Integral of real part of z around the unit circle












2












$begingroup$


question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago
















2












$begingroup$


question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago














2












2








2





$begingroup$


question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions







share|cite|improve this question









$endgroup$




question 1



What is the result of integrating the real part of z (a complex number) anti clockwise around the unit circle?



At first glance, I couldn't identify any points within the circle where analyticity breaks down. So it seemed to me that the integral should vanish and the answer should be 0. Since the real part of a complex number is differentiable everywhere right? It seems smooth.



But I tried using parameterization and got the answer to be i$pi$. So I am guessing there is a residue of 0.5 at 0 maybe?



my attempt



So my question is, can we use Cauchy residue theorem to solve this? If so how? Is there a more elegant solution for this problem?






residue-calculus complex-integration analyticity analytic-functions




residue-calculus complex-integration analyticity analytic-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









AdityaAditya

278314




278314








  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago














  • 2




    $begingroup$
    That’s the right answer, by a correct method.
    $endgroup$
    – Lubin
    1 hour ago








2




2




$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
1 hour ago




$begingroup$
That’s the right answer, by a correct method.
$endgroup$
– Lubin
1 hour ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    1 hour ago



















2












$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    1 hour ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    59 mins ago













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%2f3097060%2fintegral-of-real-part-of-z-around-the-unit-circle%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    1 hour ago
















3












$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    1 hour ago














3












3








3





$begingroup$

The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)






share|cite|improve this answer









$endgroup$



The real part of $z$ is not a complex-differentiable function anywhere. So, what can we do instead in order to apply complex analysis techniques? We can find an analytic function that agrees with it on the circle - even if it's completely different everywhere else.



First, we note that $text{Re}(z) = frac12(z+overline{z})$. Since the conjugate isn't analytic, that isn't good enough yet. But then, for $|z|=1$, $overline{z}=frac1z$. So, on the circle $|z|=1$ we care about,
$$text{Re}(z) = frac12left(z+frac1zright)$$
That's something we can work with. (Of course, it's a bit absurd to apply the residue theorem to the literal integral $frac az$ around a circle - we needed that integral to prove the theorem in the first place)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









jmerryjmerry

6,457718




6,457718












  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    1 hour ago


















  • $begingroup$
    Thanks a lot, just what I wanted.
    $endgroup$
    – Aditya
    1 hour ago
















$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
1 hour ago




$begingroup$
Thanks a lot, just what I wanted.
$endgroup$
– Aditya
1 hour ago











2












$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    1 hour ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    59 mins ago


















2












$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    1 hour ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    59 mins ago
















2












2








2





$begingroup$

${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.






share|cite|improve this answer











$endgroup$



${1 over 2 pi i} int_0^{2 pi} cos t ie^{it} dt = {1 over 2 pi} int_0^{2 pi} {1 over 2 } (e^{it} + e^{-it}) e^{it} dt = {1 over 2}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









copper.hatcopper.hat

127k559160




127k559160












  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    1 hour ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    59 mins ago




















  • $begingroup$
    Not true - $dz$ isn't real.
    $endgroup$
    – jmerry
    1 hour ago










  • $begingroup$
    I don't know what I was thinking.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    That was embarrassing.
    $endgroup$
    – copper.hat
    1 hour ago










  • $begingroup$
    Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
    $endgroup$
    – Aditya
    1 hour ago












  • $begingroup$
    I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
    $endgroup$
    – copper.hat
    59 mins ago


















$begingroup$
Not true - $dz$ isn't real.
$endgroup$
– jmerry
1 hour ago




$begingroup$
Not true - $dz$ isn't real.
$endgroup$
– jmerry
1 hour ago












$begingroup$
I don't know what I was thinking.
$endgroup$
– copper.hat
1 hour ago




$begingroup$
I don't know what I was thinking.
$endgroup$
– copper.hat
1 hour ago












$begingroup$
That was embarrassing.
$endgroup$
– copper.hat
1 hour ago




$begingroup$
That was embarrassing.
$endgroup$
– copper.hat
1 hour ago












$begingroup$
Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
$endgroup$
– Aditya
1 hour ago






$begingroup$
Here $dz = ie^{it}$ right? And we can write $Re(z) = cos t = frac{(e^{it}+ e^{-it})}{2} $ all of this makes sense. So we have $int_0^{2pi} e^{2it} + 1 = 2pi$ because $cos 0 = cos 2pi$. So I don't see why this is wrong just because $dz$ is not real? This is almost the same way I solved it above in the question details, like I did via the trigonometric route, it was much longer. This is super fast. Isn't this correct too?
$endgroup$
– Aditya
1 hour ago














$begingroup$
I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
$endgroup$
– copper.hat
59 mins ago






$begingroup$
I edited out my embarrassing mistake. The answer above is correct. For real measures we have $operatorname{re} (int f d mu) = int operatorname{re}(f) d mu$, but that does not apply here.
$endgroup$
– copper.hat
59 mins ago




















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%2f3097060%2fintegral-of-real-part-of-z-around-the-unit-circle%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

الفوسفات في المغرب

Image
قطار ينقل فوسفات من منطقة خريبكة متجه نحو الدار البيضاء. ديار عمال بمصانع الفوسفات بمدينة آسفي. المحتوى هنا ينقصه الاستشهاد بمصادر. يرجى إيراد مصادر موثوق بها. أي معلومات غير موثقة يمكن التشكيك بها وإزالتها. (أغسطس 2015) الفوسفات في المغرب هو ثالث مصدر عالمي للفوسفات، ويتوفر المغرب على %75 من الاحتياطي العالمي  [بحاجة لمصدر] ويتوزع بنسب متفاوتة بين عدة مناطق أساسية وهي: وادي زم بن جرير بوكراع خريبكة اليوسفية أولاد عبدون كنتور واد الذهب مستقلة. هذه الأخيرة لم تفتح بعد للاستغلال. الصادرات الحامض الفوسفوري P205 تمتص الهند %50 من ا لصادرات الإجمالية آسيا (دون روسيا)%59 أوروبا %30 أمريكا اللاتينية %11 أسمدة صلبة أوروبا 38 % أمريكا اللاتينية %30 آسيا (دون روسيا) 22 % أستراليا 9 % بوابة المغرب هذه بذرة مقالة عن علم الاقتصاد أو موضوع متعلق به بحاجة للتوسيع. شارك في تحريرها. This page is only for reference, If you need detailed information, please check here
Read more

Four equal circles intersect: What is the area of the small shaded portion and its height

Image
3 $begingroup$ In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this. geometry circle share | cite | improve this question edited 2 hours ago Abdulhameed asked 2 hours ago ...
Read more

جامعة ليفربول

Image
جامعة ليفربول University of Liverpool الأسماء السابقة كلية ليفربول الجامعية (1882-1884) جامعة فيكتوريا في ليفربول (1884-1903) معلومات التأسيس 1882 (منذ 137 سنة) الإنتماءات رابطة الجامعات الأوروبية مجموعة راسل جامعات القرميد الأحمر المنحة المالية 143.5 مليون £ النوع جامعة عامة الكليات 6 الموقع الجغرافي إحداثيات 53°24′22″N 2°58′01″W  /  53.406°N 2.967°W  / 53.406; -2.967   المدينة ليفربول المكان   إنجلترا البلد   المملكة المتحدة رقم الهاتف 7942000 151 (44+) إحصاءات الطلاب 17,405 (سنة 2008) عضوية مجموعة رسل [1] أورسيد [2]   طلاب الدراسات العليا 3,860 الموقع www.liverpool.ac.uk تعديل   جامعة ليفربول مبنى فكتوريا جامعة ليفربول (بالإنجليزية: The University of Liverpool )) إحدى الجامعات البريطانية العريقة، تقع في مدينة ليفربول على الساحل الشمالي الغربي من أنجلترا . وهي عضو في مجموعة راسل للجامعات وتعتبر من ضمن جامعات الطوب الأحمر الستة. محتويات ...
Read more