If $f$ is analytic and $f(z)^2$ = $bar f(z)$ then $f$ is constant












5














I'm currently stuck on the following problem.



Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.



I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.






real-analysis complex-analysis







share|cite|improve this question


















  • 3




    Maybe use $f^3 = f bar f$ is a pure real analytic function?
    – JonathanZ
    Jan 4 at 2:29






  • 1




    The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
    – Kavi Rama Murthy
    Jan 4 at 5:40
















5














I'm currently stuck on the following problem.



Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.



I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.






real-analysis complex-analysis







share|cite|improve this question


















  • 3




    Maybe use $f^3 = f bar f$ is a pure real analytic function?
    – JonathanZ
    Jan 4 at 2:29






  • 1




    The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
    – Kavi Rama Murthy
    Jan 4 at 5:40














5












5








5


2





I'm currently stuck on the following problem.



Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.



I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.






real-analysis complex-analysis







share|cite|improve this question













I'm currently stuck on the following problem.



Let f be an analytic function on a non-empty connected open set V. If $f(z)^2$=$bar f(z)$ $forall zin V$ then f is constant on V.



I think I should be working with the Maximum Modulus theorem, but I am not sure how to use it.






real-analysis complex-analysis




real-analysis complex-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 2:24









user628226user628226

525




525








  • 3




    Maybe use $f^3 = f bar f$ is a pure real analytic function?
    – JonathanZ
    Jan 4 at 2:29






  • 1




    The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
    – Kavi Rama Murthy
    Jan 4 at 5:40














  • 3




    Maybe use $f^3 = f bar f$ is a pure real analytic function?
    – JonathanZ
    Jan 4 at 2:29






  • 1




    The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
    – Kavi Rama Murthy
    Jan 4 at 5:40








3




3




Maybe use $f^3 = f bar f$ is a pure real analytic function?
– JonathanZ
Jan 4 at 2:29




Maybe use $f^3 = f bar f$ is a pure real analytic function?
– JonathanZ
Jan 4 at 2:29




1




1




The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
– Kavi Rama Murthy
Jan 4 at 5:40




The hypothesis implies that $overline {f} (z)$ is analytic and C_R equations tell you that all partial derivatives are $0$.
– Kavi Rama Murthy
Jan 4 at 5:40










2 Answers
2






active

oldest

votes


















5














No need. First solve
$$
Z^2=bar{Z}qquad (1)
$$
Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
$$
j=e^{frac{2ipi}{3}}
$$
then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.






share|cite|improve this answer































    2














    $$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
    so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      No need. First solve
      $$
      Z^2=bar{Z}qquad (1)
      $$
      Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
      $$
      j=e^{frac{2ipi}{3}}
      $$
      then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.






      share|cite|improve this answer




























        5














        No need. First solve
        $$
        Z^2=bar{Z}qquad (1)
        $$
        Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
        $$
        j=e^{frac{2ipi}{3}}
        $$
        then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.






        share|cite|improve this answer


























          5












          5








          5






          No need. First solve
          $$
          Z^2=bar{Z}qquad (1)
          $$
          Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
          $$
          j=e^{frac{2ipi}{3}}
          $$
          then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.






          share|cite|improve this answer














          No need. First solve
          $$
          Z^2=bar{Z}qquad (1)
          $$
          Eq.(1) implies $Z^3=|Z|^2$ which has $S={1,j,j^2,0}$ as set of solutions, with
          $$
          j=e^{frac{2ipi}{3}}
          $$
          then even a continuous function $f:Vto S$ cannot "jump" (due to the fact that $V$ is connected) and, as $S$ is finite, $f$ must be constant. Hope this helps.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered Jan 4 at 2:33









          Duchamp Gérard H. E.Duchamp Gérard H. E.

          2,552918




          2,552918























              2














              $$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
              so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.






              share|cite|improve this answer


























                2














                $$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
                so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  $$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
                  so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.






                  share|cite|improve this answer












                  $$|f(z)|^2 = |f(z)^2|=|overline{f}(z)|=|f(z)| $$
                  so $|f(z)|$ is either $0$ or $1$, constantly, since $|f(z)|$ is continuous. In the former case $f(z)equiv 0$, in the latter the range of $f$ is a subset of $S^1$. There are just three points on $S^1$ such that $xi^2=overline{xi}$, so our $f$ is constantly $0$, $1$, $omega$ or $omega^2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 2:51









                  Jack D'AurizioJack D'Aurizio

                  287k33280658




                  287k33280658






























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