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Can an holomorphic function have a convergent sequence of points in its domain with only real values?


An equation with Gamma Euler function in critical stripsurjective holomorphic function in a special domainSpecifying a holomorphic function by a sequence of valuesComposition of a holomorphic function with a normal family of holomorphic functions.On every simply connected domain, there exists a holomorphic function with no analytic continuation.Can a complex function be holomorphic only on a proper closed subset of $mathbbC$?Sequence of holomorphic function and its derivativeProof the continuity of a function.A sequence of holomorphic functions $f_n$ uniformly convergent on boundary of open set.Holomorphic function in a region symmetric about origin which takes real values on some non-empty interval intersection with the regionSquare Root of a Holomorphic Function













0












$begingroup$


Let $D subset mathbbC$ be a open subset of the complex plane



Let $f: D to mathbbC$ be a holomorphic function



Let $z_n_n in mathbbN subset D$ be a sequence in $D$ such that $z_n to s in D$ and we have $f(z_n) to f(s)=0$



My question is:
Is there a subsequence $z_n_k_k in mathbbN$ such that $forall k in mathbbN: Im(f(z_n_k)) neq 0$



Thanks.










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    Let $D subset mathbbC$ be a open subset of the complex plane



    Let $f: D to mathbbC$ be a holomorphic function



    Let $z_n_n in mathbbN subset D$ be a sequence in $D$ such that $z_n to s in D$ and we have $f(z_n) to f(s)=0$



    My question is:
    Is there a subsequence $z_n_k_k in mathbbN$ such that $forall k in mathbbN: Im(f(z_n_k)) neq 0$



    Thanks.










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      Let $D subset mathbbC$ be a open subset of the complex plane



      Let $f: D to mathbbC$ be a holomorphic function



      Let $z_n_n in mathbbN subset D$ be a sequence in $D$ such that $z_n to s in D$ and we have $f(z_n) to f(s)=0$



      My question is:
      Is there a subsequence $z_n_k_k in mathbbN$ such that $forall k in mathbbN: Im(f(z_n_k)) neq 0$



      Thanks.










      share|cite|improve this question









      $endgroup$




      Let $D subset mathbbC$ be a open subset of the complex plane



      Let $f: D to mathbbC$ be a holomorphic function



      Let $z_n_n in mathbbN subset D$ be a sequence in $D$ such that $z_n to s in D$ and we have $f(z_n) to f(s)=0$



      My question is:
      Is there a subsequence $z_n_k_k in mathbbN$ such that $forall k in mathbbN: Im(f(z_n_k)) neq 0$



      Thanks.







      complex-analysis holomorphic-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 12 at 10:55









      Matey MathMatey Math

      852514




      852514




















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Well... Not necessarily. Let $f(z)=z$ and $z_n=1/n$ for example.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanks for your answer @GiuseppeNegro
            $endgroup$
            – Matey Math
            Mar 12 at 11:07










          • $begingroup$
            @MateyMath: you are welcome but it is really trivial. Probably you are not asking the right question.
            $endgroup$
            – Giuseppe Negro
            Mar 12 at 11:20










          • $begingroup$
            you are right it was a trivial attempt to solve this math.stackexchange.com/questions/3142477/…
            $endgroup$
            – Matey Math
            Mar 12 at 11:42











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Well... Not necessarily. Let $f(z)=z$ and $z_n=1/n$ for example.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanks for your answer @GiuseppeNegro
            $endgroup$
            – Matey Math
            Mar 12 at 11:07










          • $begingroup$
            @MateyMath: you are welcome but it is really trivial. Probably you are not asking the right question.
            $endgroup$
            – Giuseppe Negro
            Mar 12 at 11:20










          • $begingroup$
            you are right it was a trivial attempt to solve this math.stackexchange.com/questions/3142477/…
            $endgroup$
            – Matey Math
            Mar 12 at 11:42
















          2












          $begingroup$

          Well... Not necessarily. Let $f(z)=z$ and $z_n=1/n$ for example.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanks for your answer @GiuseppeNegro
            $endgroup$
            – Matey Math
            Mar 12 at 11:07










          • $begingroup$
            @MateyMath: you are welcome but it is really trivial. Probably you are not asking the right question.
            $endgroup$
            – Giuseppe Negro
            Mar 12 at 11:20










          • $begingroup$
            you are right it was a trivial attempt to solve this math.stackexchange.com/questions/3142477/…
            $endgroup$
            – Matey Math
            Mar 12 at 11:42














          2












          2








          2





          $begingroup$

          Well... Not necessarily. Let $f(z)=z$ and $z_n=1/n$ for example.






          share|cite|improve this answer









          $endgroup$



          Well... Not necessarily. Let $f(z)=z$ and $z_n=1/n$ for example.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 12 at 11:03









          Giuseppe NegroGiuseppe Negro

          17.5k332126




          17.5k332126











          • $begingroup$
            Thanks for your answer @GiuseppeNegro
            $endgroup$
            – Matey Math
            Mar 12 at 11:07










          • $begingroup$
            @MateyMath: you are welcome but it is really trivial. Probably you are not asking the right question.
            $endgroup$
            – Giuseppe Negro
            Mar 12 at 11:20










          • $begingroup$
            you are right it was a trivial attempt to solve this math.stackexchange.com/questions/3142477/…
            $endgroup$
            – Matey Math
            Mar 12 at 11:42

















          • $begingroup$
            Thanks for your answer @GiuseppeNegro
            $endgroup$
            – Matey Math
            Mar 12 at 11:07










          • $begingroup$
            @MateyMath: you are welcome but it is really trivial. Probably you are not asking the right question.
            $endgroup$
            – Giuseppe Negro
            Mar 12 at 11:20










          • $begingroup$
            you are right it was a trivial attempt to solve this math.stackexchange.com/questions/3142477/…
            $endgroup$
            – Matey Math
            Mar 12 at 11:42
















          $begingroup$
          Thanks for your answer @GiuseppeNegro
          $endgroup$
          – Matey Math
          Mar 12 at 11:07




          $begingroup$
          Thanks for your answer @GiuseppeNegro
          $endgroup$
          – Matey Math
          Mar 12 at 11:07












          $begingroup$
          @MateyMath: you are welcome but it is really trivial. Probably you are not asking the right question.
          $endgroup$
          – Giuseppe Negro
          Mar 12 at 11:20




          $begingroup$
          @MateyMath: you are welcome but it is really trivial. Probably you are not asking the right question.
          $endgroup$
          – Giuseppe Negro
          Mar 12 at 11:20












          $begingroup$
          you are right it was a trivial attempt to solve this math.stackexchange.com/questions/3142477/…
          $endgroup$
          – Matey Math
          Mar 12 at 11:42





          $begingroup$
          you are right it was a trivial attempt to solve this math.stackexchange.com/questions/3142477/…
          $endgroup$
          – Matey Math
          Mar 12 at 11:42


















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