complex conjugate line integral
$begingroup$
i have $gamma(t)=e^{it}$, $tin[0,2pi]$ and the line integral $overline{int_{gamma} f(z)>dz}$. It is to show that
$overline{int_{gamma} f(z)dz}$=$-int_{gamma} frac{overline{f(z)}}{z^2} dz$.
If i start with $overline{int_{gamma} f(z)dz}=int_{overline{gamma}} overline{f(overline{z}} )dz=int_0^{2pi} overline{f(gamma(t)}*overline{gamma'}(t)=int_0^{2pi}overline{f(gamma(t))}*ie^{-it}$ i do not come to a solution and also do not see how i can change something on the last term.
complex-analysis
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
asked Dec 9 '18 at 13:59
tim123tim123
173
$endgroup$
add a comment |
$begingroup$
i have $gamma(t)=e^{it}$, $tin[0,2pi]$ and the line integral $overline{int_{gamma} f(z)>dz}$. It is to show that
$overline{int_{gamma} f(z)dz}$=$-int_{gamma} frac{overline{f(z)}}{z^2} dz$.
If i start with $overline{int_{gamma} f(z)dz}=int_{overline{gamma}} overline{f(overline{z}} )dz=int_0^{2pi} overline{f(gamma(t)}*overline{gamma'}(t)=int_0^{2pi}overline{f(gamma(t))}*ie^{-it}$ i do not come to a solution and also do not see how i can change something on the last term.
complex-analysis
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
asked Dec 9 '18 at 13:59
tim123tim123
173
$endgroup$
add a comment |
$begingroup$
i have $gamma(t)=e^{it}$, $tin[0,2pi]$ and the line integral $overline{int_{gamma} f(z)>dz}$. It is to show that
$overline{int_{gamma} f(z)dz}$=$-int_{gamma} frac{overline{f(z)}}{z^2} dz$.
If i start with $overline{int_{gamma} f(z)dz}=int_{overline{gamma}} overline{f(overline{z}} )dz=int_0^{2pi} overline{f(gamma(t)}*overline{gamma'}(t)=int_0^{2pi}overline{f(gamma(t))}*ie^{-it}$ i do not come to a solution and also do not see how i can change something on the last term.
complex-analysis
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
asked Dec 9 '18 at 13:59
tim123tim123
173
$endgroup$
i have $gamma(t)=e^{it}$, $tin[0,2pi]$ and the line integral $overline{int_{gamma} f(z)>dz}$. It is to show that
$overline{int_{gamma} f(z)dz}$=$-int_{gamma} frac{overline{f(z)}}{z^2} dz$.
If i start with $overline{int_{gamma} f(z)dz}=int_{overline{gamma}} overline{f(overline{z}} )dz=int_0^{2pi} overline{f(gamma(t)}*overline{gamma'}(t)=int_0^{2pi}overline{f(gamma(t))}*ie^{-it}$ i do not come to a solution and also do not see how i can change something on the last term.
complex-analysis
complex-analysis
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
asked Dec 9 '18 at 13:59
tim123tim123
173
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
asked Dec 9 '18 at 13:59
tim123tim123
173
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
edited Dec 9 '18 at 16:18
Christian Blatter
173k7113326
173k7113326
asked Dec 9 '18 at 13:59
tim123tim123
173
asked Dec 9 '18 at 13:59
tim123tim123
173
asked Dec 9 '18 at 13:59
tim123tim123
173
173
add a comment |
add a comment |
2 Answers
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$begingroup$
You are almost there. What you did wrong is $overline{gamma'(t)} = ie^{-it}$. It should be
$overline{gamma'(t)} = overline{ie^{it}}= -ie^{-it}$. Finally, from your last line,
$$
-int_0^{2pi}overline{f(gamma(t))}ie^{-it}dt = -int_0^{2pi}overline{f(gamma(t))}e^{-2it}cdot ie^{it}dt = -int_gammaoverline{f(z)}frac{1}{z^2}dz,
$$as desired.
answered Dec 9 '18 at 14:15
SongSong
12.9k631
$endgroup$
add a comment |
$begingroup$
Notice that the path us the unit circle. So the path, and it's complex conjugate are the same path in terms if their elements.
Also note that points on the unit circle have modulus=1. For such Points, their reciprocal is also their conjugate.
$bar{z}=1/z. d(bar{z})=frac{-dz}{z^2.}$
I forget what it is, but there's a relationship between f applied at z , f applied at $bar{z}$, the conjugate if f at z and the conjugate if f applied at the conjugate if z.
I think that combo should get you what you need.
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You are almost there. What you did wrong is $overline{gamma'(t)} = ie^{-it}$. It should be
$overline{gamma'(t)} = overline{ie^{it}}= -ie^{-it}$. Finally, from your last line,
$$
-int_0^{2pi}overline{f(gamma(t))}ie^{-it}dt = -int_0^{2pi}overline{f(gamma(t))}e^{-2it}cdot ie^{it}dt = -int_gammaoverline{f(z)}frac{1}{z^2}dz,
$$as desired.
answered Dec 9 '18 at 14:15
SongSong
12.9k631
$endgroup$
add a comment |
$begingroup$
You are almost there. What you did wrong is $overline{gamma'(t)} = ie^{-it}$. It should be
$overline{gamma'(t)} = overline{ie^{it}}= -ie^{-it}$. Finally, from your last line,
$$
-int_0^{2pi}overline{f(gamma(t))}ie^{-it}dt = -int_0^{2pi}overline{f(gamma(t))}e^{-2it}cdot ie^{it}dt = -int_gammaoverline{f(z)}frac{1}{z^2}dz,
$$as desired.
answered Dec 9 '18 at 14:15
SongSong
12.9k631
$endgroup$
add a comment |
$begingroup$
You are almost there. What you did wrong is $overline{gamma'(t)} = ie^{-it}$. It should be
$overline{gamma'(t)} = overline{ie^{it}}= -ie^{-it}$. Finally, from your last line,
$$
-int_0^{2pi}overline{f(gamma(t))}ie^{-it}dt = -int_0^{2pi}overline{f(gamma(t))}e^{-2it}cdot ie^{it}dt = -int_gammaoverline{f(z)}frac{1}{z^2}dz,
$$as desired.
answered Dec 9 '18 at 14:15
SongSong
12.9k631
$endgroup$
You are almost there. What you did wrong is $overline{gamma'(t)} = ie^{-it}$. It should be
$overline{gamma'(t)} = overline{ie^{it}}= -ie^{-it}$. Finally, from your last line,
$$
-int_0^{2pi}overline{f(gamma(t))}ie^{-it}dt = -int_0^{2pi}overline{f(gamma(t))}e^{-2it}cdot ie^{it}dt = -int_gammaoverline{f(z)}frac{1}{z^2}dz,
$$as desired.
answered Dec 9 '18 at 14:15
SongSong
12.9k631
answered Dec 9 '18 at 14:15
SongSong
12.9k631
answered Dec 9 '18 at 14:15
SongSong
12.9k631
answered Dec 9 '18 at 14:15
SongSong
12.9k631
12.9k631
add a comment |
add a comment |
$begingroup$
Notice that the path us the unit circle. So the path, and it's complex conjugate are the same path in terms if their elements.
Also note that points on the unit circle have modulus=1. For such Points, their reciprocal is also their conjugate.
$bar{z}=1/z. d(bar{z})=frac{-dz}{z^2.}$
I forget what it is, but there's a relationship between f applied at z , f applied at $bar{z}$, the conjugate if f at z and the conjugate if f applied at the conjugate if z.
I think that combo should get you what you need.
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
$endgroup$
add a comment |
$begingroup$
Notice that the path us the unit circle. So the path, and it's complex conjugate are the same path in terms if their elements.
Also note that points on the unit circle have modulus=1. For such Points, their reciprocal is also their conjugate.
$bar{z}=1/z. d(bar{z})=frac{-dz}{z^2.}$
I forget what it is, but there's a relationship between f applied at z , f applied at $bar{z}$, the conjugate if f at z and the conjugate if f applied at the conjugate if z.
I think that combo should get you what you need.
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
$endgroup$
add a comment |
$begingroup$
Notice that the path us the unit circle. So the path, and it's complex conjugate are the same path in terms if their elements.
Also note that points on the unit circle have modulus=1. For such Points, their reciprocal is also their conjugate.
$bar{z}=1/z. d(bar{z})=frac{-dz}{z^2.}$
I forget what it is, but there's a relationship between f applied at z , f applied at $bar{z}$, the conjugate if f at z and the conjugate if f applied at the conjugate if z.
I think that combo should get you what you need.
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
$endgroup$
Notice that the path us the unit circle. So the path, and it's complex conjugate are the same path in terms if their elements.
Also note that points on the unit circle have modulus=1. For such Points, their reciprocal is also their conjugate.
$bar{z}=1/z. d(bar{z})=frac{-dz}{z^2.}$
I forget what it is, but there's a relationship between f applied at z , f applied at $bar{z}$, the conjugate if f at z and the conjugate if f applied at the conjugate if z.
I think that combo should get you what you need.
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
answered Dec 9 '18 at 16:46
TurlocTheRedTurlocTheRed
856311
856311
add a comment |
add a comment |
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