Mixing
Fractional Linear Transformation
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Suppose that $a, b, c, dinmathbbC$ and $ad-bc=1$. Let $T$ be the fractional linear transformation $$zmapstofracaz+bcz+d.$$
Show that if $a=i, b=-i, c=1, d=i$, then the corresponding fraction linear transformation maps the upper half plane onto the unit disc.
I tried to find the image of $$w=fraciz-iz+i$$
From here, $$z=frac-i-wiw-i$$
Hence beginalign
x+iy&=z \
&= frac-i-wiw-i \
&=frac-i-i(u+iv)u+i(v-1) \
&=frac(-i-iu+v)(u-iv+i)u^2+(v-1)^2 \
&= frac1-v+uu^2+(v-1)^2+ifrac-u-u^2-v^2+vu^2+(v-1)^2 \
endalign
Thus we want the image of $y>0$, so $$frac-u-u^2-v^2+vu^2+(v-1)^2>0$$
But this does not yield the unit circle. Where have I gone wrong in my method?
complex-analysis proof-verification mobius-transformation
asked 2 days ago
Bell
543112
 |Â
show 2 more comments
up vote
0
down vote
favorite
Suppose that $a, b, c, dinmathbbC$ and $ad-bc=1$. Let $T$ be the fractional linear transformation $$zmapstofracaz+bcz+d.$$
Show that if $a=i, b=-i, c=1, d=i$, then the corresponding fraction linear transformation maps the upper half plane onto the unit disc.
I tried to find the image of $$w=fraciz-iz+i$$
From here, $$z=frac-i-wiw-i$$
Hence beginalign
x+iy&=z \
&= frac-i-wiw-i \
&=frac-i-i(u+iv)u+i(v-1) \
&=frac(-i-iu+v)(u-iv+i)u^2+(v-1)^2 \
&= frac1-v+uu^2+(v-1)^2+ifrac-u-u^2-v^2+vu^2+(v-1)^2 \
endalign
Thus we want the image of $y>0$, so $$frac-u-u^2-v^2+vu^2+(v-1)^2>0$$
But this does not yield the unit circle. Where have I gone wrong in my method?
complex-analysis proof-verification mobius-transformation
asked 2 days ago
Bell
543112
Are you familiar with properties of Mobius transformations?
– Mark
2 days ago
It is a new concept, so not entirely. I am comfortable with special cases of Mobius transformations, such as inverse maps.
– Bell
2 days ago
Note that $ad-bc=i^2+ine1$ which does not satisfy the condition.
– TheSimpliFire
2 days ago
Are you sure it shouldn't be $a=1$? Because your transformation maps 1 to 0. It means it doesn't map the boundary of the upper half plane to the boundary of the unit disk. (and hence can't map the upper half plane to the unit disk)
– Mark
2 days ago
Just double checked, $a=i$. Do you think the question is incorrect?
– Bell
2 days ago
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $a, b, c, dinmathbbC$ and $ad-bc=1$. Let $T$ be the fractional linear transformation $$zmapstofracaz+bcz+d.$$
Show that if $a=i, b=-i, c=1, d=i$, then the corresponding fraction linear transformation maps the upper half plane onto the unit disc.
I tried to find the image of $$w=fraciz-iz+i$$
From here, $$z=frac-i-wiw-i$$
Hence beginalign
x+iy&=z \
&= frac-i-wiw-i \
&=frac-i-i(u+iv)u+i(v-1) \
&=frac(-i-iu+v)(u-iv+i)u^2+(v-1)^2 \
&= frac1-v+uu^2+(v-1)^2+ifrac-u-u^2-v^2+vu^2+(v-1)^2 \
endalign
Thus we want the image of $y>0$, so $$frac-u-u^2-v^2+vu^2+(v-1)^2>0$$
But this does not yield the unit circle. Where have I gone wrong in my method?
complex-analysis proof-verification mobius-transformation
asked 2 days ago
Bell
543112
Suppose that $a, b, c, dinmathbbC$ and $ad-bc=1$. Let $T$ be the fractional linear transformation $$zmapstofracaz+bcz+d.$$
Show that if $a=i, b=-i, c=1, d=i$, then the corresponding fraction linear transformation maps the upper half plane onto the unit disc.
I tried to find the image of $$w=fraciz-iz+i$$
From here, $$z=frac-i-wiw-i$$
Hence beginalign
x+iy&=z \
&= frac-i-wiw-i \
&=frac-i-i(u+iv)u+i(v-1) \
&=frac(-i-iu+v)(u-iv+i)u^2+(v-1)^2 \
&= frac1-v+uu^2+(v-1)^2+ifrac-u-u^2-v^2+vu^2+(v-1)^2 \
endalign
Thus we want the image of $y>0$, so $$frac-u-u^2-v^2+vu^2+(v-1)^2>0$$
But this does not yield the unit circle. Where have I gone wrong in my method?
complex-analysis proof-verification mobius-transformation
asked 2 days ago
Bell
543112
asked 2 days ago
Bell
543112
asked 2 days ago
Bell
543112
asked 2 days ago
Bell
543112
543112
Are you familiar with properties of Mobius transformations?
– Mark
2 days ago
It is a new concept, so not entirely. I am comfortable with special cases of Mobius transformations, such as inverse maps.
– Bell
2 days ago
Note that $ad-bc=i^2+ine1$ which does not satisfy the condition.
– TheSimpliFire
2 days ago
Are you sure it shouldn't be $a=1$? Because your transformation maps 1 to 0. It means it doesn't map the boundary of the upper half plane to the boundary of the unit disk. (and hence can't map the upper half plane to the unit disk)
– Mark
2 days ago
Just double checked, $a=i$. Do you think the question is incorrect?
– Bell
2 days ago
 |Â
show 2 more comments
Are you familiar with properties of Mobius transformations?
– Mark
2 days ago
It is a new concept, so not entirely. I am comfortable with special cases of Mobius transformations, such as inverse maps.
– Bell
2 days ago
Note that $ad-bc=i^2+ine1$ which does not satisfy the condition.
– TheSimpliFire
2 days ago
Are you sure it shouldn't be $a=1$? Because your transformation maps 1 to 0. It means it doesn't map the boundary of the upper half plane to the boundary of the unit disk. (and hence can't map the upper half plane to the unit disk)
– Mark
2 days ago
Just double checked, $a=i$. Do you think the question is incorrect?
– Bell
2 days ago
Are you familiar with properties of Mobius transformations?
– Mark
2 days ago
Are you familiar with properties of Mobius transformations?
– Mark
2 days ago
It is a new concept, so not entirely. I am comfortable with special cases of Mobius transformations, such as inverse maps.
– Bell
2 days ago
It is a new concept, so not entirely. I am comfortable with special cases of Mobius transformations, such as inverse maps.
– Bell
2 days ago
Note that $ad-bc=i^2+ine1$ which does not satisfy the condition.
– TheSimpliFire
2 days ago
Note that $ad-bc=i^2+ine1$ which does not satisfy the condition.
– TheSimpliFire
2 days ago
Are you sure it shouldn't be $a=1$? Because your transformation maps 1 to 0. It means it doesn't map the boundary of the upper half plane to the boundary of the unit disk. (and hence can't map the upper half plane to the unit disk)
– Mark
2 days ago
Are you sure it shouldn't be $a=1$? Because your transformation maps 1 to 0. It means it doesn't map the boundary of the upper half plane to the boundary of the unit disk. (and hence can't map the upper half plane to the unit disk)
– Mark
2 days ago
Just double checked, $a=i$. Do you think the question is incorrect?
– Bell
2 days ago
Just double checked, $a=i$. Do you think the question is incorrect?
– Bell
2 days ago
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
Hint: you have to prove two things
First: if $Im(z)>0$ then $|w|leq 1$
(and if you try $z=-2+i$ you will see this is not true, so your problem is incorrect).
Second: if $|w|leq 1$ then $Im(z)>0$
answered 2 days ago
greedoid
26.1k93473
Was my method correct though? Or was it completely invalid haha
– Bell
2 days ago
Second part it is, but you didn't take into a count that $|w|leq 1$, that is $u^2+v^2leq 1$
– greedoid
2 days ago
add a comment |Â
up vote
1
down vote
If you change to $a=1, b=-i, c=1, d=i$, then the image of $zin mathbb R$ is $ $, since $z+i$ is the conjugate of $z-i$. But $i$ goes to $0$, and infinity goes to 1, so by continuity the upper half plane goes to the interior of the circle, and the lower half plane to the exterior. Since the function is a bijection of $mathbb C cup infty$ to itself, you get the result.
answered 2 days ago
xarles
83548
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint: you have to prove two things
First: if $Im(z)>0$ then $|w|leq 1$
(and if you try $z=-2+i$ you will see this is not true, so your problem is incorrect).
Second: if $|w|leq 1$ then $Im(z)>0$
answered 2 days ago
greedoid
26.1k93473
Was my method correct though? Or was it completely invalid haha
– Bell
2 days ago
Second part it is, but you didn't take into a count that $|w|leq 1$, that is $u^2+v^2leq 1$
– greedoid
2 days ago
add a comment |Â
up vote
2
down vote
Hint: you have to prove two things
First: if $Im(z)>0$ then $|w|leq 1$
(and if you try $z=-2+i$ you will see this is not true, so your problem is incorrect).
Second: if $|w|leq 1$ then $Im(z)>0$
answered 2 days ago
greedoid
26.1k93473
Was my method correct though? Or was it completely invalid haha
– Bell
2 days ago
Second part it is, but you didn't take into a count that $|w|leq 1$, that is $u^2+v^2leq 1$
– greedoid
2 days ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: you have to prove two things
First: if $Im(z)>0$ then $|w|leq 1$
(and if you try $z=-2+i$ you will see this is not true, so your problem is incorrect).
Second: if $|w|leq 1$ then $Im(z)>0$
answered 2 days ago
greedoid
26.1k93473
Hint: you have to prove two things
First: if $Im(z)>0$ then $|w|leq 1$
(and if you try $z=-2+i$ you will see this is not true, so your problem is incorrect).
Second: if $|w|leq 1$ then $Im(z)>0$
answered 2 days ago
greedoid
26.1k93473
answered 2 days ago
greedoid
26.1k93473
answered 2 days ago
greedoid
26.1k93473
answered 2 days ago
greedoid
26.1k93473
26.1k93473
Was my method correct though? Or was it completely invalid haha
– Bell
2 days ago
Second part it is, but you didn't take into a count that $|w|leq 1$, that is $u^2+v^2leq 1$
– greedoid
2 days ago
add a comment |Â
Was my method correct though? Or was it completely invalid haha
– Bell
2 days ago
Second part it is, but you didn't take into a count that $|w|leq 1$, that is $u^2+v^2leq 1$
– greedoid
2 days ago
Was my method correct though? Or was it completely invalid haha
– Bell
2 days ago
Was my method correct though? Or was it completely invalid haha
– Bell
2 days ago
Second part it is, but you didn't take into a count that $|w|leq 1$, that is $u^2+v^2leq 1$
– greedoid
2 days ago
Second part it is, but you didn't take into a count that $|w|leq 1$, that is $u^2+v^2leq 1$
– greedoid
2 days ago
add a comment |Â
up vote
1
down vote
If you change to $a=1, b=-i, c=1, d=i$, then the image of $zin mathbb R$ is $ $, since $z+i$ is the conjugate of $z-i$. But $i$ goes to $0$, and infinity goes to 1, so by continuity the upper half plane goes to the interior of the circle, and the lower half plane to the exterior. Since the function is a bijection of $mathbb C cup infty$ to itself, you get the result.
answered 2 days ago
xarles
83548
add a comment |Â
up vote
1
down vote
If you change to $a=1, b=-i, c=1, d=i$, then the image of $zin mathbb R$ is $ $, since $z+i$ is the conjugate of $z-i$. But $i$ goes to $0$, and infinity goes to 1, so by continuity the upper half plane goes to the interior of the circle, and the lower half plane to the exterior. Since the function is a bijection of $mathbb C cup infty$ to itself, you get the result.
answered 2 days ago
xarles
83548
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you change to $a=1, b=-i, c=1, d=i$, then the image of $zin mathbb R$ is $ $, since $z+i$ is the conjugate of $z-i$. But $i$ goes to $0$, and infinity goes to 1, so by continuity the upper half plane goes to the interior of the circle, and the lower half plane to the exterior. Since the function is a bijection of $mathbb C cup infty$ to itself, you get the result.
answered 2 days ago
xarles
83548
If you change to $a=1, b=-i, c=1, d=i$, then the image of $zin mathbb R$ is $ $, since $z+i$ is the conjugate of $z-i$. But $i$ goes to $0$, and infinity goes to 1, so by continuity the upper half plane goes to the interior of the circle, and the lower half plane to the exterior. Since the function is a bijection of $mathbb C cup infty$ to itself, you get the result.
answered 2 days ago
xarles
83548
answered 2 days ago
xarles
83548
answered 2 days ago
xarles
83548
answered 2 days ago
xarles
83548
83548
add a comment |Â
add a comment |Â
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Are you familiar with properties of Mobius transformations?
– Mark
2 days ago
It is a new concept, so not entirely. I am comfortable with special cases of Mobius transformations, such as inverse maps.
– Bell
2 days ago
Note that $ad-bc=i^2+ine1$ which does not satisfy the condition.
– TheSimpliFire
2 days ago
Are you sure it shouldn't be $a=1$? Because your transformation maps 1 to 0. It means it doesn't map the boundary of the upper half plane to the boundary of the unit disk. (and hence can't map the upper half plane to the unit disk)
– Mark
2 days ago
Just double checked, $a=i$. Do you think the question is incorrect?
– Bell
2 days ago