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Transform circle to $mathbb R$: Will any 3 distinct points on the circle work?
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.15
$C[i+1,1] = $ is the circle with centre (1,1) and radius 1
(Part I) I'm going to try $g=[z,(2,1),(1,2),(0,1)], -g=[z,(2,1),(1,0),(0,1)]$. Please point out and explain any errors.
I observe that that $pm g$ map $C[i+1,1] to mathbb R$. By plugging in $z=(1+i)+e^i phi$, we can show that $pm g((1+i)+e^i phi) in mathbb R$. Then we extend $pm g$ to be defined for the singularity $z=z_3=i=(0,1)$.
Finally, the images are (WA: plus, minus)
$$pm g(1+i) = pm i$$
Did I go wrong anywhere, and why?
(Part II) Can I actually pick any 3 points on $C[i+1,1]$ to form a required cross-ratio?
By inputting in the 3 points used to make up a cross ratio, into the cross ratio, it seems that the cross ratio will output $0,1,infty$, soooo it looks like all cross ratios will map 3 distinct points on a circle to the real line.
$therefore,$ although the question asks for 2 Möbius transformations, there are uncountably $infty$ly many Möbius transformations/cross-ratios that work, so it seems?
So how will we compute the images $[i+1,z_1,z_2,z_3]$? Is it simply that with no particular pattern?
complex-analysis geometry transformation mobius-transformation cross-ratio
asked Jul 31 at 11:30
BCLC
7,00821973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-15 14:08:05Z">in 6 days.
This question has not received enough attention.
 |Â
show 2 more comments
up vote
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down vote
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.15
$C[i+1,1] = $ is the circle with centre (1,1) and radius 1
(Part I) I'm going to try $g=[z,(2,1),(1,2),(0,1)], -g=[z,(2,1),(1,0),(0,1)]$. Please point out and explain any errors.
I observe that that $pm g$ map $C[i+1,1] to mathbb R$. By plugging in $z=(1+i)+e^i phi$, we can show that $pm g((1+i)+e^i phi) in mathbb R$. Then we extend $pm g$ to be defined for the singularity $z=z_3=i=(0,1)$.
Finally, the images are (WA: plus, minus)
$$pm g(1+i) = pm i$$
Did I go wrong anywhere, and why?
(Part II) Can I actually pick any 3 points on $C[i+1,1]$ to form a required cross-ratio?
By inputting in the 3 points used to make up a cross ratio, into the cross ratio, it seems that the cross ratio will output $0,1,infty$, soooo it looks like all cross ratios will map 3 distinct points on a circle to the real line.
$therefore,$ although the question asks for 2 Möbius transformations, there are uncountably $infty$ly many Möbius transformations/cross-ratios that work, so it seems?
So how will we compute the images $[i+1,z_1,z_2,z_3]$? Is it simply that with no particular pattern?
complex-analysis geometry transformation mobius-transformation cross-ratio
asked Jul 31 at 11:30
BCLC
7,00821973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-15 14:08:05Z">in 6 days.
This question has not received enough attention.
1
What is $C[i+1,1]$?
– Somos
Jul 31 at 13:28
@Somos circle with centre (1,1) and radius 1
– BCLC
Jul 31 at 13:28
1
Please put that vital information into your question.
– Somos
Jul 31 at 13:30
@Somos Done. Thanks.
– BCLC
Jul 31 at 13:35
1
Why do you think you have gone wrong?
– Somos
Jul 31 at 13:39
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.15
$C[i+1,1] = $ is the circle with centre (1,1) and radius 1
(Part I) I'm going to try $g=[z,(2,1),(1,2),(0,1)], -g=[z,(2,1),(1,0),(0,1)]$. Please point out and explain any errors.
I observe that that $pm g$ map $C[i+1,1] to mathbb R$. By plugging in $z=(1+i)+e^i phi$, we can show that $pm g((1+i)+e^i phi) in mathbb R$. Then we extend $pm g$ to be defined for the singularity $z=z_3=i=(0,1)$.
Finally, the images are (WA: plus, minus)
$$pm g(1+i) = pm i$$
Did I go wrong anywhere, and why?
(Part II) Can I actually pick any 3 points on $C[i+1,1]$ to form a required cross-ratio?
By inputting in the 3 points used to make up a cross ratio, into the cross ratio, it seems that the cross ratio will output $0,1,infty$, soooo it looks like all cross ratios will map 3 distinct points on a circle to the real line.
$therefore,$ although the question asks for 2 Möbius transformations, there are uncountably $infty$ly many Möbius transformations/cross-ratios that work, so it seems?
So how will we compute the images $[i+1,z_1,z_2,z_3]$? Is it simply that with no particular pattern?
complex-analysis geometry transformation mobius-transformation cross-ratio
asked Jul 31 at 11:30
BCLC
7,00821973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.15
$C[i+1,1] = $ is the circle with centre (1,1) and radius 1
(Part I) I'm going to try $g=[z,(2,1),(1,2),(0,1)], -g=[z,(2,1),(1,0),(0,1)]$. Please point out and explain any errors.
I observe that that $pm g$ map $C[i+1,1] to mathbb R$. By plugging in $z=(1+i)+e^i phi$, we can show that $pm g((1+i)+e^i phi) in mathbb R$. Then we extend $pm g$ to be defined for the singularity $z=z_3=i=(0,1)$.
Finally, the images are (WA: plus, minus)
$$pm g(1+i) = pm i$$
Did I go wrong anywhere, and why?
(Part II) Can I actually pick any 3 points on $C[i+1,1]$ to form a required cross-ratio?
By inputting in the 3 points used to make up a cross ratio, into the cross ratio, it seems that the cross ratio will output $0,1,infty$, soooo it looks like all cross ratios will map 3 distinct points on a circle to the real line.
$therefore,$ although the question asks for 2 Möbius transformations, there are uncountably $infty$ly many Möbius transformations/cross-ratios that work, so it seems?
So how will we compute the images $[i+1,z_1,z_2,z_3]$? Is it simply that with no particular pattern?
complex-analysis geometry transformation mobius-transformation cross-ratio
asked Jul 31 at 11:30
BCLC
7,00821973
edited 1 hour ago
asked Jul 31 at 11:30
BCLC
7,00821973
asked Jul 31 at 11:30
BCLC
7,00821973
asked Jul 31 at 11:30
BCLC
7,00821973
7,00821973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-15 14:08:05Z">in 6 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-15 14:08:05Z">in 6 days.
This question has not received enough attention.
1
What is $C[i+1,1]$?
– Somos
Jul 31 at 13:28
@Somos circle with centre (1,1) and radius 1
– BCLC
Jul 31 at 13:28
1
Please put that vital information into your question.
– Somos
Jul 31 at 13:30
@Somos Done. Thanks.
– BCLC
Jul 31 at 13:35
1
Why do you think you have gone wrong?
– Somos
Jul 31 at 13:39
 |Â
show 2 more comments
1
What is $C[i+1,1]$?
– Somos
Jul 31 at 13:28
@Somos circle with centre (1,1) and radius 1
– BCLC
Jul 31 at 13:28
1
Please put that vital information into your question.
– Somos
Jul 31 at 13:30
@Somos Done. Thanks.
– BCLC
Jul 31 at 13:35
1
Why do you think you have gone wrong?
– Somos
Jul 31 at 13:39
1
1
What is $C[i+1,1]$?
– Somos
Jul 31 at 13:28
What is $C[i+1,1]$?
– Somos
Jul 31 at 13:28
@Somos circle with centre (1,1) and radius 1
– BCLC
Jul 31 at 13:28
@Somos circle with centre (1,1) and radius 1
– BCLC
Jul 31 at 13:28
1
1
Please put that vital information into your question.
– Somos
Jul 31 at 13:30
Please put that vital information into your question.
– Somos
Jul 31 at 13:30
@Somos Done. Thanks.
– BCLC
Jul 31 at 13:35
@Somos Done. Thanks.
– BCLC
Jul 31 at 13:35
1
1
Why do you think you have gone wrong?
– Somos
Jul 31 at 13:39
Why do you think you have gone wrong?
– Somos
Jul 31 at 13:39
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
Write the identity stating that the cross-ratio is preserved:
$$frac z - z_1 z - z_3 frac z_2 - z_3 z_2 - z_1 =
frac w - w_1 w - w_3 frac w_2 - w_3 w_2 - w_1.$$
Take $z_1, z_2, z_3$ on the circle and $w_1, w_2, w_3$ on the real line. If one of the $w_i$ is infinity, cancel the two factors containing this $w_i$. The function $w(z)$ is the answer.
answered 11 hours ago
Maxim
1,975112
Maxim, edited question. In re your answer 1. Will any 3 distinct points on the circle work? 2. How will we compute the images? Please and thank you!
– BCLC
2 hours ago
The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$.
– Maxim
1 hour ago
Maxim, it seems you answered (2). How about (1) please? My first attempt was to use specific points on the circle. But actually any 3 distinct will work?
– BCLC
1 hour ago
Take $w_1 = 0, w_2 = pm 1, w_3 = infty$. The rhs becomes $pm w$, therefore $w = pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice.
– Maxim
1 hour ago
and $ pm w$ is on the real line and thus $pm$ cross ratio works indeed for any 2 distinct points?
– BCLC
1 hour ago
 |Â
show 6 more comments
up vote
0
down vote
Part I is right.
Part II No! They have to be distinct. Otherwise, we don't have that they are Möbius transformations. Other than that, yes. Also, I don't see any particular pattern with the images.
answered 1 hour ago
BCLC
7,00821973
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Write the identity stating that the cross-ratio is preserved:
$$frac z - z_1 z - z_3 frac z_2 - z_3 z_2 - z_1 =
frac w - w_1 w - w_3 frac w_2 - w_3 w_2 - w_1.$$
Take $z_1, z_2, z_3$ on the circle and $w_1, w_2, w_3$ on the real line. If one of the $w_i$ is infinity, cancel the two factors containing this $w_i$. The function $w(z)$ is the answer.
answered 11 hours ago
Maxim
1,975112
Maxim, edited question. In re your answer 1. Will any 3 distinct points on the circle work? 2. How will we compute the images? Please and thank you!
– BCLC
2 hours ago
The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$.
– Maxim
1 hour ago
Maxim, it seems you answered (2). How about (1) please? My first attempt was to use specific points on the circle. But actually any 3 distinct will work?
– BCLC
1 hour ago
Take $w_1 = 0, w_2 = pm 1, w_3 = infty$. The rhs becomes $pm w$, therefore $w = pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice.
– Maxim
1 hour ago
and $ pm w$ is on the real line and thus $pm$ cross ratio works indeed for any 2 distinct points?
– BCLC
1 hour ago
 |Â
show 6 more comments
up vote
1
down vote
Write the identity stating that the cross-ratio is preserved:
$$frac z - z_1 z - z_3 frac z_2 - z_3 z_2 - z_1 =
frac w - w_1 w - w_3 frac w_2 - w_3 w_2 - w_1.$$
Take $z_1, z_2, z_3$ on the circle and $w_1, w_2, w_3$ on the real line. If one of the $w_i$ is infinity, cancel the two factors containing this $w_i$. The function $w(z)$ is the answer.
answered 11 hours ago
Maxim
1,975112
Maxim, edited question. In re your answer 1. Will any 3 distinct points on the circle work? 2. How will we compute the images? Please and thank you!
– BCLC
2 hours ago
The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$.
– Maxim
1 hour ago
Maxim, it seems you answered (2). How about (1) please? My first attempt was to use specific points on the circle. But actually any 3 distinct will work?
– BCLC
1 hour ago
Take $w_1 = 0, w_2 = pm 1, w_3 = infty$. The rhs becomes $pm w$, therefore $w = pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice.
– Maxim
1 hour ago
and $ pm w$ is on the real line and thus $pm$ cross ratio works indeed for any 2 distinct points?
– BCLC
1 hour ago
 |Â
show 6 more comments
up vote
1
down vote
up vote
1
down vote
Write the identity stating that the cross-ratio is preserved:
$$frac z - z_1 z - z_3 frac z_2 - z_3 z_2 - z_1 =
frac w - w_1 w - w_3 frac w_2 - w_3 w_2 - w_1.$$
Take $z_1, z_2, z_3$ on the circle and $w_1, w_2, w_3$ on the real line. If one of the $w_i$ is infinity, cancel the two factors containing this $w_i$. The function $w(z)$ is the answer.
answered 11 hours ago
Maxim
1,975112
Write the identity stating that the cross-ratio is preserved:
$$frac z - z_1 z - z_3 frac z_2 - z_3 z_2 - z_1 =
frac w - w_1 w - w_3 frac w_2 - w_3 w_2 - w_1.$$
Take $z_1, z_2, z_3$ on the circle and $w_1, w_2, w_3$ on the real line. If one of the $w_i$ is infinity, cancel the two factors containing this $w_i$. The function $w(z)$ is the answer.
answered 11 hours ago
Maxim
1,975112
answered 11 hours ago
Maxim
1,975112
answered 11 hours ago
Maxim
1,975112
answered 11 hours ago
Maxim
1,975112
1,975112
Maxim, edited question. In re your answer 1. Will any 3 distinct points on the circle work? 2. How will we compute the images? Please and thank you!
– BCLC
2 hours ago
The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$.
– Maxim
1 hour ago
Maxim, it seems you answered (2). How about (1) please? My first attempt was to use specific points on the circle. But actually any 3 distinct will work?
– BCLC
1 hour ago
Take $w_1 = 0, w_2 = pm 1, w_3 = infty$. The rhs becomes $pm w$, therefore $w = pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice.
– Maxim
1 hour ago
and $ pm w$ is on the real line and thus $pm$ cross ratio works indeed for any 2 distinct points?
– BCLC
1 hour ago
 |Â
show 6 more comments
Maxim, edited question. In re your answer 1. Will any 3 distinct points on the circle work? 2. How will we compute the images? Please and thank you!
– BCLC
2 hours ago
The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$.
– Maxim
1 hour ago
Maxim, it seems you answered (2). How about (1) please? My first attempt was to use specific points on the circle. But actually any 3 distinct will work?
– BCLC
1 hour ago
Take $w_1 = 0, w_2 = pm 1, w_3 = infty$. The rhs becomes $pm w$, therefore $w = pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice.
– Maxim
1 hour ago
and $ pm w$ is on the real line and thus $pm$ cross ratio works indeed for any 2 distinct points?
– BCLC
1 hour ago
Maxim, edited question. In re your answer 1. Will any 3 distinct points on the circle work? 2. How will we compute the images? Please and thank you!
– BCLC
2 hours ago
Maxim, edited question. In re your answer 1. Will any 3 distinct points on the circle work? 2. How will we compute the images? Please and thank you!
– BCLC
2 hours ago
The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$.
– Maxim
1 hour ago
The formula gives the (unique) linear-fractional transform mapping the three distinct points $z_i$ to the three distinct points $w_i$, which you can verify by substituting $z = z_i$.
– Maxim
1 hour ago
Maxim, it seems you answered (2). How about (1) please? My first attempt was to use specific points on the circle. But actually any 3 distinct will work?
– BCLC
1 hour ago
Maxim, it seems you answered (2). How about (1) please? My first attempt was to use specific points on the circle. But actually any 3 distinct will work?
– BCLC
1 hour ago
Take $w_1 = 0, w_2 = pm 1, w_3 = infty$. The rhs becomes $pm w$, therefore $w = pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice.
– Maxim
1 hour ago
Take $w_1 = 0, w_2 = pm 1, w_3 = infty$. The rhs becomes $pm w$, therefore $w = pm (z - z_1)/(z - z_3) (z_2 - z_3)/(z_2 - z_1)$ is a valid choice.
– Maxim
1 hour ago
and $ pm w$ is on the real line and thus $pm$ cross ratio works indeed for any 2 distinct points?
– BCLC
1 hour ago
and $ pm w$ is on the real line and thus $pm$ cross ratio works indeed for any 2 distinct points?
– BCLC
1 hour ago
 |Â
show 6 more comments
up vote
0
down vote
Part I is right.
Part II No! They have to be distinct. Otherwise, we don't have that they are Möbius transformations. Other than that, yes. Also, I don't see any particular pattern with the images.
answered 1 hour ago
BCLC
7,00821973
add a comment |Â
up vote
0
down vote
Part I is right.
Part II No! They have to be distinct. Otherwise, we don't have that they are Möbius transformations. Other than that, yes. Also, I don't see any particular pattern with the images.
answered 1 hour ago
BCLC
7,00821973
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Part I is right.
Part II No! They have to be distinct. Otherwise, we don't have that they are Möbius transformations. Other than that, yes. Also, I don't see any particular pattern with the images.
answered 1 hour ago
BCLC
7,00821973
Part I is right.
Part II No! They have to be distinct. Otherwise, we don't have that they are Möbius transformations. Other than that, yes. Also, I don't see any particular pattern with the images.
answered 1 hour ago
BCLC
7,00821973
answered 1 hour ago
BCLC
7,00821973
answered 1 hour ago
BCLC
7,00821973
answered 1 hour ago
BCLC
7,00821973
7,00821973
add a comment |Â
add a comment |Â
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1
What is $C[i+1,1]$?
– Somos
Jul 31 at 13:28
@Somos circle with centre (1,1) and radius 1
– BCLC
Jul 31 at 13:28
1
Please put that vital information into your question.
– Somos
Jul 31 at 13:30
@Somos Done. Thanks.
– BCLC
Jul 31 at 13:35
1
Why do you think you have gone wrong?
– Somos
Jul 31 at 13:39