Two Bijective Holomorphic Functions

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Let $f,g : mathbbE to mathbbE$ be bijective, holomorphic functions with $f(0) = g(0)$ and $f'(0) = g'(0)$. Also, $f'$ and $g'$ don't have zeros.
Prove $f(z) = g(z)$ for all $z in mathbbE$. (Note: $mathbbE := B_1(0)$.)



It seemed like an easy problem but somehow I can't find an argument for this statement. I guess bijectivity is the hard part of this problem.




complex-analysis




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    Let $f,g : mathbbE to mathbbE$ be bijective, holomorphic functions with $f(0) = g(0)$ and $f'(0) = g'(0)$. Also, $f'$ and $g'$ don't have zeros.
    Prove $f(z) = g(z)$ for all $z in mathbbE$. (Note: $mathbbE := B_1(0)$.)



    It seemed like an easy problem but somehow I can't find an argument for this statement. I guess bijectivity is the hard part of this problem.




    complex-analysis




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $f,g : mathbbE to mathbbE$ be bijective, holomorphic functions with $f(0) = g(0)$ and $f'(0) = g'(0)$. Also, $f'$ and $g'$ don't have zeros.
      Prove $f(z) = g(z)$ for all $z in mathbbE$. (Note: $mathbbE := B_1(0)$.)



      It seemed like an easy problem but somehow I can't find an argument for this statement. I guess bijectivity is the hard part of this problem.




      complex-analysis




      share|cite|improve this question











      Let $f,g : mathbbE to mathbbE$ be bijective, holomorphic functions with $f(0) = g(0)$ and $f'(0) = g'(0)$. Also, $f'$ and $g'$ don't have zeros.
      Prove $f(z) = g(z)$ for all $z in mathbbE$. (Note: $mathbbE := B_1(0)$.)



      It seemed like an easy problem but somehow I can't find an argument for this statement. I guess bijectivity is the hard part of this problem.





      complex-analysis





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 14 at 22:53









      Kezer

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          Since these functions are bijective and holomorphic, they are also biholomorphic. Define $h(z) = f^-1circ g(z)$. We have $h$ is a biholomorphism of the disk, $h(0) = 0$ and $h'(0) = 1$. By the Schwarz lemma we conclude $h(z)=z$.






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          • 2




            Karl Hermann Amandus Schwarz, not Schwartz :)
            – Martin R
            Jul 14 at 23:02











          • Thank you! I corrected it.
            – Hugocito
            Jul 14 at 23:03










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          up vote
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          Since these functions are bijective and holomorphic, they are also biholomorphic. Define $h(z) = f^-1circ g(z)$. We have $h$ is a biholomorphism of the disk, $h(0) = 0$ and $h'(0) = 1$. By the Schwarz lemma we conclude $h(z)=z$.






          share|cite|improve this answer



















          • 2




            Karl Hermann Amandus Schwarz, not Schwartz :)
            – Martin R
            Jul 14 at 23:02











          • Thank you! I corrected it.
            – Hugocito
            Jul 14 at 23:03














          up vote
          1
          down vote



          accepted










          Since these functions are bijective and holomorphic, they are also biholomorphic. Define $h(z) = f^-1circ g(z)$. We have $h$ is a biholomorphism of the disk, $h(0) = 0$ and $h'(0) = 1$. By the Schwarz lemma we conclude $h(z)=z$.






          share|cite|improve this answer



















          • 2




            Karl Hermann Amandus Schwarz, not Schwartz :)
            – Martin R
            Jul 14 at 23:02











          • Thank you! I corrected it.
            – Hugocito
            Jul 14 at 23:03












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Since these functions are bijective and holomorphic, they are also biholomorphic. Define $h(z) = f^-1circ g(z)$. We have $h$ is a biholomorphism of the disk, $h(0) = 0$ and $h'(0) = 1$. By the Schwarz lemma we conclude $h(z)=z$.






          share|cite|improve this answer















          Since these functions are bijective and holomorphic, they are also biholomorphic. Define $h(z) = f^-1circ g(z)$. We have $h$ is a biholomorphism of the disk, $h(0) = 0$ and $h'(0) = 1$. By the Schwarz lemma we conclude $h(z)=z$.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 14 at 23:03


























          answered Jul 14 at 23:01









          Hugocito

          1,6451019




          1,6451019







          • 2




            Karl Hermann Amandus Schwarz, not Schwartz :)
            – Martin R
            Jul 14 at 23:02











          • Thank you! I corrected it.
            – Hugocito
            Jul 14 at 23:03












          • 2




            Karl Hermann Amandus Schwarz, not Schwartz :)
            – Martin R
            Jul 14 at 23:02











          • Thank you! I corrected it.
            – Hugocito
            Jul 14 at 23:03







          2




          2




          Karl Hermann Amandus Schwarz, not Schwartz :)
          – Martin R
          Jul 14 at 23:02





          Karl Hermann Amandus Schwarz, not Schwartz :)
          – Martin R
          Jul 14 at 23:02













          Thank you! I corrected it.
          – Hugocito
          Jul 14 at 23:03




          Thank you! I corrected it.
          – Hugocito
          Jul 14 at 23:03












           

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