Proof that, if $f$ and $|f|$ are analytic, then $f$ is constant












1












$begingroup$


I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12
















1












$begingroup$


I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12














1












1








1





$begingroup$


I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$




I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions




complex-analysis analytic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 19:36









Did

248k23224462




248k23224462










asked Dec 2 '18 at 16:14









user601175user601175

114




114








  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12














  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12








1




1




$begingroup$
Analytic where? Connectivity has to enter into this somehow.
$endgroup$
– zhw.
Dec 2 '18 at 19:12




$begingroup$
Analytic where? Connectivity has to enter into this somehow.
$endgroup$
– zhw.
Dec 2 '18 at 19:12










3 Answers
3






active

oldest

votes


















0












$begingroup$

$|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



$implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



$U_x=V_yimplies uu_x+vv_x=0$



$U_y=-V_ximplies uu_y+vv_y=0$



From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



$implies uu_x-vu_y=0$



$implies uu_y+vu_x=0$



$implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



For non-trivial solution, $u^2+v^2neq0$



$implies u_x=u_y=v_x=v_y=0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
    $endgroup$
    – user601175
    Dec 2 '18 at 17:19










  • $begingroup$
    I edited the answer for the complete solution.
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 17:59



















3












$begingroup$

If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you know that g is constant. Are all analytic functions with modulus <1 constant?
    $endgroup$
    – user601175
    Dec 2 '18 at 17:08






  • 1




    $begingroup$
    Any bounded entire function is constant (Liouville's Theorem)
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:10



















0












$begingroup$

Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






share|cite|improve this answer









$endgroup$













    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%2f3022828%2fproof-that-if-f-and-f-are-analytic-then-f-is-constant%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59
















    0












    $begingroup$

    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59














    0












    0








    0





    $begingroup$

    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$






    share|cite|improve this answer











    $endgroup$



    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 2 '18 at 18:06

























    answered Dec 2 '18 at 16:35









    Shubham JohriShubham Johri

    5,192717




    5,192717












    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59


















    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59
















    $begingroup$
    I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
    $endgroup$
    – user601175
    Dec 2 '18 at 17:19




    $begingroup$
    I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
    $endgroup$
    – user601175
    Dec 2 '18 at 17:19












    $begingroup$
    I edited the answer for the complete solution.
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 17:59




    $begingroup$
    I edited the answer for the complete solution.
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 17:59











    3












    $begingroup$

    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10
















    3












    $begingroup$

    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10














    3












    3








    3





    $begingroup$

    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






    share|cite|improve this answer









    $endgroup$



    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 16:35









    Tito EliatronTito Eliatron

    1,601622




    1,601622












    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10


















    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10
















    $begingroup$
    How do you know that g is constant. Are all analytic functions with modulus <1 constant?
    $endgroup$
    – user601175
    Dec 2 '18 at 17:08




    $begingroup$
    How do you know that g is constant. Are all analytic functions with modulus <1 constant?
    $endgroup$
    – user601175
    Dec 2 '18 at 17:08




    1




    1




    $begingroup$
    Any bounded entire function is constant (Liouville's Theorem)
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:10




    $begingroup$
    Any bounded entire function is constant (Liouville's Theorem)
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:10











    0












    $begingroup$

    Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






        share|cite|improve this answer









        $endgroup$



        Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 19:33









        zhw.zhw.

        73.3k43175




        73.3k43175






























            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%2f3022828%2fproof-that-if-f-and-f-are-analytic-then-f-is-constant%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

            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

            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