A question on simply connected domains $U,V subset mathbb{C}$, not isomorphic to $mathbb{C}$ and conformal...
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I'm given simply connected domains $U,V subset mathbb{C}$, not isomorphic to $mathbb{C}$. Fixing points $a in U, b in V$. I have to show that $|f'(a)|$ is the same for all conformal mappings $f:U rightarrow V$ such that $f(a) = b$
Now, by Riemann Mapping Theorem we have $U,V cong D$, where $D$ is the open disk. via the Riemann maps $g$ and $h$, such that $g(a) = 0$ and $h(b) = 0$, alongwith $g'(a), h'(b) >0$
So, we get $k=h circ f circ g^{-1}:D rightarrow D$, with $k(0)=0$. We can now apply Schwarz Lemma on this map, to obtain the condition $|k'(0)|leq1$. Now, since $l = k^{-1}$, also satisfies the conditions of Schwarz Lemma, we get $|l'(0)|leq 1 implies |k'(0)|geq 1 implies |k'(0)|=1$.
Therefore, $|h'(b)|cdot|f'(a)|cdot1/|g'(a)| = 1 implies f'(a) = |g'(a)/h'(b)|$
Am I right? Or am I going terribly wrong somewhere? If so, could anyone point me towards the right direction?
complex-analysis
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
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add a comment |
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I'm given simply connected domains $U,V subset mathbb{C}$, not isomorphic to $mathbb{C}$. Fixing points $a in U, b in V$. I have to show that $|f'(a)|$ is the same for all conformal mappings $f:U rightarrow V$ such that $f(a) = b$
Now, by Riemann Mapping Theorem we have $U,V cong D$, where $D$ is the open disk. via the Riemann maps $g$ and $h$, such that $g(a) = 0$ and $h(b) = 0$, alongwith $g'(a), h'(b) >0$
So, we get $k=h circ f circ g^{-1}:D rightarrow D$, with $k(0)=0$. We can now apply Schwarz Lemma on this map, to obtain the condition $|k'(0)|leq1$. Now, since $l = k^{-1}$, also satisfies the conditions of Schwarz Lemma, we get $|l'(0)|leq 1 implies |k'(0)|geq 1 implies |k'(0)|=1$.
Therefore, $|h'(b)|cdot|f'(a)|cdot1/|g'(a)| = 1 implies f'(a) = |g'(a)/h'(b)|$
Am I right? Or am I going terribly wrong somewhere? If so, could anyone point me towards the right direction?
complex-analysis
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
$endgroup$
add a comment |
$begingroup$
I'm given simply connected domains $U,V subset mathbb{C}$, not isomorphic to $mathbb{C}$. Fixing points $a in U, b in V$. I have to show that $|f'(a)|$ is the same for all conformal mappings $f:U rightarrow V$ such that $f(a) = b$
Now, by Riemann Mapping Theorem we have $U,V cong D$, where $D$ is the open disk. via the Riemann maps $g$ and $h$, such that $g(a) = 0$ and $h(b) = 0$, alongwith $g'(a), h'(b) >0$
So, we get $k=h circ f circ g^{-1}:D rightarrow D$, with $k(0)=0$. We can now apply Schwarz Lemma on this map, to obtain the condition $|k'(0)|leq1$. Now, since $l = k^{-1}$, also satisfies the conditions of Schwarz Lemma, we get $|l'(0)|leq 1 implies |k'(0)|geq 1 implies |k'(0)|=1$.
Therefore, $|h'(b)|cdot|f'(a)|cdot1/|g'(a)| = 1 implies f'(a) = |g'(a)/h'(b)|$
Am I right? Or am I going terribly wrong somewhere? If so, could anyone point me towards the right direction?
complex-analysis
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
$endgroup$
I'm given simply connected domains $U,V subset mathbb{C}$, not isomorphic to $mathbb{C}$. Fixing points $a in U, b in V$. I have to show that $|f'(a)|$ is the same for all conformal mappings $f:U rightarrow V$ such that $f(a) = b$
Now, by Riemann Mapping Theorem we have $U,V cong D$, where $D$ is the open disk. via the Riemann maps $g$ and $h$, such that $g(a) = 0$ and $h(b) = 0$, alongwith $g'(a), h'(b) >0$
So, we get $k=h circ f circ g^{-1}:D rightarrow D$, with $k(0)=0$. We can now apply Schwarz Lemma on this map, to obtain the condition $|k'(0)|leq1$. Now, since $l = k^{-1}$, also satisfies the conditions of Schwarz Lemma, we get $|l'(0)|leq 1 implies |k'(0)|geq 1 implies |k'(0)|=1$.
Therefore, $|h'(b)|cdot|f'(a)|cdot1/|g'(a)| = 1 implies f'(a) = |g'(a)/h'(b)|$
Am I right? Or am I going terribly wrong somewhere? If so, could anyone point me towards the right direction?
complex-analysis
complex-analysis
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
edited Dec 10 '18 at 21:11
Naweed G. Seldon
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
asked Dec 10 '18 at 18:02
Naweed G. SeldonNaweed G. Seldon
1,304419
1,304419
add a comment |
add a comment |
1 Answer
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Yes you are right, except in the last sentence of the third paragraph: $k'(0)=1$ should be $|k'(0)|=1$.
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
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Corrected, thanks!
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– Naweed G. Seldon
Dec 10 '18 at 21:12
add a comment |
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Yes you are right, except in the last sentence of the third paragraph: $k'(0)=1$ should be $|k'(0)|=1$.
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
$endgroup$
$begingroup$
Corrected, thanks!
$endgroup$
– Naweed G. Seldon
Dec 10 '18 at 21:12
add a comment |
$begingroup$
Yes you are right, except in the last sentence of the third paragraph: $k'(0)=1$ should be $|k'(0)|=1$.
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
$endgroup$
$begingroup$
Corrected, thanks!
$endgroup$
– Naweed G. Seldon
Dec 10 '18 at 21:12
add a comment |
$begingroup$
Yes you are right, except in the last sentence of the third paragraph: $k'(0)=1$ should be $|k'(0)|=1$.
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
$endgroup$
Yes you are right, except in the last sentence of the third paragraph: $k'(0)=1$ should be $|k'(0)|=1$.
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
answered Dec 10 '18 at 21:10
Eclipse SunEclipse Sun
7,2941437
7,2941437
$begingroup$
Corrected, thanks!
$endgroup$
– Naweed G. Seldon
Dec 10 '18 at 21:12
add a comment |
$begingroup$
Corrected, thanks!
$endgroup$
– Naweed G. Seldon
Dec 10 '18 at 21:12
$begingroup$
Corrected, thanks!
$endgroup$
– Naweed G. Seldon
Dec 10 '18 at 21:12
$begingroup$
Corrected, thanks!
$endgroup$
– Naweed G. Seldon
Dec 10 '18 at 21:12
add a comment |
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