Mapping the upper half plane to unit disc
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: $M(z)=frac{z-i}{z+i}$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^{itheta}frac{z - beta}{1 - {beta}z}.$$
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
93.4k1061132
asked Nov 10 '16 at 18:22
B.tom
166
bumped to the homepage by Community♦ 15 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
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: $M(z)=frac{z-i}{z+i}$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^{itheta}frac{z - beta}{1 - {beta}z}.$$
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
93.4k1061132
asked Nov 10 '16 at 18:22
B.tom
166
bumped to the homepage by Community♦ 15 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
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: $M(z)=frac{z-i}{z+i}$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^{itheta}frac{z - beta}{1 - {beta}z}.$$
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
93.4k1061132
asked Nov 10 '16 at 18:22
B.tom
166
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: $M(z)=frac{z-i}{z+i}$ maps the upper half plane to the unit circle,
I also have a mapping from the unit circle to unit disk as
$$f(z) = e^{itheta}frac{z - beta}{1 - {beta}z}.$$
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
93.4k1061132
asked Nov 10 '16 at 18:22
B.tom
166
edited Nov 10 '16 at 18:55
Robert Z
93.4k1061132
asked Nov 10 '16 at 18:22
B.tom
166
edited Nov 10 '16 at 18:55
Robert Z
93.4k1061132
edited Nov 10 '16 at 18:55
Robert Z
93.4k1061132
edited Nov 10 '16 at 18:55
Robert Z
93.4k1061132
93.4k1061132
asked Nov 10 '16 at 18:22
B.tom
166
asked Nov 10 '16 at 18:22
B.tom
166
asked Nov 10 '16 at 18:22
B.tom
166
166
bumped to the homepage by Community♦ 15 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 15 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
1 Answer
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Hint. You should consider the composition $f(M(z))$ with
$$M(z)=frac{z-i}{z+i}quadmbox{and}quad f(z) = e^{itheta}frac{z - beta}{1 - overline{beta}z}.$$
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
93.4k1061132
add a comment |
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1 Answer
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Hint. You should consider the composition $f(M(z))$ with
$$M(z)=frac{z-i}{z+i}quadmbox{and}quad f(z) = e^{itheta}frac{z - beta}{1 - overline{beta}z}.$$
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
93.4k1061132
add a comment |
Hint. You should consider the composition $f(M(z))$ with
$$M(z)=frac{z-i}{z+i}quadmbox{and}quad f(z) = e^{itheta}frac{z - beta}{1 - overline{beta}z}.$$
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
93.4k1061132
add a comment |
Hint. You should consider the composition $f(M(z))$ with
$$M(z)=frac{z-i}{z+i}quadmbox{and}quad f(z) = e^{itheta}frac{z - beta}{1 - overline{beta}z}.$$
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
93.4k1061132
Hint. You should consider the composition $f(M(z))$ with
$$M(z)=frac{z-i}{z+i}quadmbox{and}quad f(z) = e^{itheta}frac{z - beta}{1 - overline{beta}z}.$$
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
93.4k1061132
edited Nov 10 '16 at 18:59
answered Nov 10 '16 at 18:38
Robert Z
93.4k1061132
answered Nov 10 '16 at 18:38
Robert Z
93.4k1061132
answered Nov 10 '16 at 18:38
Robert Z
93.4k1061132
93.4k1061132
add a comment |
add a comment |
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