{overline{zeta}})} + f(0),
$$as desired.
One of the quickest way to show this is to use power series. If $|zeta|<1$, we may write
$frac{1}{z-zeta} = sum_{j=0}^infty frac{zeta^j}{z^{j+1}}$, and if $|zeta|>1$, then we have $frac{1}{z-zeta} = -sum_{j=0}^infty frac{z^j}{zeta^{j+1}}.$ Note that $overline{z} = z^{-1}$ on the unit circle and
$$frac{1}{2pi i}int_gamma z^k dz = 1_{{k=-1}}.$$
Let $f(z) = sum_{i=0}^infty a_i z^i$. In case $|zeta|<1$,
$$
frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i-j-1}zeta^jright) dz = overline{a_0} = overline{f(0)}.
$$
Otherwise,
$$
-frac{1}{2pi i} int_gamma left(sum_{i,jgeq 0} overline{a_i}z^{-i+j}zeta^{-j-1}right) dz = -sum_{j=0}^infty overline{a_{j+1}}zeta^{-j-1} = -overline{f(frac{1}{overline{zeta}})} + f(0),
$$as desired.
answered Nov 18 at 12:10
Song
84011
84011
We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time.
– MathWolf
Nov 18 at 14:05
add a comment |
We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time.
– MathWolf
Nov 18 at 14:05
We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time.
– MathWolf
Nov 18 at 14:05
We've only just started doing power series in complex analysis so this is a little too advanced for me at this point. But thank you for taking the time.
– MathWolf
Nov 18 at 14:05
add a comment |
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