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






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      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

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          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





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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            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






























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