Doubt in passing in the Riemann mapping theorem
I have a question, maybe silly in the passage marked in red. I understood everything up to this part. Why does $H'(0) > 0$ imply $e^{i theta} = 1$? Is the Schwarz Lemma being used?
complex-analysis
asked Nov 29 at 5:22
Ricardo Freire
392110
add a comment |
I have a question, maybe silly in the passage marked in red. I understood everything up to this part. Why does $H'(0) > 0$ imply $e^{i theta} = 1$? Is the Schwarz Lemma being used?
complex-analysis
asked Nov 29 at 5:22
Ricardo Freire
392110
The derivative of $H(z)$ is the constant function $e^{itheta}$.
– Lord Shark the Unknown
Nov 29 at 5:24
@LordSharktheUnknown if I understood correctly, just note that $|H'(z)| = 1$, but $H$ has positive derivative in $0$ because $F$ and $G$ have positive derivatives in zero. Then $e^{i theta}=H'(0) = |H'(0)| = 1$.
– Ricardo Freire
Nov 29 at 5:37
add a comment |
I have a question, maybe silly in the passage marked in red. I understood everything up to this part. Why does $H'(0) > 0$ imply $e^{i theta} = 1$? Is the Schwarz Lemma being used?
complex-analysis
asked Nov 29 at 5:22
Ricardo Freire
392110
I have a question, maybe silly in the passage marked in red. I understood everything up to this part. Why does $H'(0) > 0$ imply $e^{i theta} = 1$? Is the Schwarz Lemma being used?
complex-analysis
complex-analysis
asked Nov 29 at 5:22
Ricardo Freire
392110
asked Nov 29 at 5:22
Ricardo Freire
392110
asked Nov 29 at 5:22
Ricardo Freire
392110
asked Nov 29 at 5:22
Ricardo Freire
392110
asked Nov 29 at 5:22
Ricardo Freire
392110
392110
The derivative of $H(z)$ is the constant function $e^{itheta}$.
– Lord Shark the Unknown
Nov 29 at 5:24
@LordSharktheUnknown if I understood correctly, just note that $|H'(z)| = 1$, but $H$ has positive derivative in $0$ because $F$ and $G$ have positive derivatives in zero. Then $e^{i theta}=H'(0) = |H'(0)| = 1$.
– Ricardo Freire
Nov 29 at 5:37
add a comment |
The derivative of $H(z)$ is the constant function $e^{itheta}$.
– Lord Shark the Unknown
Nov 29 at 5:24
@LordSharktheUnknown if I understood correctly, just note that $|H'(z)| = 1$, but $H$ has positive derivative in $0$ because $F$ and $G$ have positive derivatives in zero. Then $e^{i theta}=H'(0) = |H'(0)| = 1$.
– Ricardo Freire
Nov 29 at 5:37
The derivative of $H(z)$ is the constant function $e^{itheta}$.
– Lord Shark the Unknown
Nov 29 at 5:24
The derivative of $H(z)$ is the constant function $e^{itheta}$.
– Lord Shark the Unknown
Nov 29 at 5:24
@LordSharktheUnknown if I understood correctly, just note that $|H'(z)| = 1$, but $H$ has positive derivative in $0$ because $F$ and $G$ have positive derivatives in zero. Then $e^{i theta}=H'(0) = |H'(0)| = 1$.
– Ricardo Freire
Nov 29 at 5:37
@LordSharktheUnknown if I understood correctly, just note that $|H'(z)| = 1$, but $H$ has positive derivative in $0$ because $F$ and $G$ have positive derivatives in zero. Then $e^{i theta}=H'(0) = |H'(0)| = 1$.
– Ricardo Freire
Nov 29 at 5:37
add a comment |
1 Answer
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From $H=F circ G^{-1}$ we get with the chain rule and the rule for the derivative of $G^{-1}$ that
$$ H'(0)= frac{F'(z_0)}{G'(z_0)}.$$
Since $F'(z_0), G'(z_0)>0$, it follows that $H'(0)>0$.
From $H(z)=e^{i theta}z$, we get $H'(0)=e^{i theta}.$
Furthermore we have: $e^{i theta}>0 iff e^{i theta}=1.$
answered Nov 29 at 5:53
Fred
44.2k1645
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1 Answer
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1 Answer
1
active
oldest
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votes
From $H=F circ G^{-1}$ we get with the chain rule and the rule for the derivative of $G^{-1}$ that
$$ H'(0)= frac{F'(z_0)}{G'(z_0)}.$$
Since $F'(z_0), G'(z_0)>0$, it follows that $H'(0)>0$.
From $H(z)=e^{i theta}z$, we get $H'(0)=e^{i theta}.$
Furthermore we have: $e^{i theta}>0 iff e^{i theta}=1.$
answered Nov 29 at 5:53
Fred
44.2k1645
add a comment |
From $H=F circ G^{-1}$ we get with the chain rule and the rule for the derivative of $G^{-1}$ that
$$ H'(0)= frac{F'(z_0)}{G'(z_0)}.$$
Since $F'(z_0), G'(z_0)>0$, it follows that $H'(0)>0$.
From $H(z)=e^{i theta}z$, we get $H'(0)=e^{i theta}.$
Furthermore we have: $e^{i theta}>0 iff e^{i theta}=1.$
answered Nov 29 at 5:53
Fred
44.2k1645
add a comment |
From $H=F circ G^{-1}$ we get with the chain rule and the rule for the derivative of $G^{-1}$ that
$$ H'(0)= frac{F'(z_0)}{G'(z_0)}.$$
Since $F'(z_0), G'(z_0)>0$, it follows that $H'(0)>0$.
From $H(z)=e^{i theta}z$, we get $H'(0)=e^{i theta}.$
Furthermore we have: $e^{i theta}>0 iff e^{i theta}=1.$
answered Nov 29 at 5:53
Fred
44.2k1645
From $H=F circ G^{-1}$ we get with the chain rule and the rule for the derivative of $G^{-1}$ that
$$ H'(0)= frac{F'(z_0)}{G'(z_0)}.$$
Since $F'(z_0), G'(z_0)>0$, it follows that $H'(0)>0$.
From $H(z)=e^{i theta}z$, we get $H'(0)=e^{i theta}.$
Furthermore we have: $e^{i theta}>0 iff e^{i theta}=1.$
answered Nov 29 at 5:53
Fred
44.2k1645
answered Nov 29 at 5:53
Fred
44.2k1645
answered Nov 29 at 5:53
Fred
44.2k1645
answered Nov 29 at 5:53
Fred
44.2k1645
44.2k1645
add a comment |
add a comment |
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The derivative of $H(z)$ is the constant function $e^{itheta}$.
– Lord Shark the Unknown
Nov 29 at 5:24
@LordSharktheUnknown if I understood correctly, just note that $|H'(z)| = 1$, but $H$ has positive derivative in $0$ because $F$ and $G$ have positive derivatives in zero. Then $e^{i theta}=H'(0) = |H'(0)| = 1$.
– Ricardo Freire
Nov 29 at 5:37