$1$-$1$ holomorphic function on a simple closed contour is $1$-$1$ inside the contour?












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Suppose that $f(z)$ is holomorphic inside and on a simple closed contour $C.$



I'd like to prove the following:



If $f(z)$ is a one-to-one function on $C$, then $f(z)$ is one-to-one function inside $C.$



I don't know how to prove it. The only thing I know is that $f(C)$ is a simple closed contour, since $f$ is one-to-one on a simple closed contour $C$.



Please let me know if you have any idea or comment for this question.



Thanks in advance!










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    1














    Suppose that $f(z)$ is holomorphic inside and on a simple closed contour $C.$



    I'd like to prove the following:



    If $f(z)$ is a one-to-one function on $C$, then $f(z)$ is one-to-one function inside $C.$



    I don't know how to prove it. The only thing I know is that $f(C)$ is a simple closed contour, since $f$ is one-to-one on a simple closed contour $C$.



    Please let me know if you have any idea or comment for this question.



    Thanks in advance!










    share|cite|improve this question



























      1












      1








      1


      1





      Suppose that $f(z)$ is holomorphic inside and on a simple closed contour $C.$



      I'd like to prove the following:



      If $f(z)$ is a one-to-one function on $C$, then $f(z)$ is one-to-one function inside $C.$



      I don't know how to prove it. The only thing I know is that $f(C)$ is a simple closed contour, since $f$ is one-to-one on a simple closed contour $C$.



      Please let me know if you have any idea or comment for this question.



      Thanks in advance!










      share|cite|improve this question















      Suppose that $f(z)$ is holomorphic inside and on a simple closed contour $C.$



      I'd like to prove the following:



      If $f(z)$ is a one-to-one function on $C$, then $f(z)$ is one-to-one function inside $C.$



      I don't know how to prove it. The only thing I know is that $f(C)$ is a simple closed contour, since $f$ is one-to-one on a simple closed contour $C$.



      Please let me know if you have any idea or comment for this question.



      Thanks in advance!







      complex-analysis






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      share|cite|improve this question













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      edited Nov 22 '18 at 13:21







      0706

















      asked Nov 22 '18 at 2:32









      07060706

      414110




      414110






















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