If $f$ is holomorphic on a domain $D$, and satisfies $aoperatorname{Re}(f(z))+boperatorname{Im}(f(z))=c$ for...












3












$begingroup$


The proposition that I am required to prove is the following




Let $D subset mathbb{C}$ be a domain and suppose $f(x, y)=u(x,
y)+iv(x, y)$ is holomorphic on $D$. Let $a, b, cin mathbb{R}$ such
that $a^2+b^2>0$ and $$au(x, y)+bv(x, y)=c quad forall z=(x, y)in D$$



Prove $f$ is constant




I will write $u(x, y)$ as $u$ in short, and similarly for $v(x, y)$.
My attempt is as follows:
Since f is holomorphic we have:
$$dfrac{partial u}{partial x}=dfrac{partial v}{partial y}qquad dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}$$
If $a=0neq b$, then $bv=c implies v=Kinmathbb{R}$



$$implies dfrac{partial u}{partial x}=dfrac{partial v}{partial y}=dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}=0$$
$$implies f'(z)=0 quad forall zin D$$
Similarly for $aneq 0=b$
Now the remaining case to check is if $aneq 0neq b$



Taking partial derivatives both sides of the given relation we get:



$$adfrac{partial u}{partial x}+bdfrac{partial v}{partial x}=0$$
$$implies dfrac{partial u}{partial x}=dfrac{b}{a}bigg(-dfrac{partial v}{partial x}bigg)=dfrac{b}{a}dfrac{partial u}{partial y}$$
Similarly taking partial derivatives w.r.t $y$ we get:
$$dfrac{partial u}{partial x}=-dfrac{a}{b}dfrac{partial u}{partial y}$$
$$implies dfrac{a}{b}=dfrac{-b}{a}$$
$$implies a^2+b^2=0$$
This is a contradiction, so it is not possible that $aneq 0neq b$. Therefore for all possible values of $a$ and $b$, the function $f$ is constant.






complex-analysis proof-verification holomorphic-functions







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    The proposition that I am required to prove is the following




    Let $D subset mathbb{C}$ be a domain and suppose $f(x, y)=u(x,
    y)+iv(x, y)$     is holomorphic on $D$. Let $a, b, cin mathbb{R}$ such
    that $a^2+b^2>0$ and $$au(x, y)+bv(x, y)=c quad forall z=(x, y)in D$$



    Prove $f$ is constant




    I will write $u(x, y)$ as $u$ in short, and similarly for $v(x, y)$.
    My attempt is as follows:
    Since f is holomorphic we have:
    $$dfrac{partial u}{partial x}=dfrac{partial v}{partial y}qquad dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}$$
    If $a=0neq b$, then $bv=c implies v=Kinmathbb{R}$



    $$implies dfrac{partial u}{partial x}=dfrac{partial v}{partial y}=dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}=0$$
    $$implies f'(z)=0 quad forall zin D$$
    Similarly for $aneq 0=b$
    Now the remaining case to check is if $aneq 0neq b$



    Taking partial derivatives both sides of the given relation we get:



    $$adfrac{partial u}{partial x}+bdfrac{partial v}{partial x}=0$$
    $$implies dfrac{partial u}{partial x}=dfrac{b}{a}bigg(-dfrac{partial v}{partial x}bigg)=dfrac{b}{a}dfrac{partial u}{partial y}$$
    Similarly taking partial derivatives w.r.t $y$ we get:
    $$dfrac{partial u}{partial x}=-dfrac{a}{b}dfrac{partial u}{partial y}$$
    $$implies dfrac{a}{b}=dfrac{-b}{a}$$
    $$implies a^2+b^2=0$$
    This is a contradiction, so it is not possible that $aneq 0neq b$. Therefore for all possible values of $a$ and $b$, the function $f$ is constant.






    complex-analysis proof-verification holomorphic-functions







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      The proposition that I am required to prove is the following




      Let $D subset mathbb{C}$ be a domain and suppose $f(x, y)=u(x,
      y)+iv(x, y)$       is holomorphic on $D$. Let $a, b, cin mathbb{R}$ such
      that $a^2+b^2>0$ and $$au(x, y)+bv(x, y)=c quad forall z=(x, y)in D$$



      Prove $f$ is constant




      I will write $u(x, y)$ as $u$ in short, and similarly for $v(x, y)$.
      My attempt is as follows:
      Since f is holomorphic we have:
      $$dfrac{partial u}{partial x}=dfrac{partial v}{partial y}qquad dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}$$
      If $a=0neq b$, then $bv=c implies v=Kinmathbb{R}$



      $$implies dfrac{partial u}{partial x}=dfrac{partial v}{partial y}=dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}=0$$
      $$implies f'(z)=0 quad forall zin D$$
      Similarly for $aneq 0=b$
      Now the remaining case to check is if $aneq 0neq b$



      Taking partial derivatives both sides of the given relation we get:



      $$adfrac{partial u}{partial x}+bdfrac{partial v}{partial x}=0$$
      $$implies dfrac{partial u}{partial x}=dfrac{b}{a}bigg(-dfrac{partial v}{partial x}bigg)=dfrac{b}{a}dfrac{partial u}{partial y}$$
      Similarly taking partial derivatives w.r.t $y$ we get:
      $$dfrac{partial u}{partial x}=-dfrac{a}{b}dfrac{partial u}{partial y}$$
      $$implies dfrac{a}{b}=dfrac{-b}{a}$$
      $$implies a^2+b^2=0$$
      This is a contradiction, so it is not possible that $aneq 0neq b$. Therefore for all possible values of $a$ and $b$, the function $f$ is constant.






      complex-analysis proof-verification holomorphic-functions







      share|cite|improve this question











      $endgroup$




      The proposition that I am required to prove is the following




      Let $D subset mathbb{C}$ be a domain and suppose $f(x, y)=u(x,
      y)+iv(x, y)$       is holomorphic on $D$. Let $a, b, cin mathbb{R}$ such
      that $a^2+b^2>0$ and $$au(x, y)+bv(x, y)=c quad forall z=(x, y)in D$$



      Prove $f$ is constant




      I will write $u(x, y)$ as $u$ in short, and similarly for $v(x, y)$.
      My attempt is as follows:
      Since f is holomorphic we have:
      $$dfrac{partial u}{partial x}=dfrac{partial v}{partial y}qquad dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}$$
      If $a=0neq b$, then $bv=c implies v=Kinmathbb{R}$



      $$implies dfrac{partial u}{partial x}=dfrac{partial v}{partial y}=dfrac{partial u}{partial y}=-dfrac{partial v}{partial x}=0$$
      $$implies f'(z)=0 quad forall zin D$$
      Similarly for $aneq 0=b$
      Now the remaining case to check is if $aneq 0neq b$



      Taking partial derivatives both sides of the given relation we get:



      $$adfrac{partial u}{partial x}+bdfrac{partial v}{partial x}=0$$
      $$implies dfrac{partial u}{partial x}=dfrac{b}{a}bigg(-dfrac{partial v}{partial x}bigg)=dfrac{b}{a}dfrac{partial u}{partial y}$$
      Similarly taking partial derivatives w.r.t $y$ we get:
      $$dfrac{partial u}{partial x}=-dfrac{a}{b}dfrac{partial u}{partial y}$$
      $$implies dfrac{a}{b}=dfrac{-b}{a}$$
      $$implies a^2+b^2=0$$
      This is a contradiction, so it is not possible that $aneq 0neq b$. Therefore for all possible values of $a$ and $b$, the function $f$ is constant.






      complex-analysis proof-verification holomorphic-functions




      complex-analysis proof-verification holomorphic-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 9:32









      Asaf Karagila♦

      303k32429760




      303k32429760










      asked Dec 2 '18 at 8:33









      Dylan ZammitDylan Zammit

      8751416




      8751416






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Yes, it is correct.



          Note that, after having proved it for the case in which $a=0$ or $b=0$, you could use that to prove it in the general case, since$$au+bv=operatorname{Re}bigl((a-bi)(u+vibigr).$$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Interesting observation, Thanks!
            $endgroup$
            – Dylan Zammit
            Dec 2 '18 at 8:41



















          1












          $begingroup$

          In the case $a ne 0 ne b$ you obtained the identities
          $$
          frac{partial u}{partial x} = frac ba frac{partial u}{partial y} \
          frac{partial u}{partial x} = -frac ab frac{partial u}{partial y}
          $$
          At that point you should conclude that
          $ frac{a}{b}=frac{-b}{a}$ or $frac{partial u}{partial x} = frac{partial v}{partial x} = 0$. The first case leads to a contradiction, therefore $f'(z)=0$.




          You don't have to consider the cases $a=0$ or $b=0$ separately if you proceed
          as follows: Differentiating with respect to $x$ gives



          $$
          afrac{partial u}{partial x}+bfrac{partial v}{partial x}=0
          $$



          Differentiating with respect to $y$ and applying the Cauchy-Riemann
          equations gives
          $$
          bfrac{partial u}{partial x}-afrac{partial v}{partial x}=0
          $$



          That is a linear equation system, and its coefficient determinant is $-a^2-b^2 ne 0$. It follows that it only has the trivial solution
          $$
          frac{partial u}{partial x} = frac{partial v}{partial x} = 0
          $$






          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%2f3022402%2fif-f-is-holomorphic-on-a-domain-d-and-satisfies-a-operatornamerefzb%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









            4












            $begingroup$

            Yes, it is correct.



            Note that, after having proved it for the case in which $a=0$ or $b=0$, you could use that to prove it in the general case, since$$au+bv=operatorname{Re}bigl((a-bi)(u+vibigr).$$






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Interesting observation, Thanks!
              $endgroup$
              – Dylan Zammit
              Dec 2 '18 at 8:41
















            4












            $begingroup$

            Yes, it is correct.



            Note that, after having proved it for the case in which $a=0$ or $b=0$, you could use that to prove it in the general case, since$$au+bv=operatorname{Re}bigl((a-bi)(u+vibigr).$$






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Interesting observation, Thanks!
              $endgroup$
              – Dylan Zammit
              Dec 2 '18 at 8:41














            4












            4








            4





            $begingroup$

            Yes, it is correct.



            Note that, after having proved it for the case in which $a=0$ or $b=0$, you could use that to prove it in the general case, since$$au+bv=operatorname{Re}bigl((a-bi)(u+vibigr).$$






            share|cite|improve this answer











            $endgroup$



            Yes, it is correct.



            Note that, after having proved it for the case in which $a=0$ or $b=0$, you could use that to prove it in the general case, since$$au+bv=operatorname{Re}bigl((a-bi)(u+vibigr).$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 '18 at 14:22

























            answered Dec 2 '18 at 8:39









            José Carlos SantosJosé Carlos Santos

            156k22126227




            156k22126227








            • 2




              $begingroup$
              Interesting observation, Thanks!
              $endgroup$
              – Dylan Zammit
              Dec 2 '18 at 8:41














            • 2




              $begingroup$
              Interesting observation, Thanks!
              $endgroup$
              – Dylan Zammit
              Dec 2 '18 at 8:41








            2




            2




            $begingroup$
            Interesting observation, Thanks!
            $endgroup$
            – Dylan Zammit
            Dec 2 '18 at 8:41




            $begingroup$
            Interesting observation, Thanks!
            $endgroup$
            – Dylan Zammit
            Dec 2 '18 at 8:41











            1












            $begingroup$

            In the case $a ne 0 ne b$ you obtained the identities
            $$
            frac{partial u}{partial x} = frac ba frac{partial u}{partial y} \
            frac{partial u}{partial x} = -frac ab frac{partial u}{partial y}
            $$
            At that point you should conclude that
            $ frac{a}{b}=frac{-b}{a}$ or $frac{partial u}{partial x} = frac{partial v}{partial x} = 0$. The first case leads to a contradiction, therefore $f'(z)=0$.




            You don't have to consider the cases $a=0$ or $b=0$ separately if you proceed
            as follows: Differentiating with respect to $x$ gives



            $$
            afrac{partial u}{partial x}+bfrac{partial v}{partial x}=0
            $$



            Differentiating with respect to $y$ and applying the Cauchy-Riemann
            equations gives
            $$
            bfrac{partial u}{partial x}-afrac{partial v}{partial x}=0
            $$



            That is a linear equation system, and its coefficient determinant is $-a^2-b^2 ne 0$. It follows that it only has the trivial solution
            $$
            frac{partial u}{partial x} = frac{partial v}{partial x} = 0
            $$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              In the case $a ne 0 ne b$ you obtained the identities
              $$
              frac{partial u}{partial x} = frac ba frac{partial u}{partial y} \
              frac{partial u}{partial x} = -frac ab frac{partial u}{partial y}
              $$
              At that point you should conclude that
              $ frac{a}{b}=frac{-b}{a}$ or $frac{partial u}{partial x} = frac{partial v}{partial x} = 0$. The first case leads to a contradiction, therefore $f'(z)=0$.




              You don't have to consider the cases $a=0$ or $b=0$ separately if you proceed
              as follows: Differentiating with respect to $x$ gives



              $$
              afrac{partial u}{partial x}+bfrac{partial v}{partial x}=0
              $$



              Differentiating with respect to $y$ and applying the Cauchy-Riemann
              equations gives
              $$
              bfrac{partial u}{partial x}-afrac{partial v}{partial x}=0
              $$



              That is a linear equation system, and its coefficient determinant is $-a^2-b^2 ne 0$. It follows that it only has the trivial solution
              $$
              frac{partial u}{partial x} = frac{partial v}{partial x} = 0
              $$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                In the case $a ne 0 ne b$ you obtained the identities
                $$
                frac{partial u}{partial x} = frac ba frac{partial u}{partial y} \
                frac{partial u}{partial x} = -frac ab frac{partial u}{partial y}
                $$
                At that point you should conclude that
                $ frac{a}{b}=frac{-b}{a}$ or $frac{partial u}{partial x} = frac{partial v}{partial x} = 0$. The first case leads to a contradiction, therefore $f'(z)=0$.




                You don't have to consider the cases $a=0$ or $b=0$ separately if you proceed
                as follows: Differentiating with respect to $x$ gives



                $$
                afrac{partial u}{partial x}+bfrac{partial v}{partial x}=0
                $$



                Differentiating with respect to $y$ and applying the Cauchy-Riemann
                equations gives
                $$
                bfrac{partial u}{partial x}-afrac{partial v}{partial x}=0
                $$



                That is a linear equation system, and its coefficient determinant is $-a^2-b^2 ne 0$. It follows that it only has the trivial solution
                $$
                frac{partial u}{partial x} = frac{partial v}{partial x} = 0
                $$






                share|cite|improve this answer











                $endgroup$



                In the case $a ne 0 ne b$ you obtained the identities
                $$
                frac{partial u}{partial x} = frac ba frac{partial u}{partial y} \
                frac{partial u}{partial x} = -frac ab frac{partial u}{partial y}
                $$
                At that point you should conclude that
                $ frac{a}{b}=frac{-b}{a}$ or $frac{partial u}{partial x} = frac{partial v}{partial x} = 0$. The first case leads to a contradiction, therefore $f'(z)=0$.




                You don't have to consider the cases $a=0$ or $b=0$ separately if you proceed
                as follows: Differentiating with respect to $x$ gives



                $$
                afrac{partial u}{partial x}+bfrac{partial v}{partial x}=0
                $$



                Differentiating with respect to $y$ and applying the Cauchy-Riemann
                equations gives
                $$
                bfrac{partial u}{partial x}-afrac{partial v}{partial x}=0
                $$



                That is a linear equation system, and its coefficient determinant is $-a^2-b^2 ne 0$. It follows that it only has the trivial solution
                $$
                frac{partial u}{partial x} = frac{partial v}{partial x} = 0
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 2 '18 at 9:00

























                answered Dec 2 '18 at 8:48









                Martin RMartin R

                27.9k33255




                27.9k33255






























                    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%2f3022402%2fif-f-is-holomorphic-on-a-domain-d-and-satisfies-a-operatornamerefzb%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

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                    Read more

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                    Read more

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                    Image
                    1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                    Read more