Mapping the upper half plane to unit disc
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Find a Mobius transformation from the closed upper half plane onto the closed unit disc taking $1 + i$ to $0$ and $1$ to $−i$.
So far I have the Cayley map: $M(z)=fracz-iz+i$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^ithetafracz - beta1 - betaz.$$
I then thought of doing $M(z)circ f(z)$ however when I input the values $1 + i$ and 1 they do not get the required values $0$ and $-i$.
Where have I gone wrong?
Thanks!
complex-analysis conformal-geometry mobius-transformation
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
asked Nov 10 '16 at 18:22
B.tom
165
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up vote
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down vote
favorite
Find a Mobius transformation from the closed upper half plane onto the closed unit disc taking $1 + i$ to $0$ and $1$ to $−i$.
So far I have the Cayley map: $M(z)=fracz-iz+i$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^ithetafracz - beta1 - betaz.$$
I then thought of doing $M(z)circ f(z)$ however when I input the values $1 + i$ and 1 they do not get the required values $0$ and $-i$.
Where have I gone wrong?
Thanks!
complex-analysis conformal-geometry mobius-transformation
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
asked Nov 10 '16 at 18:22
B.tom
165
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find a Mobius transformation from the closed upper half plane onto the closed unit disc taking $1 + i$ to $0$ and $1$ to $−i$.
So far I have the Cayley map: $M(z)=fracz-iz+i$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^ithetafracz - beta1 - betaz.$$
I then thought of doing $M(z)circ f(z)$ however when I input the values $1 + i$ and 1 they do not get the required values $0$ and $-i$.
Where have I gone wrong?
Thanks!
complex-analysis conformal-geometry mobius-transformation
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
asked Nov 10 '16 at 18:22
B.tom
165
Find a Mobius transformation from the closed upper half plane onto the closed unit disc taking $1 + i$ to $0$ and $1$ to $−i$.
So far I have the Cayley map: $M(z)=fracz-iz+i$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^ithetafracz - beta1 - betaz.$$
I then thought of doing $M(z)circ f(z)$ however when I input the values $1 + i$ and 1 they do not get the required values $0$ and $-i$.
Where have I gone wrong?
Thanks!
complex-analysis conformal-geometry mobius-transformation
complex-analysis conformal-geometry mobius-transformation
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
asked Nov 10 '16 at 18:22
B.tom
165
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
asked Nov 10 '16 at 18:22
B.tom
165
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
edited Nov 10 '16 at 18:55
Robert Z
85.6k1055123
85.6k1055123
asked Nov 10 '16 at 18:22
B.tom
165
asked Nov 10 '16 at 18:22
B.tom
165
asked Nov 10 '16 at 18:22
B.tom
165
165
add a comment |Â
add a comment |Â
1 Answer
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Hint. You should consider the composition $f(M(z))$ with
$$M(z)=fracz-iz+iquadmboxandquad f(z) = e^ithetafracz - beta1 - overlinebetaz.$$
We have that $beta:=M(1+i)=1/(1+2i)$ (note that $|beta|<1$).
Finally use $f(M(1))=f((1-i)/(1+i))=-i$ to find $e^itheta$. It turns out that $e^itheta=(4+3i)/5$.
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint. You should consider the composition $f(M(z))$ with
$$M(z)=fracz-iz+iquadmboxandquad f(z) = e^ithetafracz - beta1 - overlinebetaz.$$
We have that $beta:=M(1+i)=1/(1+2i)$ (note that $|beta|<1$).
Finally use $f(M(1))=f((1-i)/(1+i))=-i$ to find $e^itheta$. It turns out that $e^itheta=(4+3i)/5$.
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
add a comment |Â
up vote
0
down vote
Hint. You should consider the composition $f(M(z))$ with
$$M(z)=fracz-iz+iquadmboxandquad f(z) = e^ithetafracz - beta1 - overlinebetaz.$$
We have that $beta:=M(1+i)=1/(1+2i)$ (note that $|beta|<1$).
Finally use $f(M(1))=f((1-i)/(1+i))=-i$ to find $e^itheta$. It turns out that $e^itheta=(4+3i)/5$.
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint. You should consider the composition $f(M(z))$ with
$$M(z)=fracz-iz+iquadmboxandquad f(z) = e^ithetafracz - beta1 - overlinebetaz.$$
We have that $beta:=M(1+i)=1/(1+2i)$ (note that $|beta|<1$).
Finally use $f(M(1))=f((1-i)/(1+i))=-i$ to find $e^itheta$. It turns out that $e^itheta=(4+3i)/5$.
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
Hint. You should consider the composition $f(M(z))$ with
$$M(z)=fracz-iz+iquadmboxandquad f(z) = e^ithetafracz - beta1 - overlinebetaz.$$
We have that $beta:=M(1+i)=1/(1+2i)$ (note that $|beta|<1$).
Finally use $f(M(1))=f((1-i)/(1+i))=-i$ to find $e^itheta$. It turns out that $e^itheta=(4+3i)/5$.
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
edited Nov 10 '16 at 18:59
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
answered Nov 10 '16 at 18:38
Robert Z
85.6k1055123
85.6k1055123
add a comment |Â
add a comment |Â
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