Computing a potential with a complex integration
$begingroup$
I'm trying to understand the link between $ mathbb R^2 $ and $ mathbb C $.
I know how to compute potentials when I know the gradient of it :
$$ f(x,y) = left( frac {-x}{(x^2+y^2)^2 } , frac {-y}{(x^2+y^2)^2 } right) $$
gives
$$ F(x,y) = frac {1}{2(x^2+y^2) } $$
My question is,
How can you find the same result with complex integration ?
My attempt :
using this formula (and my calculator I admit)
$$int f(z) dz = int u dx - v dy + i int v dx + u dy $$
$$ int int frac {-x}{(x^2+y^2)^2 } +i frac {-y}{(x^2+y^2)^2 } dx dy
= frac { arctan (frac y x) } { 2x} - i frac { 2y ln(y) + ln( frac 1 { y^2 }) y - 2 arctan (frac x y ) x } { 4 x y }$$
which looks like something not that afar from my result, but I'm not convinced.
So is it possible and how do you do this?
real-analysis complex-analysis complex-integration
edited Jan 4 at 11:21
Larry
2,41831129
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to understand the link between $ mathbb R^2 $ and $ mathbb C $.
I know how to compute potentials when I know the gradient of it :
$$ f(x,y) = left( frac {-x}{(x^2+y^2)^2 } , frac {-y}{(x^2+y^2)^2 } right) $$
gives
$$ F(x,y) = frac {1}{2(x^2+y^2) } $$
My question is,
How can you find the same result with complex integration ?
My attempt :
using this formula (and my calculator I admit)
$$int f(z) dz = int u dx - v dy + i int v dx + u dy $$
$$ int int frac {-x}{(x^2+y^2)^2 } +i frac {-y}{(x^2+y^2)^2 } dx dy
= frac { arctan (frac y x) } { 2x} - i frac { 2y ln(y) + ln( frac 1 { y^2 }) y - 2 arctan (frac x y ) x } { 4 x y }$$
which looks like something not that afar from my result, but I'm not convinced.
So is it possible and how do you do this?
real-analysis complex-analysis complex-integration
edited Jan 4 at 11:21
Larry
2,41831129
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
$endgroup$
$begingroup$
You should start by finding a complex function $f(z=x+i y)$ such that the first component of $f(x,y)$ is the real part and the second component is the imaginary part...
$endgroup$
– Fabian
Jan 4 at 11:01
$begingroup$
isn't it done ? Isn't what you're saying the last line in Latex?
$endgroup$
– Marine Galantin
Jan 4 at 11:07
$begingroup$
Fabian was suggesting to express $f$ as a function of $z$, not $F$.
$endgroup$
– N74
Jan 4 at 14:47
$begingroup$
I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
$endgroup$
– Marine Galantin
Jan 4 at 23:21
$begingroup$
I don't understand what do you mean by "artificially", but letting $z=x+iy$ can be a good substitution to find an expression for $f(z)$.
$endgroup$
– N74
Jan 5 at 18:51
|
show 1 more comment
$begingroup$
I'm trying to understand the link between $ mathbb R^2 $ and $ mathbb C $.
I know how to compute potentials when I know the gradient of it :
$$ f(x,y) = left( frac {-x}{(x^2+y^2)^2 } , frac {-y}{(x^2+y^2)^2 } right) $$
gives
$$ F(x,y) = frac {1}{2(x^2+y^2) } $$
My question is,
How can you find the same result with complex integration ?
My attempt :
using this formula (and my calculator I admit)
$$int f(z) dz = int u dx - v dy + i int v dx + u dy $$
$$ int int frac {-x}{(x^2+y^2)^2 } +i frac {-y}{(x^2+y^2)^2 } dx dy
= frac { arctan (frac y x) } { 2x} - i frac { 2y ln(y) + ln( frac 1 { y^2 }) y - 2 arctan (frac x y ) x } { 4 x y }$$
which looks like something not that afar from my result, but I'm not convinced.
So is it possible and how do you do this?
real-analysis complex-analysis complex-integration
edited Jan 4 at 11:21
Larry
2,41831129
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
$endgroup$
I'm trying to understand the link between $ mathbb R^2 $ and $ mathbb C $.
I know how to compute potentials when I know the gradient of it :
$$ f(x,y) = left( frac {-x}{(x^2+y^2)^2 } , frac {-y}{(x^2+y^2)^2 } right) $$
gives
$$ F(x,y) = frac {1}{2(x^2+y^2) } $$
My question is,
How can you find the same result with complex integration ?
My attempt :
using this formula (and my calculator I admit)
$$int f(z) dz = int u dx - v dy + i int v dx + u dy $$
$$ int int frac {-x}{(x^2+y^2)^2 } +i frac {-y}{(x^2+y^2)^2 } dx dy
= frac { arctan (frac y x) } { 2x} - i frac { 2y ln(y) + ln( frac 1 { y^2 }) y - 2 arctan (frac x y ) x } { 4 x y }$$
which looks like something not that afar from my result, but I'm not convinced.
So is it possible and how do you do this?
real-analysis complex-analysis complex-integration
real-analysis complex-analysis complex-integration
edited Jan 4 at 11:21
Larry
2,41831129
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
edited Jan 4 at 11:21
Larry
2,41831129
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
edited Jan 4 at 11:21
Larry
2,41831129
edited Jan 4 at 11:21
Larry
2,41831129
edited Jan 4 at 11:21
Larry
2,41831129
2,41831129
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
asked Jan 4 at 10:54
Marine GalantinMarine Galantin
860316
860316
$begingroup$
You should start by finding a complex function $f(z=x+i y)$ such that the first component of $f(x,y)$ is the real part and the second component is the imaginary part...
$endgroup$
– Fabian
Jan 4 at 11:01
$begingroup$
isn't it done ? Isn't what you're saying the last line in Latex?
$endgroup$
– Marine Galantin
Jan 4 at 11:07
$begingroup$
Fabian was suggesting to express $f$ as a function of $z$, not $F$.
$endgroup$
– N74
Jan 4 at 14:47
$begingroup$
I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
$endgroup$
– Marine Galantin
Jan 4 at 23:21
$begingroup$
I don't understand what do you mean by "artificially", but letting $z=x+iy$ can be a good substitution to find an expression for $f(z)$.
$endgroup$
– N74
Jan 5 at 18:51
|
show 1 more comment
$begingroup$
You should start by finding a complex function $f(z=x+i y)$ such that the first component of $f(x,y)$ is the real part and the second component is the imaginary part...
$endgroup$
– Fabian
Jan 4 at 11:01
$begingroup$
isn't it done ? Isn't what you're saying the last line in Latex?
$endgroup$
– Marine Galantin
Jan 4 at 11:07
$begingroup$
Fabian was suggesting to express $f$ as a function of $z$, not $F$.
$endgroup$
– N74
Jan 4 at 14:47
$begingroup$
I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
$endgroup$
– Marine Galantin
Jan 4 at 23:21
$begingroup$
I don't understand what do you mean by "artificially", but letting $z=x+iy$ can be a good substitution to find an expression for $f(z)$.
$endgroup$
– N74
Jan 5 at 18:51
$begingroup$
You should start by finding a complex function $f(z=x+i y)$ such that the first component of $f(x,y)$ is the real part and the second component is the imaginary part...
$endgroup$
– Fabian
Jan 4 at 11:01
$begingroup$
You should start by finding a complex function $f(z=x+i y)$ such that the first component of $f(x,y)$ is the real part and the second component is the imaginary part...
$endgroup$
– Fabian
Jan 4 at 11:01
$begingroup$
isn't it done ? Isn't what you're saying the last line in Latex?
$endgroup$
– Marine Galantin
Jan 4 at 11:07
$begingroup$
isn't it done ? Isn't what you're saying the last line in Latex?
$endgroup$
– Marine Galantin
Jan 4 at 11:07
$begingroup$
Fabian was suggesting to express $f$ as a function of $z$, not $F$.
$endgroup$
– N74
Jan 4 at 14:47
$begingroup$
Fabian was suggesting to express $f$ as a function of $z$, not $F$.
$endgroup$
– N74
Jan 4 at 14:47
$begingroup$
I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
$endgroup$
– Marine Galantin
Jan 4 at 23:21
$begingroup$
I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
$endgroup$
– Marine Galantin
Jan 4 at 23:21
$begingroup$
I don't understand what do you mean by "artificially", but letting $z=x+iy$ can be a good substitution to find an expression for $f(z)$.
$endgroup$
– N74
Jan 5 at 18:51
$begingroup$
I don't understand what do you mean by "artificially", but letting $z=x+iy$ can be a good substitution to find an expression for $f(z)$.
$endgroup$
– N74
Jan 5 at 18:51
|
show 1 more comment
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$begingroup$
You should start by finding a complex function $f(z=x+i y)$ such that the first component of $f(x,y)$ is the real part and the second component is the imaginary part...
$endgroup$
– Fabian
Jan 4 at 11:01
$begingroup$
isn't it done ? Isn't what you're saying the last line in Latex?
$endgroup$
– Marine Galantin
Jan 4 at 11:07
$begingroup$
Fabian was suggesting to express $f$ as a function of $z$, not $F$.
$endgroup$
– N74
Jan 4 at 14:47
$begingroup$
I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
$endgroup$
– Marine Galantin
Jan 4 at 23:21
$begingroup$
I don't understand what do you mean by "artificially", but letting $z=x+iy$ can be a good substitution to find an expression for $f(z)$.
$endgroup$
– N74
Jan 5 at 18:51