Mixing
Two Bijective Holomorphic Functions
Clash Royale CLAN TAG#URR8PPP
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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
asked Jul 14 at 22:53
Kezer
1,135316
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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
asked Jul 14 at 22:53
Kezer
1,135316
add a comment |Â
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
asked Jul 14 at 22:53
Kezer
1,135316
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
asked Jul 14 at 22:53
Kezer
1,135316
asked Jul 14 at 22:53
Kezer
1,135316
asked Jul 14 at 22:53
Kezer
1,135316
asked Jul 14 at 22:53
Kezer
1,135316
1,135316
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
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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 $h(z)=z$.
answered Jul 14 at 23:01
Hugocito
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
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $h(z)=z$.
answered Jul 14 at 23:01
Hugocito
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
add a comment |Â
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 $h(z)=z$.
answered Jul 14 at 23:01
Hugocito
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
add a comment |Â
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 $h(z)=z$.
answered Jul 14 at 23:01
Hugocito
1,6451019
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 $h(z)=z$.
answered Jul 14 at 23:01
Hugocito
1,6451019
edited Jul 14 at 23:03
answered Jul 14 at 23:01
Hugocito
1,6451019
answered Jul 14 at 23:01
Hugocito
1,6451019
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
add a comment |Â
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
add a comment |Â
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