Computing a potential with a complex integration












2














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?










share|cite|improve this question
























  • 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...
    – Fabian
    Jan 4 at 11:01










  • isn't it done ? Isn't what you're saying the last line in Latex?
    – Marine Galantin
    Jan 4 at 11:07












  • Fabian was suggesting to express $f$ as a function of $z$, not $F$.
    – N74
    Jan 4 at 14:47










  • I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
    – Marine Galantin
    Jan 4 at 23:21










  • 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 $f(z)$.
    – N74
    Jan 5 at 18:51
















2














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?










share|cite|improve this question
























  • 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...
    – Fabian
    Jan 4 at 11:01










  • isn't it done ? Isn't what you're saying the last line in Latex?
    – Marine Galantin
    Jan 4 at 11:07












  • Fabian was suggesting to express $f$ as a function of $z$, not $F$.
    – N74
    Jan 4 at 14:47










  • I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
    – Marine Galantin
    Jan 4 at 23:21










  • 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 $f(z)$.
    – N74
    Jan 5 at 18:51














2












2








2







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?










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:21









Larry

1,9712823




1,9712823










asked Jan 4 at 10:54









Marine GalantinMarine Galantin

795215




795215












  • 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...
    – Fabian
    Jan 4 at 11:01










  • isn't it done ? Isn't what you're saying the last line in Latex?
    – Marine Galantin
    Jan 4 at 11:07












  • Fabian was suggesting to express $f$ as a function of $z$, not $F$.
    – N74
    Jan 4 at 14:47










  • I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
    – Marine Galantin
    Jan 4 at 23:21










  • 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 $f(z)$.
    – N74
    Jan 5 at 18:51


















  • 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...
    – Fabian
    Jan 4 at 11:01










  • isn't it done ? Isn't what you're saying the last line in Latex?
    – Marine Galantin
    Jan 4 at 11:07












  • Fabian was suggesting to express $f$ as a function of $z$, not $F$.
    – N74
    Jan 4 at 14:47










  • I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
    – Marine Galantin
    Jan 4 at 23:21










  • 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 $f(z)$.
    – N74
    Jan 5 at 18:51
















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...
– Fabian
Jan 4 at 11:01




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...
– Fabian
Jan 4 at 11:01












isn't it done ? Isn't what you're saying the last line in Latex?
– Marine Galantin
Jan 4 at 11:07






isn't it done ? Isn't what you're saying the last line in Latex?
– Marine Galantin
Jan 4 at 11:07














Fabian was suggesting to express $f$ as a function of $z$, not $F$.
– N74
Jan 4 at 14:47




Fabian was suggesting to express $f$ as a function of $z$, not $F$.
– N74
Jan 4 at 14:47












I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
– Marine Galantin
Jan 4 at 23:21




I'm not sure what that means @N74, Am I supposed to create artificially x + iy ?
– Marine Galantin
Jan 4 at 23:21












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 $f(z)$.
– N74
Jan 5 at 18:51




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 $f(z)$.
– N74
Jan 5 at 18:51










0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061519%2fcomputing-a-potential-with-a-complex-integration%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061519%2fcomputing-a-potential-with-a-complex-integration%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

1300-talet

1300-talet

Display a custom attribute below product name in the front-end Magento 1.9.3.8