Computation of contour integral without winding numbers.












0














I am trying to solve problem $3$ on page 108 in Ahlfors. The problem asks to compute $int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}$ so I am trying to do so without the use of winding numbers, which isn't introduced until later sections.



I first used partial fraction decomposition to write $frac{1}{z^2 - 1}$ as $frac{1}{2} cdot (frac{1}{z-1} - frac{1}{z+1})$. From here, it's clear that the integral is $0$ if we resort to winding numbers. Since I was unable to compute the integral using elementary methods, I resorted to reading this solution. However, I don't understand how they changed the path of integration from $leftvert z rightvert = 2$ to $leftvert z - 1 rightvert = 1$ because when $z = -2$, $leftvert z - 1 rightvert = leftvert -3 rightvert = 3 ne 1$.






complex-analysis







share|cite|improve this question






















  • I don't have the text. Can you use the Cauchy theorem?
    – Sean Roberson
    Dec 1 at 1:02










  • @SeanRoberson Cauchy's integral formula and Cauchy's theorem haven't been introduced at this point. All we know is what a line integral is and that the integral of $frac{1}{z-a}dz$ taken over a circle centered at $a$ is $2pi i$
    – Yunus Syed
    Dec 1 at 1:04






  • 1




    So you aren't allowed to use the fact that $oint_C 1/z, dz = 2pi i$ for any positively oriented Jordan curve about $0$? If not, why not just evaluate each integral directly, e.g. use substitution $z = 2e^{it}$ and so $oint_{|z|=2} frac{1}{z-1} dz = 2iint_0^{2pi}frac{e^{it}}{2e^{it}-1}, dt = left[ilog(2e^{it}-1)right]_0^{2pi} = 2pi i.$ I don't know how to account for the manipulation used in the solutions with just the information you provided - it looks like the solutions are homework solutions? They are probably using extra information not yet introduced in Ahlfors.
    – Timothy Hedgeworth
    Dec 1 at 2:09












  • Ah I didn't think to use the derivative of log. Thank you!
    – Yunus Syed
    Dec 1 at 2:21
















0














I am trying to solve problem $3$ on page 108 in Ahlfors. The problem asks to compute $int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}$ so I am trying to do so without the use of winding numbers, which isn't introduced until later sections.



I first used partial fraction decomposition to write $frac{1}{z^2 - 1}$ as $frac{1}{2} cdot (frac{1}{z-1} - frac{1}{z+1})$. From here, it's clear that the integral is $0$ if we resort to winding numbers. Since I was unable to compute the integral using elementary methods, I resorted to reading this solution. However, I don't understand how they changed the path of integration from $leftvert z rightvert = 2$ to $leftvert z - 1 rightvert = 1$ because when $z = -2$, $leftvert z - 1 rightvert = leftvert -3 rightvert = 3 ne 1$.






complex-analysis







share|cite|improve this question






















  • I don't have the text. Can you use the Cauchy theorem?
    – Sean Roberson
    Dec 1 at 1:02










  • @SeanRoberson Cauchy's integral formula and Cauchy's theorem haven't been introduced at this point. All we know is what a line integral is and that the integral of $frac{1}{z-a}dz$ taken over a circle centered at $a$ is $2pi i$
    – Yunus Syed
    Dec 1 at 1:04






  • 1




    So you aren't allowed to use the fact that $oint_C 1/z, dz = 2pi i$ for any positively oriented Jordan curve about $0$? If not, why not just evaluate each integral directly, e.g. use substitution $z = 2e^{it}$ and so $oint_{|z|=2} frac{1}{z-1} dz = 2iint_0^{2pi}frac{e^{it}}{2e^{it}-1}, dt = left[ilog(2e^{it}-1)right]_0^{2pi} = 2pi i.$ I don't know how to account for the manipulation used in the solutions with just the information you provided - it looks like the solutions are homework solutions? They are probably using extra information not yet introduced in Ahlfors.
    – Timothy Hedgeworth
    Dec 1 at 2:09












  • Ah I didn't think to use the derivative of log. Thank you!
    – Yunus Syed
    Dec 1 at 2:21














0












0








0







I am trying to solve problem $3$ on page 108 in Ahlfors. The problem asks to compute $int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}$ so I am trying to do so without the use of winding numbers, which isn't introduced until later sections.



I first used partial fraction decomposition to write $frac{1}{z^2 - 1}$ as $frac{1}{2} cdot (frac{1}{z-1} - frac{1}{z+1})$. From here, it's clear that the integral is $0$ if we resort to winding numbers. Since I was unable to compute the integral using elementary methods, I resorted to reading this solution. However, I don't understand how they changed the path of integration from $leftvert z rightvert = 2$ to $leftvert z - 1 rightvert = 1$ because when $z = -2$, $leftvert z - 1 rightvert = leftvert -3 rightvert = 3 ne 1$.






complex-analysis







share|cite|improve this question













I am trying to solve problem $3$ on page 108 in Ahlfors. The problem asks to compute $int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}$ so I am trying to do so without the use of winding numbers, which isn't introduced until later sections.



I first used partial fraction decomposition to write $frac{1}{z^2 - 1}$ as $frac{1}{2} cdot (frac{1}{z-1} - frac{1}{z+1})$. From here, it's clear that the integral is $0$ if we resort to winding numbers. Since I was unable to compute the integral using elementary methods, I resorted to reading this solution. However, I don't understand how they changed the path of integration from $leftvert z rightvert = 2$ to $leftvert z - 1 rightvert = 1$ because when $z = -2$, $leftvert z - 1 rightvert = leftvert -3 rightvert = 3 ne 1$.






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 Dec 1 at 1:00









Yunus Syed

1,117218




1,117218












  • I don't have the text. Can you use the Cauchy theorem?
    – Sean Roberson
    Dec 1 at 1:02










  • @SeanRoberson Cauchy's integral formula and Cauchy's theorem haven't been introduced at this point. All we know is what a line integral is and that the integral of $frac{1}{z-a}dz$ taken over a circle centered at $a$ is $2pi i$
    – Yunus Syed
    Dec 1 at 1:04






  • 1




    So you aren't allowed to use the fact that $oint_C 1/z, dz = 2pi i$ for any positively oriented Jordan curve about $0$? If not, why not just evaluate each integral directly, e.g. use substitution $z = 2e^{it}$ and so $oint_{|z|=2} frac{1}{z-1} dz = 2iint_0^{2pi}frac{e^{it}}{2e^{it}-1}, dt = left[ilog(2e^{it}-1)right]_0^{2pi} = 2pi i.$ I don't know how to account for the manipulation used in the solutions with just the information you provided - it looks like the solutions are homework solutions? They are probably using extra information not yet introduced in Ahlfors.
    – Timothy Hedgeworth
    Dec 1 at 2:09












  • Ah I didn't think to use the derivative of log. Thank you!
    – Yunus Syed
    Dec 1 at 2:21


















  • I don't have the text. Can you use the Cauchy theorem?
    – Sean Roberson
    Dec 1 at 1:02










  • @SeanRoberson Cauchy's integral formula and Cauchy's theorem haven't been introduced at this point. All we know is what a line integral is and that the integral of $frac{1}{z-a}dz$ taken over a circle centered at $a$ is $2pi i$
    – Yunus Syed
    Dec 1 at 1:04






  • 1




    So you aren't allowed to use the fact that $oint_C 1/z, dz = 2pi i$ for any positively oriented Jordan curve about $0$? If not, why not just evaluate each integral directly, e.g. use substitution $z = 2e^{it}$ and so $oint_{|z|=2} frac{1}{z-1} dz = 2iint_0^{2pi}frac{e^{it}}{2e^{it}-1}, dt = left[ilog(2e^{it}-1)right]_0^{2pi} = 2pi i.$ I don't know how to account for the manipulation used in the solutions with just the information you provided - it looks like the solutions are homework solutions? They are probably using extra information not yet introduced in Ahlfors.
    – Timothy Hedgeworth
    Dec 1 at 2:09












  • Ah I didn't think to use the derivative of log. Thank you!
    – Yunus Syed
    Dec 1 at 2:21
















I don't have the text. Can you use the Cauchy theorem?
– Sean Roberson
Dec 1 at 1:02




I don't have the text. Can you use the Cauchy theorem?
– Sean Roberson
Dec 1 at 1:02












@SeanRoberson Cauchy's integral formula and Cauchy's theorem haven't been introduced at this point. All we know is what a line integral is and that the integral of $frac{1}{z-a}dz$ taken over a circle centered at $a$ is $2pi i$
– Yunus Syed
Dec 1 at 1:04




@SeanRoberson Cauchy's integral formula and Cauchy's theorem haven't been introduced at this point. All we know is what a line integral is and that the integral of $frac{1}{z-a}dz$ taken over a circle centered at $a$ is $2pi i$
– Yunus Syed
Dec 1 at 1:04




1




1




So you aren't allowed to use the fact that $oint_C 1/z, dz = 2pi i$ for any positively oriented Jordan curve about $0$? If not, why not just evaluate each integral directly, e.g. use substitution $z = 2e^{it}$ and so $oint_{|z|=2} frac{1}{z-1} dz = 2iint_0^{2pi}frac{e^{it}}{2e^{it}-1}, dt = left[ilog(2e^{it}-1)right]_0^{2pi} = 2pi i.$ I don't know how to account for the manipulation used in the solutions with just the information you provided - it looks like the solutions are homework solutions? They are probably using extra information not yet introduced in Ahlfors.
– Timothy Hedgeworth
Dec 1 at 2:09






So you aren't allowed to use the fact that $oint_C 1/z, dz = 2pi i$ for any positively oriented Jordan curve about $0$? If not, why not just evaluate each integral directly, e.g. use substitution $z = 2e^{it}$ and so $oint_{|z|=2} frac{1}{z-1} dz = 2iint_0^{2pi}frac{e^{it}}{2e^{it}-1}, dt = left[ilog(2e^{it}-1)right]_0^{2pi} = 2pi i.$ I don't know how to account for the manipulation used in the solutions with just the information you provided - it looks like the solutions are homework solutions? They are probably using extra information not yet introduced in Ahlfors.
– Timothy Hedgeworth
Dec 1 at 2:09














Ah I didn't think to use the derivative of log. Thank you!
– Yunus Syed
Dec 1 at 2:21




Ah I didn't think to use the derivative of log. Thank you!
– Yunus Syed
Dec 1 at 2:21










2 Answers
2






active

oldest

votes


















2














You have
begin{align}
int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}&=frac12int_0^{2pi}left(frac{2ie^{it} }{2e^{it}-1}-frac{2ie^{it} }{2e^{it}+1}right),dt\ \
&=frac12left.left(log(2e^{it} -1)-log(2e^{it}+1)right)right|_0^{2pi}\ \
&=0-0=0
end{align}






share|cite|improve this answer





























    2














    This is a fairly general argument you will probably find useful.



    Note that the function $frac{1}{z^2-1}$ is analytic everywhere except the points $1$ and $-1$. This means that we can move the contour so long as the singular points ($1$ and $-1$) remain inside the domain. We can deform the contour into an infinity-looking domain, where the touch point is the origin, and the center of the two circles located in the infinity are $1$ and $-1$. Because those are essentially two different circles, the integral around the contour is equal to the sum of the integrals around the two circles. These two circle's new contours are $|z-1| = 1$ and $|z+1| = 1$ (note they still intersect at the origin). Now, we can use the formula you know regarding $2 pi i$, which would then yield the answer of $0$.






    share|cite|improve this answer





















    • And this leads to the homotopic version of the Cauchy theorem.
      – Sean Roberson
      Dec 1 at 4:20











    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%2f3020874%2fcomputation-of-contour-integral-without-winding-numbers%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









    2














    You have
    begin{align}
    int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}&=frac12int_0^{2pi}left(frac{2ie^{it} }{2e^{it}-1}-frac{2ie^{it} }{2e^{it}+1}right),dt\ \
    &=frac12left.left(log(2e^{it} -1)-log(2e^{it}+1)right)right|_0^{2pi}\ \
    &=0-0=0
    end{align}






    share|cite|improve this answer


























      2














      You have
      begin{align}
      int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}&=frac12int_0^{2pi}left(frac{2ie^{it} }{2e^{it}-1}-frac{2ie^{it} }{2e^{it}+1}right),dt\ \
      &=frac12left.left(log(2e^{it} -1)-log(2e^{it}+1)right)right|_0^{2pi}\ \
      &=0-0=0
      end{align}






      share|cite|improve this answer
























        2












        2








        2






        You have
        begin{align}
        int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}&=frac12int_0^{2pi}left(frac{2ie^{it} }{2e^{it}-1}-frac{2ie^{it} }{2e^{it}+1}right),dt\ \
        &=frac12left.left(log(2e^{it} -1)-log(2e^{it}+1)right)right|_0^{2pi}\ \
        &=0-0=0
        end{align}






        share|cite|improve this answer












        You have
        begin{align}
        int_{leftvert z rightvert = 2} frac{dz}{z^2 - 1}&=frac12int_0^{2pi}left(frac{2ie^{it} }{2e^{it}-1}-frac{2ie^{it} }{2e^{it}+1}right),dt\ \
        &=frac12left.left(log(2e^{it} -1)-log(2e^{it}+1)right)right|_0^{2pi}\ \
        &=0-0=0
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 at 1:52









        Martin Argerami

        123k1176174




        123k1176174























            2














            This is a fairly general argument you will probably find useful.



            Note that the function $frac{1}{z^2-1}$ is analytic everywhere except the points $1$ and $-1$. This means that we can move the contour so long as the singular points ($1$ and $-1$) remain inside the domain. We can deform the contour into an infinity-looking domain, where the touch point is the origin, and the center of the two circles located in the infinity are $1$ and $-1$. Because those are essentially two different circles, the integral around the contour is equal to the sum of the integrals around the two circles. These two circle's new contours are $|z-1| = 1$ and $|z+1| = 1$ (note they still intersect at the origin). Now, we can use the formula you know regarding $2 pi i$, which would then yield the answer of $0$.






            share|cite|improve this answer





















            • And this leads to the homotopic version of the Cauchy theorem.
              – Sean Roberson
              Dec 1 at 4:20
















            2














            This is a fairly general argument you will probably find useful.



            Note that the function $frac{1}{z^2-1}$ is analytic everywhere except the points $1$ and $-1$. This means that we can move the contour so long as the singular points ($1$ and $-1$) remain inside the domain. We can deform the contour into an infinity-looking domain, where the touch point is the origin, and the center of the two circles located in the infinity are $1$ and $-1$. Because those are essentially two different circles, the integral around the contour is equal to the sum of the integrals around the two circles. These two circle's new contours are $|z-1| = 1$ and $|z+1| = 1$ (note they still intersect at the origin). Now, we can use the formula you know regarding $2 pi i$, which would then yield the answer of $0$.






            share|cite|improve this answer





















            • And this leads to the homotopic version of the Cauchy theorem.
              – Sean Roberson
              Dec 1 at 4:20














            2












            2








            2






            This is a fairly general argument you will probably find useful.



            Note that the function $frac{1}{z^2-1}$ is analytic everywhere except the points $1$ and $-1$. This means that we can move the contour so long as the singular points ($1$ and $-1$) remain inside the domain. We can deform the contour into an infinity-looking domain, where the touch point is the origin, and the center of the two circles located in the infinity are $1$ and $-1$. Because those are essentially two different circles, the integral around the contour is equal to the sum of the integrals around the two circles. These two circle's new contours are $|z-1| = 1$ and $|z+1| = 1$ (note they still intersect at the origin). Now, we can use the formula you know regarding $2 pi i$, which would then yield the answer of $0$.






            share|cite|improve this answer












            This is a fairly general argument you will probably find useful.



            Note that the function $frac{1}{z^2-1}$ is analytic everywhere except the points $1$ and $-1$. This means that we can move the contour so long as the singular points ($1$ and $-1$) remain inside the domain. We can deform the contour into an infinity-looking domain, where the touch point is the origin, and the center of the two circles located in the infinity are $1$ and $-1$. Because those are essentially two different circles, the integral around the contour is equal to the sum of the integrals around the two circles. These two circle's new contours are $|z-1| = 1$ and $|z+1| = 1$ (note they still intersect at the origin). Now, we can use the formula you know regarding $2 pi i$, which would then yield the answer of $0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 2:31









            Neeyanth Kopparapu

            1665




            1665












            • And this leads to the homotopic version of the Cauchy theorem.
              – Sean Roberson
              Dec 1 at 4:20


















            • And this leads to the homotopic version of the Cauchy theorem.
              – Sean Roberson
              Dec 1 at 4:20
















            And this leads to the homotopic version of the Cauchy theorem.
            – Sean Roberson
            Dec 1 at 4:20




            And this leads to the homotopic version of the Cauchy theorem.
            – Sean Roberson
            Dec 1 at 4:20


















            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%2f3020874%2fcomputation-of-contour-integral-without-winding-numbers%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

            Berounka

            Image
            Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
            Read more

            Sphinx de Gizeh

            Image
            Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
            Read more

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

            Image
            3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
            Read more