Integral with differential is purely imaginary












3












$begingroup$


Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.



I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?






complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:27










  • $begingroup$
    @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $begingroup$
    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:34










  • $begingroup$
    @achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:44










  • $begingroup$
    Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:50


















3












$begingroup$


Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.



I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?






complex-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:27










  • $begingroup$
    @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $begingroup$
    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:34










  • $begingroup$
    @achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:44










  • $begingroup$
    Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:50
















3












3








3





$begingroup$


Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.



I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?






complex-analysis







share|cite|improve this question









$endgroup$




Prove that if $f(z)$ is analytic and $f'(z)$ is continuous on a closed curve $gamma$, then $int_gammaoverline{f(z)}f'(z)dz$ is purely imaginary.



I'm not so sure where to start. Maybe parametrize $z$ by $z(t)$, so that the integral becomes $int_a^boverline{f(z(t))}f'(z(t))z'(t)dt$. Why will this be purely imaginary?






complex-analysis




complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 5 '13 at 19:15









Paul S.Paul S.

1,40211540




1,40211540












  • $begingroup$
    Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:27










  • $begingroup$
    @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $begingroup$
    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:34










  • $begingroup$
    @achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:44










  • $begingroup$
    Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:50




















  • $begingroup$
    Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:27










  • $begingroup$
    @achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:30






  • 1




    $begingroup$
    $2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
    $endgroup$
    – achille hui
    Oct 5 '13 at 19:34










  • $begingroup$
    @achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:44










  • $begingroup$
    Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
    $endgroup$
    – Paul S.
    Oct 5 '13 at 19:50


















$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27




$begingroup$
Hint: The question you should ask is why the real part of the integral vanishes. Can you write it down and simplify it?
$endgroup$
– achille hui
Oct 5 '13 at 19:27












$begingroup$
@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30




$begingroup$
@achillehui Write the real part down? I'm not sure how to simplify the integral from where it is.
$endgroup$
– Paul S.
Oct 5 '13 at 19:30




1




1




$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
$endgroup$
– achille hui
Oct 5 '13 at 19:34




$begingroup$
$2^{nd}$ Hint: what is $frac{d}{dt} |f(z(t))|^2$?
$endgroup$
– achille hui
Oct 5 '13 at 19:34












$begingroup$
@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44




$begingroup$
@achillehui It is $overline{f(z(t))}f'(z(t))z'(t)+f(z(t))overline{f'(z(t))}overline{z'(t)}$. This can be written as $2Re[overline{f(z(t))}f'(z(t))z'(t)]$. It's still not clear how this helps, since I want to integrate the term (not just the real part).
$endgroup$
– Paul S.
Oct 5 '13 at 19:44












$begingroup$
Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50






$begingroup$
Well okay, so the integral of the real part is $dfrac12|f(z(t))|^2$. And since $z(a)=z(b)$, this vanishes!
$endgroup$
– Paul S.
Oct 5 '13 at 19:50












1 Answer
1






active

oldest

votes


















0












$begingroup$

How about this solution



Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.



$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$



The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
    $endgroup$
    – Sambo
    Oct 2 '17 at 2:16











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%2f515828%2fintegral-with-differential-is-purely-imaginary%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

How about this solution



Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.



$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$



The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
    $endgroup$
    – Sambo
    Oct 2 '17 at 2:16
















0












$begingroup$

How about this solution



Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.



$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$



The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
    $endgroup$
    – Sambo
    Oct 2 '17 at 2:16














0












0








0





$begingroup$

How about this solution



Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.



$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$



The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.






share|cite|improve this answer











$endgroup$



How about this solution



Let $r(z)=|f(z)|$ and $theta(z)=arg(f(z))$. Start with simplifying.



$$int_gammaoverline{f(z)}f'(z) dz=int_gamma|f(z)|^2frac{f'(z)}{f(z)} dz=int_gamma e^{2ln|f(z)|}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}e^{2ln f(z)}frac{f'(z)}{f(z)} dz=int_gamma e^{-2iarg(f(z))}f(z)f'(z) dz=int_gamma e^{-2itheta(z)}r(z)e^{itheta(z)}[r'(z)e^{itheta(z)}+ir(z)theta'(z)e^{itheta(z)}] dz=int_gamma r(z)r'(z) dz+iint_gamma r^2(z)theta'(z) dz=frac{1}{2}r^2(z)Big|_gamma+iint_gamma r^2(z) d[theta(z)]$$



The term for the real part is clearly 0 as $gamma$ is closed. The integral for the imaginary part is clearly real since both $r(z)$ and $theta(z)$ are real. Thus the original integral's value must be purely imaginary.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 5 '14 at 21:31

























answered May 5 '14 at 19:44









Roman ChoklerRoman Chokler

82258




82258












  • $begingroup$
    Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
    $endgroup$
    – Sambo
    Oct 2 '17 at 2:16


















  • $begingroup$
    Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
    $endgroup$
    – Sambo
    Oct 2 '17 at 2:16
















$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16




$begingroup$
Aren't you taking the log of a complex number in the fourth expression? I thought that wasn't allowed because it has many possible values.
$endgroup$
– Sambo
Oct 2 '17 at 2:16


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f515828%2fintegral-with-differential-is-purely-imaginary%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

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more