Mixing
Explicit calculation of the center of a circle, image of a circle by a Möbius transformation
$begingroup$
It's a warm up calculation I decided to carry out while reading "PCT,Spin and statistics, and all that" by Streater and Wightmann. However I do not find what they have.
p.79 within the proof of Thm 2-14 p.77 (the calculation has not much to do with the proof, at least at this point. But if you are reading the book, notice that in the figure 2-7 p.79 they consider a function of z and $|u|neq 1$ while just above, it was a function of u on the unit circle...): let's consider the following Möbius transformation
$$ T: z mapsto frac{u+z}{1+uz} ,quad |u|neq 1$$
(otherwise the unit circle is mapped to $mathbb{R}$, as can be seen by a calculation analogous to the following)
The unit circle is mapped to another circle, whose center I wish to find. I recall that the inverse of a Möbius transformation (in particular, such maps are invertible...)
$$ S: z mapsto frac{az+b}{cz+d} , a,b,c,d in mathbb{C} quad text{is}quad S^{-1}: z mapsto frac{dz-b}{-cz+a}$$
so in our case (as can also be checked directly)
$$ T^{-1}: w mapsto frac{w-u}{1-uw}$$
Let's now write the condition for $z$ to be on the unit circle and see what conditions its image $w:=T(z)$ will then satisfy:
$$ |z|²= 1 quad Leftrightarrow quad |T^{-1}(w)|^2 =1 quad Leftrightarrow quad left(frac{w-u}{1-uw}right) overline{left(frac{w-u}{1-uw}right)}=1$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}(woverline{u}) + |u|^2 = 1 - 2, mathop{Re}(w u) + |uw|^2$$
$$ Leftrightarrow quad (1-|u|²)|w|^2 - 2, mathop{Re}(w(overline{u}-u)) + |u|^2 -1 = 0$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}left(w frac{2,i mathop{Im}(u)}{1-|u|²}right) -1 = 0$$
Identifying with the equation of a circle of center $cin mathbb{C}$ and radius $r in mathbb{R}_+$:
$$ |w-c|^2=r² quad Leftrightarrow quad |w|^2 - 2, mathop{Re}(w overline{c}) +|c|^2 - r² = 0$$
one obtains
$$ c=- frac{2,i mathop{Im}(u)}{1-|u|²} quad text{and}quad r= sqrt{1 + |c|²}$$
However, in the book it seems that they find
$$ c= frac{4 left[ u(1+|u|²)- (1+|u|²) mathop{Re}(u) right]}{left[ (1+|u|²)(1+u²) - 4 u mathop{Re}(u) right]}$$
So if a benevolent mind double checks the present calculation (or does something of its own), I'll be happy to discuss the result.
complex-numbers mobius-transformation
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
$endgroup$
add a comment |
$begingroup$
It's a warm up calculation I decided to carry out while reading "PCT,Spin and statistics, and all that" by Streater and Wightmann. However I do not find what they have.
p.79 within the proof of Thm 2-14 p.77 (the calculation has not much to do with the proof, at least at this point. But if you are reading the book, notice that in the figure 2-7 p.79 they consider a function of z and $|u|neq 1$ while just above, it was a function of u on the unit circle...): let's consider the following Möbius transformation
$$ T: z mapsto frac{u+z}{1+uz} ,quad |u|neq 1$$
(otherwise the unit circle is mapped to $mathbb{R}$, as can be seen by a calculation analogous to the following)
The unit circle is mapped to another circle, whose center I wish to find. I recall that the inverse of a Möbius transformation (in particular, such maps are invertible...)
$$ S: z mapsto frac{az+b}{cz+d} , a,b,c,d in mathbb{C} quad text{is}quad S^{-1}: z mapsto frac{dz-b}{-cz+a}$$
so in our case (as can also be checked directly)
$$ T^{-1}: w mapsto frac{w-u}{1-uw}$$
Let's now write the condition for $z$ to be on the unit circle and see what conditions its image $w:=T(z)$ will then satisfy:
$$ |z|²= 1 quad Leftrightarrow quad |T^{-1}(w)|^2 =1 quad Leftrightarrow quad left(frac{w-u}{1-uw}right) overline{left(frac{w-u}{1-uw}right)}=1$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}(woverline{u}) + |u|^2 = 1 - 2, mathop{Re}(w u) + |uw|^2$$
$$ Leftrightarrow quad (1-|u|²)|w|^2 - 2, mathop{Re}(w(overline{u}-u)) + |u|^2 -1 = 0$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}left(w frac{2,i mathop{Im}(u)}{1-|u|²}right) -1 = 0$$
Identifying with the equation of a circle of center $cin mathbb{C}$ and radius $r in mathbb{R}_+$:
$$ |w-c|^2=r² quad Leftrightarrow quad |w|^2 - 2, mathop{Re}(w overline{c}) +|c|^2 - r² = 0$$
one obtains
$$ c=- frac{2,i mathop{Im}(u)}{1-|u|²} quad text{and}quad r= sqrt{1 + |c|²}$$
However, in the book it seems that they find
$$ c= frac{4 left[ u(1+|u|²)- (1+|u|²) mathop{Re}(u) right]}{left[ (1+|u|²)(1+u²) - 4 u mathop{Re}(u) right]}$$
So if a benevolent mind double checks the present calculation (or does something of its own), I'll be happy to discuss the result.
complex-numbers mobius-transformation
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
$endgroup$
add a comment |
$begingroup$
It's a warm up calculation I decided to carry out while reading "PCT,Spin and statistics, and all that" by Streater and Wightmann. However I do not find what they have.
p.79 within the proof of Thm 2-14 p.77 (the calculation has not much to do with the proof, at least at this point. But if you are reading the book, notice that in the figure 2-7 p.79 they consider a function of z and $|u|neq 1$ while just above, it was a function of u on the unit circle...): let's consider the following Möbius transformation
$$ T: z mapsto frac{u+z}{1+uz} ,quad |u|neq 1$$
(otherwise the unit circle is mapped to $mathbb{R}$, as can be seen by a calculation analogous to the following)
The unit circle is mapped to another circle, whose center I wish to find. I recall that the inverse of a Möbius transformation (in particular, such maps are invertible...)
$$ S: z mapsto frac{az+b}{cz+d} , a,b,c,d in mathbb{C} quad text{is}quad S^{-1}: z mapsto frac{dz-b}{-cz+a}$$
so in our case (as can also be checked directly)
$$ T^{-1}: w mapsto frac{w-u}{1-uw}$$
Let's now write the condition for $z$ to be on the unit circle and see what conditions its image $w:=T(z)$ will then satisfy:
$$ |z|²= 1 quad Leftrightarrow quad |T^{-1}(w)|^2 =1 quad Leftrightarrow quad left(frac{w-u}{1-uw}right) overline{left(frac{w-u}{1-uw}right)}=1$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}(woverline{u}) + |u|^2 = 1 - 2, mathop{Re}(w u) + |uw|^2$$
$$ Leftrightarrow quad (1-|u|²)|w|^2 - 2, mathop{Re}(w(overline{u}-u)) + |u|^2 -1 = 0$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}left(w frac{2,i mathop{Im}(u)}{1-|u|²}right) -1 = 0$$
Identifying with the equation of a circle of center $cin mathbb{C}$ and radius $r in mathbb{R}_+$:
$$ |w-c|^2=r² quad Leftrightarrow quad |w|^2 - 2, mathop{Re}(w overline{c}) +|c|^2 - r² = 0$$
one obtains
$$ c=- frac{2,i mathop{Im}(u)}{1-|u|²} quad text{and}quad r= sqrt{1 + |c|²}$$
However, in the book it seems that they find
$$ c= frac{4 left[ u(1+|u|²)- (1+|u|²) mathop{Re}(u) right]}{left[ (1+|u|²)(1+u²) - 4 u mathop{Re}(u) right]}$$
So if a benevolent mind double checks the present calculation (or does something of its own), I'll be happy to discuss the result.
complex-numbers mobius-transformation
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
$endgroup$
It's a warm up calculation I decided to carry out while reading "PCT,Spin and statistics, and all that" by Streater and Wightmann. However I do not find what they have.
p.79 within the proof of Thm 2-14 p.77 (the calculation has not much to do with the proof, at least at this point. But if you are reading the book, notice that in the figure 2-7 p.79 they consider a function of z and $|u|neq 1$ while just above, it was a function of u on the unit circle...): let's consider the following Möbius transformation
$$ T: z mapsto frac{u+z}{1+uz} ,quad |u|neq 1$$
(otherwise the unit circle is mapped to $mathbb{R}$, as can be seen by a calculation analogous to the following)
The unit circle is mapped to another circle, whose center I wish to find. I recall that the inverse of a Möbius transformation (in particular, such maps are invertible...)
$$ S: z mapsto frac{az+b}{cz+d} , a,b,c,d in mathbb{C} quad text{is}quad S^{-1}: z mapsto frac{dz-b}{-cz+a}$$
so in our case (as can also be checked directly)
$$ T^{-1}: w mapsto frac{w-u}{1-uw}$$
Let's now write the condition for $z$ to be on the unit circle and see what conditions its image $w:=T(z)$ will then satisfy:
$$ |z|²= 1 quad Leftrightarrow quad |T^{-1}(w)|^2 =1 quad Leftrightarrow quad left(frac{w-u}{1-uw}right) overline{left(frac{w-u}{1-uw}right)}=1$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}(woverline{u}) + |u|^2 = 1 - 2, mathop{Re}(w u) + |uw|^2$$
$$ Leftrightarrow quad (1-|u|²)|w|^2 - 2, mathop{Re}(w(overline{u}-u)) + |u|^2 -1 = 0$$
$$ Leftrightarrow quad |w|^2 - 2, mathop{Re}left(w frac{2,i mathop{Im}(u)}{1-|u|²}right) -1 = 0$$
Identifying with the equation of a circle of center $cin mathbb{C}$ and radius $r in mathbb{R}_+$:
$$ |w-c|^2=r² quad Leftrightarrow quad |w|^2 - 2, mathop{Re}(w overline{c}) +|c|^2 - r² = 0$$
one obtains
$$ c=- frac{2,i mathop{Im}(u)}{1-|u|²} quad text{and}quad r= sqrt{1 + |c|²}$$
However, in the book it seems that they find
$$ c= frac{4 left[ u(1+|u|²)- (1+|u|²) mathop{Re}(u) right]}{left[ (1+|u|²)(1+u²) - 4 u mathop{Re}(u) right]}$$
So if a benevolent mind double checks the present calculation (or does something of its own), I'll be happy to discuss the result.
complex-numbers mobius-transformation
complex-numbers mobius-transformation
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
asked Nov 8 '18 at 16:09
Noix07Noix07
1,242922
1,242922
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I almost agree with your answer, except you made a mistake in the sign: $bar u-u=-2ioperatorname{Im}u$, not $2ioperatorname{Im}u$, so
$$
c=frac{2ioperatorname{Im}u}{1-lvert urvert^2}.
$$
Here is an alternative method.
Since $uneqpm 1$, the fixed points $z=T(z)$ is easily seen to be $z=pm 1$. Hence $T$ maps the unit circle to a circle containing $pm 1$, so the centre has to lie on the imaginary axis. Moreover, we compute derivative
$$
T'(z)=frac{1-u^2}{(1+uz)^2},quad T'(pm1)=frac{1mp u}{1pm u}
$$
So the unit circle (with normal direction $1$ at $pm 1$) is mapped to a circle whose normal direction at $pm 1$ is given by $frac{1mp u}{1pm u}$. Now
$$
frac{1mp u}{1pm u}=frac{(1mp u)(1pmbar u)}{(1pm u)(1pmbar u)}=
frac{(1-lvert urvert^2)mp (u-bar u)}{lvert 1pm urvert^2}
$$
So both normals intersect the imaginary axis at
$$
c=frac{(u-bar u)}{1-lvert urvert^2}.
$$
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
$endgroup$
add a comment |
$begingroup$
$-1/u$ is mapped to $infty$, therefore its conjugate point wrt the unit circle, which is $-overline u$, is mapped to the center of the image circle.
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
$endgroup$
$begingroup$
I didn't know this "rule", insteresting...
$endgroup$
– Noix07
Nov 12 '18 at 10:39
1
$begingroup$
It's not about the point at infinity specifically; Mobius transformations preserve conjugation.
$endgroup$
– Maxim
Nov 12 '18 at 12:14
$begingroup$
Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line"
$endgroup$
– Noix07
Nov 16 '18 at 13:32
add a comment |
$begingroup$
As shown in this answer, given the LFT $frac{z+u}{uz+1}$ and the circle with radius $1$ and center $0$, we find the antipodal points
$$
0pmfrac{0+1/u}{|0+1/u|}cdot1=pmfrac{|u|}{u}
$$
These points get mapped to antipodal points in the image of $frac{z+u}{uz+1}$:
$$
frac{u+frac{|u|}{u}}{1+|u|}qquadfrac{u-frac{|u|}{u}}{1-|u|}
$$
Therefore, the center is
$$newcommand{Im}{operatorname{Im}}
frac12left(frac{u+frac{|u|}{u}}{1+|u|}+frac{u-frac{|u|}{u}}{1-|u|}right)
=frac{2iIm(u)}{1-|u|^2}
$$
and the radius is
$$
frac12left|frac{u+frac{|u|}{u}}{1+|u|}-frac{u-frac{|u|}{u}}{1-|u|}right|
=left|frac{1-u^2}{1-|u|^2}right|
$$
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2990177%2fexplicit-calculation-of-the-center-of-a-circle-image-of-a-circle-by-a-m%25c3%25b6bius-tr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I almost agree with your answer, except you made a mistake in the sign: $bar u-u=-2ioperatorname{Im}u$, not $2ioperatorname{Im}u$, so
$$
c=frac{2ioperatorname{Im}u}{1-lvert urvert^2}.
$$
Here is an alternative method.
Since $uneqpm 1$, the fixed points $z=T(z)$ is easily seen to be $z=pm 1$. Hence $T$ maps the unit circle to a circle containing $pm 1$, so the centre has to lie on the imaginary axis. Moreover, we compute derivative
$$
T'(z)=frac{1-u^2}{(1+uz)^2},quad T'(pm1)=frac{1mp u}{1pm u}
$$
So the unit circle (with normal direction $1$ at $pm 1$) is mapped to a circle whose normal direction at $pm 1$ is given by $frac{1mp u}{1pm u}$. Now
$$
frac{1mp u}{1pm u}=frac{(1mp u)(1pmbar u)}{(1pm u)(1pmbar u)}=
frac{(1-lvert urvert^2)mp (u-bar u)}{lvert 1pm urvert^2}
$$
So both normals intersect the imaginary axis at
$$
c=frac{(u-bar u)}{1-lvert urvert^2}.
$$
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
$endgroup$
add a comment |
$begingroup$
I almost agree with your answer, except you made a mistake in the sign: $bar u-u=-2ioperatorname{Im}u$, not $2ioperatorname{Im}u$, so
$$
c=frac{2ioperatorname{Im}u}{1-lvert urvert^2}.
$$
Here is an alternative method.
Since $uneqpm 1$, the fixed points $z=T(z)$ is easily seen to be $z=pm 1$. Hence $T$ maps the unit circle to a circle containing $pm 1$, so the centre has to lie on the imaginary axis. Moreover, we compute derivative
$$
T'(z)=frac{1-u^2}{(1+uz)^2},quad T'(pm1)=frac{1mp u}{1pm u}
$$
So the unit circle (with normal direction $1$ at $pm 1$) is mapped to a circle whose normal direction at $pm 1$ is given by $frac{1mp u}{1pm u}$. Now
$$
frac{1mp u}{1pm u}=frac{(1mp u)(1pmbar u)}{(1pm u)(1pmbar u)}=
frac{(1-lvert urvert^2)mp (u-bar u)}{lvert 1pm urvert^2}
$$
So both normals intersect the imaginary axis at
$$
c=frac{(u-bar u)}{1-lvert urvert^2}.
$$
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
$endgroup$
add a comment |
$begingroup$
I almost agree with your answer, except you made a mistake in the sign: $bar u-u=-2ioperatorname{Im}u$, not $2ioperatorname{Im}u$, so
$$
c=frac{2ioperatorname{Im}u}{1-lvert urvert^2}.
$$
Here is an alternative method.
Since $uneqpm 1$, the fixed points $z=T(z)$ is easily seen to be $z=pm 1$. Hence $T$ maps the unit circle to a circle containing $pm 1$, so the centre has to lie on the imaginary axis. Moreover, we compute derivative
$$
T'(z)=frac{1-u^2}{(1+uz)^2},quad T'(pm1)=frac{1mp u}{1pm u}
$$
So the unit circle (with normal direction $1$ at $pm 1$) is mapped to a circle whose normal direction at $pm 1$ is given by $frac{1mp u}{1pm u}$. Now
$$
frac{1mp u}{1pm u}=frac{(1mp u)(1pmbar u)}{(1pm u)(1pmbar u)}=
frac{(1-lvert urvert^2)mp (u-bar u)}{lvert 1pm urvert^2}
$$
So both normals intersect the imaginary axis at
$$
c=frac{(u-bar u)}{1-lvert urvert^2}.
$$
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
$endgroup$
I almost agree with your answer, except you made a mistake in the sign: $bar u-u=-2ioperatorname{Im}u$, not $2ioperatorname{Im}u$, so
$$
c=frac{2ioperatorname{Im}u}{1-lvert urvert^2}.
$$
Here is an alternative method.
Since $uneqpm 1$, the fixed points $z=T(z)$ is easily seen to be $z=pm 1$. Hence $T$ maps the unit circle to a circle containing $pm 1$, so the centre has to lie on the imaginary axis. Moreover, we compute derivative
$$
T'(z)=frac{1-u^2}{(1+uz)^2},quad T'(pm1)=frac{1mp u}{1pm u}
$$
So the unit circle (with normal direction $1$ at $pm 1$) is mapped to a circle whose normal direction at $pm 1$ is given by $frac{1mp u}{1pm u}$. Now
$$
frac{1mp u}{1pm u}=frac{(1mp u)(1pmbar u)}{(1pm u)(1pmbar u)}=
frac{(1-lvert urvert^2)mp (u-bar u)}{lvert 1pm urvert^2}
$$
So both normals intersect the imaginary axis at
$$
c=frac{(u-bar u)}{1-lvert urvert^2}.
$$
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
edited Nov 9 '18 at 16:36
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
answered Nov 8 '18 at 20:24
user10354138user10354138
7,4422925
7,4422925
add a comment |
add a comment |
$begingroup$
$-1/u$ is mapped to $infty$, therefore its conjugate point wrt the unit circle, which is $-overline u$, is mapped to the center of the image circle.
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
$endgroup$
$begingroup$
I didn't know this "rule", insteresting...
$endgroup$
– Noix07
Nov 12 '18 at 10:39
1
$begingroup$
It's not about the point at infinity specifically; Mobius transformations preserve conjugation.
$endgroup$
– Maxim
Nov 12 '18 at 12:14
$begingroup$
Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line"
$endgroup$
– Noix07
Nov 16 '18 at 13:32
add a comment |
$begingroup$
$-1/u$ is mapped to $infty$, therefore its conjugate point wrt the unit circle, which is $-overline u$, is mapped to the center of the image circle.
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
$endgroup$
$begingroup$
I didn't know this "rule", insteresting...
$endgroup$
– Noix07
Nov 12 '18 at 10:39
1
$begingroup$
It's not about the point at infinity specifically; Mobius transformations preserve conjugation.
$endgroup$
– Maxim
Nov 12 '18 at 12:14
$begingroup$
Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line"
$endgroup$
– Noix07
Nov 16 '18 at 13:32
add a comment |
$begingroup$
$-1/u$ is mapped to $infty$, therefore its conjugate point wrt the unit circle, which is $-overline u$, is mapped to the center of the image circle.
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
$endgroup$
$-1/u$ is mapped to $infty$, therefore its conjugate point wrt the unit circle, which is $-overline u$, is mapped to the center of the image circle.
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
answered Nov 11 '18 at 16:09
MaximMaxim
5,5431219
5,5431219
$begingroup$
I didn't know this "rule", insteresting...
$endgroup$
– Noix07
Nov 12 '18 at 10:39
1
$begingroup$
It's not about the point at infinity specifically; Mobius transformations preserve conjugation.
$endgroup$
– Maxim
Nov 12 '18 at 12:14
$begingroup$
Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line"
$endgroup$
– Noix07
Nov 16 '18 at 13:32
add a comment |
$begingroup$
I didn't know this "rule", insteresting...
$endgroup$
– Noix07
Nov 12 '18 at 10:39
1
$begingroup$
It's not about the point at infinity specifically; Mobius transformations preserve conjugation.
$endgroup$
– Maxim
Nov 12 '18 at 12:14
$begingroup$
Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line"
$endgroup$
– Noix07
Nov 16 '18 at 13:32
$begingroup$
I didn't know this "rule", insteresting...
$endgroup$
– Noix07
Nov 12 '18 at 10:39
$begingroup$
I didn't know this "rule", insteresting...
$endgroup$
– Noix07
Nov 12 '18 at 10:39
1
1
$begingroup$
It's not about the point at infinity specifically; Mobius transformations preserve conjugation.
$endgroup$
– Maxim
Nov 12 '18 at 12:14
$begingroup$
It's not about the point at infinity specifically; Mobius transformations preserve conjugation.
$endgroup$
– Maxim
Nov 12 '18 at 12:14
$begingroup$
Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line"
$endgroup$
– Noix07
Nov 16 '18 at 13:32
$begingroup$
Thanks, got it. To make a useful comment, for those like me who did not know about these stuffs, "preserve conjugation" means "if two points are conjugate w.r.t. a circle or a line, then there image are conjugate w.r.t. the image circle/line"
$endgroup$
– Noix07
Nov 16 '18 at 13:32
add a comment |
$begingroup$
As shown in this answer, given the LFT $frac{z+u}{uz+1}$ and the circle with radius $1$ and center $0$, we find the antipodal points
$$
0pmfrac{0+1/u}{|0+1/u|}cdot1=pmfrac{|u|}{u}
$$
These points get mapped to antipodal points in the image of $frac{z+u}{uz+1}$:
$$
frac{u+frac{|u|}{u}}{1+|u|}qquadfrac{u-frac{|u|}{u}}{1-|u|}
$$
Therefore, the center is
$$newcommand{Im}{operatorname{Im}}
frac12left(frac{u+frac{|u|}{u}}{1+|u|}+frac{u-frac{|u|}{u}}{1-|u|}right)
=frac{2iIm(u)}{1-|u|^2}
$$
and the radius is
$$
frac12left|frac{u+frac{|u|}{u}}{1+|u|}-frac{u-frac{|u|}{u}}{1-|u|}right|
=left|frac{1-u^2}{1-|u|^2}right|
$$
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
$begingroup$
As shown in this answer, given the LFT $frac{z+u}{uz+1}$ and the circle with radius $1$ and center $0$, we find the antipodal points
$$
0pmfrac{0+1/u}{|0+1/u|}cdot1=pmfrac{|u|}{u}
$$
These points get mapped to antipodal points in the image of $frac{z+u}{uz+1}$:
$$
frac{u+frac{|u|}{u}}{1+|u|}qquadfrac{u-frac{|u|}{u}}{1-|u|}
$$
Therefore, the center is
$$newcommand{Im}{operatorname{Im}}
frac12left(frac{u+frac{|u|}{u}}{1+|u|}+frac{u-frac{|u|}{u}}{1-|u|}right)
=frac{2iIm(u)}{1-|u|^2}
$$
and the radius is
$$
frac12left|frac{u+frac{|u|}{u}}{1+|u|}-frac{u-frac{|u|}{u}}{1-|u|}right|
=left|frac{1-u^2}{1-|u|^2}right|
$$
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
$endgroup$
add a comment |
$begingroup$
As shown in this answer, given the LFT $frac{z+u}{uz+1}$ and the circle with radius $1$ and center $0$, we find the antipodal points
$$
0pmfrac{0+1/u}{|0+1/u|}cdot1=pmfrac{|u|}{u}
$$
These points get mapped to antipodal points in the image of $frac{z+u}{uz+1}$:
$$
frac{u+frac{|u|}{u}}{1+|u|}qquadfrac{u-frac{|u|}{u}}{1-|u|}
$$
Therefore, the center is
$$newcommand{Im}{operatorname{Im}}
frac12left(frac{u+frac{|u|}{u}}{1+|u|}+frac{u-frac{|u|}{u}}{1-|u|}right)
=frac{2iIm(u)}{1-|u|^2}
$$
and the radius is
$$
frac12left|frac{u+frac{|u|}{u}}{1+|u|}-frac{u-frac{|u|}{u}}{1-|u|}right|
=left|frac{1-u^2}{1-|u|^2}right|
$$
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
$endgroup$
As shown in this answer, given the LFT $frac{z+u}{uz+1}$ and the circle with radius $1$ and center $0$, we find the antipodal points
$$
0pmfrac{0+1/u}{|0+1/u|}cdot1=pmfrac{|u|}{u}
$$
These points get mapped to antipodal points in the image of $frac{z+u}{uz+1}$:
$$
frac{u+frac{|u|}{u}}{1+|u|}qquadfrac{u-frac{|u|}{u}}{1-|u|}
$$
Therefore, the center is
$$newcommand{Im}{operatorname{Im}}
frac12left(frac{u+frac{|u|}{u}}{1+|u|}+frac{u-frac{|u|}{u}}{1-|u|}right)
=frac{2iIm(u)}{1-|u|^2}
$$
and the radius is
$$
frac12left|frac{u+frac{|u|}{u}}{1+|u|}-frac{u-frac{|u|}{u}}{1-|u|}right|
=left|frac{1-u^2}{1-|u|^2}right|
$$
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
answered Jan 12 at 12:29
robjohn♦robjohn
267k27308632
267k27308632
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2990177%2fexplicit-calculation-of-the-center-of-a-circle-image-of-a-circle-by-a-m%25c3%25b6bius-tr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
