Mixing
What is the radius and center of the image of $|z|=1$ under $ f(z) = frac{3z+2}{4z+3}$?
$begingroup$
I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.
The Möbius transformation can be turned into inversion as well:
- $C_1= 4|z|^2+3overline{z}-3z-2 $
- $C_2 =|z|^2 - 1$
Or we could turn the second circle into a fractional lineartrasnforation $g(z) = - frac{1}{z}$. Then I could multiply the two transformations:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
left[ begin{array}{cc} 0 & 1 \ 1 & 0 end{array} right]
=
left[ begin{array}{cc} 2 & 3 \ 3 & 4 end{array} right]
$$
and this could turn back into a circle:
- $ C_1C_2 = 3|z|^2 + 4 overline{z} + 3z + 2 $
I found this technique in an somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and Circles can be identified.
complex-numbers conformal-geometry mobius-transformation inversive-geometry geometric-transformation
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
$endgroup$
add a comment |
$begingroup$
I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.
The Möbius transformation can be turned into inversion as well:
- $C_1= 4|z|^2+3overline{z}-3z-2 $
- $C_2 =|z|^2 - 1$
Or we could turn the second circle into a fractional lineartrasnforation $g(z) = - frac{1}{z}$. Then I could multiply the two transformations:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
left[ begin{array}{cc} 0 & 1 \ 1 & 0 end{array} right]
=
left[ begin{array}{cc} 2 & 3 \ 3 & 4 end{array} right]
$$
and this could turn back into a circle:
- $ C_1C_2 = 3|z|^2 + 4 overline{z} + 3z + 2 $
I found this technique in an somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and Circles can be identified.
complex-numbers conformal-geometry mobius-transformation inversive-geometry geometric-transformation
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
$endgroup$
$begingroup$
Why is accepted solution more usefull than mine?
$endgroup$
– greedoid
Jan 12 at 9:58
add a comment |
$begingroup$
I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.
The Möbius transformation can be turned into inversion as well:
- $C_1= 4|z|^2+3overline{z}-3z-2 $
- $C_2 =|z|^2 - 1$
Or we could turn the second circle into a fractional lineartrasnforation $g(z) = - frac{1}{z}$. Then I could multiply the two transformations:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
left[ begin{array}{cc} 0 & 1 \ 1 & 0 end{array} right]
=
left[ begin{array}{cc} 2 & 3 \ 3 & 4 end{array} right]
$$
and this could turn back into a circle:
- $ C_1C_2 = 3|z|^2 + 4 overline{z} + 3z + 2 $
I found this technique in an somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and Circles can be identified.
complex-numbers conformal-geometry mobius-transformation inversive-geometry geometric-transformation
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
$endgroup$
I would like to compute the image of the circle $|z|=1$ about the fractional linear transformation:
$$ f(z) = frac{3z+2}{4z+3} $$
In particular, I'd like to compute the new center and radius.
The Möbius transformation can be turned into inversion as well:
- $C_1= 4|z|^2+3overline{z}-3z-2 $
- $C_2 =|z|^2 - 1$
Or we could turn the second circle into a fractional lineartrasnforation $g(z) = - frac{1}{z}$. Then I could multiply the two transformations:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
left[ begin{array}{cc} 0 & 1 \ 1 & 0 end{array} right]
=
left[ begin{array}{cc} 2 & 3 \ 3 & 4 end{array} right]
$$
and this could turn back into a circle:
- $ C_1C_2 = 3|z|^2 + 4 overline{z} + 3z + 2 $
I found this technique in an somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and Circles can be identified.
complex-numbers conformal-geometry mobius-transformation inversive-geometry geometric-transformation
complex-numbers conformal-geometry mobius-transformation inversive-geometry geometric-transformation
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
edited Jan 11 at 6:02
Eric Wofsey
185k13213339
185k13213339
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
asked Jan 10 at 14:56
cactus314cactus314
15.4k42269
15.4k42269
$begingroup$
Why is accepted solution more usefull than mine?
$endgroup$
– greedoid
Jan 12 at 9:58
add a comment |
$begingroup$
Why is accepted solution more usefull than mine?
$endgroup$
– greedoid
Jan 12 at 9:58
$begingroup$
Why is accepted solution more usefull than mine?
$endgroup$
– greedoid
Jan 12 at 9:58
$begingroup$
Why is accepted solution more usefull than mine?
$endgroup$
– greedoid
Jan 12 at 9:58
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
I would take three points on $mid zmid=1$ and see where they go. As noted in @greedoid's answer, we have $f(1)=frac57,,f(-1)=1$ and $f(i)=frac{18+i}{25}$.
Since these points are not colinear, the image is indeed a circle.
So, if $z$ is the center, we have: $mid z-1mid=mid z-frac57mid=mid z-frac{18+i}{25}mid=r$.
This leads via a little algebra to $z=frac67$. Thus $r=frac17$.
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
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add a comment |
$begingroup$
Here is an automatic procedure: first invert the relation $w=f(z)$, then apply the condition $|z|=1$ to the inverse formula $z=g(w)$ to deduce an equation of the image set.
In the present case, $w=f(z)$ means that $$w=frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=frac{2-3w}{4w-3}$$
Thus, the image of the circle has equation $$left|frac{2-3w}{4w-3}right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+bar w)+9|w|^2=16|w|^2-12(w+bar w)+9$$ that is, $$7|w|^2-6(w+bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.
Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...
answered Jan 10 at 16:13
DidDid
247k23223460
$endgroup$
3
$begingroup$
+1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-frac67(w+bar{w})+frac57=0$. Now "complete the square" to get $|w|^2-frac67(w+bar{w})+frac{36}{49}=frac{36}{49}-frac57=frac1{49}$. That's $left(w-frac67right)left(bar{w}-frac67right)=frac1{49}$. Now take the square root: $|w-frac67|=frac17$.
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– alex.jordan
Jan 11 at 6:34
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@alex.jordan Indeed. Well done and thanks.
$endgroup$
– Did
Jan 11 at 9:02
add a comment |
$begingroup$
Let see where this transformation takes $1,-1$ and $i$:
begin{eqnarray}
1&longmapsto &{5over 7}\-1&longmapsto &1\i&longmapsto &{18+iover 25}\
end{eqnarray}
Now calculate the center and radius of a triangle on $alpha ={5over 7}$, $beta =1$ and $gamma ={18+iover 25}$.
Since this triangle is right at $gamma$ we see that midpoint of segment $alpha beta$, that is $sigma = {6over 7}$ is a center of new circle with $r = {1over 7}$.
edited Jan 10 at 22:41
Botond
5,7282732
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
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add a comment |
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Using Inversive Geometry
For a given LFT $frac{az+b}{cz+d}$ and circle of radius $r$ centered at $k$, the antipodal points of that source circle
$$
kpmfrac{k+d/c}{|k+d/c|}rtag1
$$
get mapped by the LFT to antipodal points of the image circle.
This is because these points are on the line containing the center of the circle, $k$, and the center of the inversion, $-d/c$. Any line through the center of the inversion is mapped to a line, and since that line is perpendicular to the source circle at the points of intersection, the image line is perpendicular to the image circle; that is, they intersect at antipodal points.
If $c=0$ (the LFT is simply affine) or $k+d/c=0$ (the center of the source circle is the center of the inversion), then any two antipodal points get mapped to antipodal points, so replace $frac{k+d/c}{|k+d/c|}$ with any point on the unit circle in $mathbb{C}$.
If one of the points computed in $(1)$ equals $-frac dc$ (that is, that point is mapped to $infty$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line.
Given a pair of antipodal points on a circle, ${p_1,p_2}$, the radius, r, and center, k, of that circle are given by
$$
r=frac{|p_1-p_2|}2qquad k=frac{p_1+p_2}2tag2
$$
Application
In this case, we have $frac{3z+2}{4z+3}$, $k=0$, and $r=1$. Therefore, $(1)$ says that the antipodal points of the source circle
$$
0pmfrac{0+3/4}{|0+3/4|}cdot1={-1,1}tag3
$$
get mapped by the LFT to the antipodal points of the image circle
$$
left{1,frac57right}tag4
$$
then $(2)$ says that the radius, $r$, and the center, $k$, of the image circle are
$$
bbox[5px,border:2px solid #C0A000]{r=frac17qquad k=frac67}tag5
$$
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
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I've applied this method to this answer and this answer.
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– robjohn♦
Jan 12 at 4:32
add a comment |
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The image circle is symmetric wrt the real axis, therefore $[f(1), f(-1)] = [5/7, 1]$ is a diameter.
answered Jan 12 at 7:41
MaximMaxim
5,1981219
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Here's another solution I was able to find. Notice the matrix factorization:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
=
left[ begin{array}{cc} frac{1}{5} & 0\ 0 & 5 end{array} right] times
left[ begin{array}{cc} 1 & 18 \ 0 & 1 end{array} right] times
left[ begin{array}{cr} frac{3}{5}& -frac{4}{5}\ frac{4}{5}& frac{3}{5} end{array} right] = A times B times C
$$
The geometry behind this is that we have a Möbius transformation that factors into three parts:
$$ text{Möbius} = rotation times translation times dilation $$
Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = pm 1$ and $ z = i$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:
$$ (-1,1) stackrel{C}{to} (7, - frac{1}{7}) stackrel{B}{to} (25,frac{125}{7})stackrel{A}{to} (1, frac{5}{7}) $$
This corresponds to a circle centered at $z = frac{6}{7}$ with radius $frac{1}{7}$.
One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, vec{u}) = (i, (1,0)) in T_1(mathbb{H}) $.
A Möbius transformation on $mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(mathbb{H})$ like this:
$$
left[ z mapsto frac{az+b}{cz+d} right] to left[
(z, vec{u}) mapsto left( frac{az+b}{cz+d} , frac{vec{u}}{(cz+d)^2} right) right] $$
Let's see what happens when I try the previous example here:
$$ big(i, (1,0)big) mapsto left( frac{3i+2}{4i+3}, frac{(1,0)}{(4i+3)^2}right) = left( frac{18+i}{25} , frac{1}{25}(24,-7) right) $$
The factor of $frac{1}{25}$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $vec{u}$ would be tangent to a semi-circle with radius in the direction $vec{u}_perp$ passing through the point $f(z)=(frac{18}{25}, frac{1}{25})$. Therefore the center would be:
$$ (frac{18}{25}, frac{1}{25}) + frac{1}{7 times 25}(24,-7) = (frac{1}{7},0) $$
agreeing with the previous answer.
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
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6 Answers
6
active
oldest
votes
6 Answers
6
active
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$begingroup$
I would take three points on $mid zmid=1$ and see where they go. As noted in @greedoid's answer, we have $f(1)=frac57,,f(-1)=1$ and $f(i)=frac{18+i}{25}$.
Since these points are not colinear, the image is indeed a circle.
So, if $z$ is the center, we have: $mid z-1mid=mid z-frac57mid=mid z-frac{18+i}{25}mid=r$.
This leads via a little algebra to $z=frac67$. Thus $r=frac17$.
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
$endgroup$
add a comment |
$begingroup$
I would take three points on $mid zmid=1$ and see where they go. As noted in @greedoid's answer, we have $f(1)=frac57,,f(-1)=1$ and $f(i)=frac{18+i}{25}$.
Since these points are not colinear, the image is indeed a circle.
So, if $z$ is the center, we have: $mid z-1mid=mid z-frac57mid=mid z-frac{18+i}{25}mid=r$.
This leads via a little algebra to $z=frac67$. Thus $r=frac17$.
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
$endgroup$
add a comment |
$begingroup$
I would take three points on $mid zmid=1$ and see where they go. As noted in @greedoid's answer, we have $f(1)=frac57,,f(-1)=1$ and $f(i)=frac{18+i}{25}$.
Since these points are not colinear, the image is indeed a circle.
So, if $z$ is the center, we have: $mid z-1mid=mid z-frac57mid=mid z-frac{18+i}{25}mid=r$.
This leads via a little algebra to $z=frac67$. Thus $r=frac17$.
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
$endgroup$
I would take three points on $mid zmid=1$ and see where they go. As noted in @greedoid's answer, we have $f(1)=frac57,,f(-1)=1$ and $f(i)=frac{18+i}{25}$.
Since these points are not colinear, the image is indeed a circle.
So, if $z$ is the center, we have: $mid z-1mid=mid z-frac57mid=mid z-frac{18+i}{25}mid=r$.
This leads via a little algebra to $z=frac67$. Thus $r=frac17$.
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
answered Jan 11 at 7:48
Chris CusterChris Custer
12.7k3825
12.7k3825
add a comment |
add a comment |
$begingroup$
Here is an automatic procedure: first invert the relation $w=f(z)$, then apply the condition $|z|=1$ to the inverse formula $z=g(w)$ to deduce an equation of the image set.
In the present case, $w=f(z)$ means that $$w=frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=frac{2-3w}{4w-3}$$
Thus, the image of the circle has equation $$left|frac{2-3w}{4w-3}right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+bar w)+9|w|^2=16|w|^2-12(w+bar w)+9$$ that is, $$7|w|^2-6(w+bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.
Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...
answered Jan 10 at 16:13
DidDid
247k23223460
$endgroup$
3
$begingroup$
+1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-frac67(w+bar{w})+frac57=0$. Now "complete the square" to get $|w|^2-frac67(w+bar{w})+frac{36}{49}=frac{36}{49}-frac57=frac1{49}$. That's $left(w-frac67right)left(bar{w}-frac67right)=frac1{49}$. Now take the square root: $|w-frac67|=frac17$.
$endgroup$
– alex.jordan
Jan 11 at 6:34
$begingroup$
@alex.jordan Indeed. Well done and thanks.
$endgroup$
– Did
Jan 11 at 9:02
add a comment |
$begingroup$
Here is an automatic procedure: first invert the relation $w=f(z)$, then apply the condition $|z|=1$ to the inverse formula $z=g(w)$ to deduce an equation of the image set.
In the present case, $w=f(z)$ means that $$w=frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=frac{2-3w}{4w-3}$$
Thus, the image of the circle has equation $$left|frac{2-3w}{4w-3}right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+bar w)+9|w|^2=16|w|^2-12(w+bar w)+9$$ that is, $$7|w|^2-6(w+bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.
Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...
answered Jan 10 at 16:13
DidDid
247k23223460
$endgroup$
3
$begingroup$
+1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-frac67(w+bar{w})+frac57=0$. Now "complete the square" to get $|w|^2-frac67(w+bar{w})+frac{36}{49}=frac{36}{49}-frac57=frac1{49}$. That's $left(w-frac67right)left(bar{w}-frac67right)=frac1{49}$. Now take the square root: $|w-frac67|=frac17$.
$endgroup$
– alex.jordan
Jan 11 at 6:34
$begingroup$
@alex.jordan Indeed. Well done and thanks.
$endgroup$
– Did
Jan 11 at 9:02
add a comment |
$begingroup$
Here is an automatic procedure: first invert the relation $w=f(z)$, then apply the condition $|z|=1$ to the inverse formula $z=g(w)$ to deduce an equation of the image set.
In the present case, $w=f(z)$ means that $$w=frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=frac{2-3w}{4w-3}$$
Thus, the image of the circle has equation $$left|frac{2-3w}{4w-3}right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+bar w)+9|w|^2=16|w|^2-12(w+bar w)+9$$ that is, $$7|w|^2-6(w+bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.
Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...
answered Jan 10 at 16:13
DidDid
247k23223460
$endgroup$
Here is an automatic procedure: first invert the relation $w=f(z)$, then apply the condition $|z|=1$ to the inverse formula $z=g(w)$ to deduce an equation of the image set.
In the present case, $w=f(z)$ means that $$w=frac{3z+2}{4z+3}$$ that is, $(4z+3)w=3z+2$, that is, $(4w-3)z=2-3w$, that is, $$z=frac{2-3w}{4w-3}$$
Thus, the image of the circle has equation $$left|frac{2-3w}{4w-3}right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+bar w)+9|w|^2=16|w|^2-12(w+bar w)+9$$ that is, $$7|w|^2-6(w+bar w)+5=0$$ and finally, if $w=x+iy$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.
Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...
answered Jan 10 at 16:13
DidDid
247k23223460
edited Jan 12 at 9:41
answered Jan 10 at 16:13
DidDid
247k23223460
answered Jan 10 at 16:13
DidDid
247k23223460
answered Jan 10 at 16:13
DidDid
247k23223460
247k23223460
3
$begingroup$
+1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-frac67(w+bar{w})+frac57=0$. Now "complete the square" to get $|w|^2-frac67(w+bar{w})+frac{36}{49}=frac{36}{49}-frac57=frac1{49}$. That's $left(w-frac67right)left(bar{w}-frac67right)=frac1{49}$. Now take the square root: $|w-frac67|=frac17$.
$endgroup$
– alex.jordan
Jan 11 at 6:34
$begingroup$
@alex.jordan Indeed. Well done and thanks.
$endgroup$
– Did
Jan 11 at 9:02
add a comment |
3
$begingroup$
+1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-frac67(w+bar{w})+frac57=0$. Now "complete the square" to get $|w|^2-frac67(w+bar{w})+frac{36}{49}=frac{36}{49}-frac57=frac1{49}$. That's $left(w-frac67right)left(bar{w}-frac67right)=frac1{49}$. Now take the square root: $|w-frac67|=frac17$.
$endgroup$
– alex.jordan
Jan 11 at 6:34
$begingroup$
@alex.jordan Indeed. Well done and thanks.
$endgroup$
– Did
Jan 11 at 9:02
3
3
$begingroup$
+1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-frac67(w+bar{w})+frac57=0$. Now "complete the square" to get $|w|^2-frac67(w+bar{w})+frac{36}{49}=frac{36}{49}-frac57=frac1{49}$. That's $left(w-frac67right)left(bar{w}-frac67right)=frac1{49}$. Now take the square root: $|w-frac67|=frac17$.
$endgroup$
– alex.jordan
Jan 11 at 6:34
$begingroup$
+1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-frac67(w+bar{w})+frac57=0$. Now "complete the square" to get $|w|^2-frac67(w+bar{w})+frac{36}{49}=frac{36}{49}-frac57=frac1{49}$. That's $left(w-frac67right)left(bar{w}-frac67right)=frac1{49}$. Now take the square root: $|w-frac67|=frac17$.
$endgroup$
– alex.jordan
Jan 11 at 6:34
$begingroup$
@alex.jordan Indeed. Well done and thanks.
$endgroup$
– Did
Jan 11 at 9:02
$begingroup$
@alex.jordan Indeed. Well done and thanks.
$endgroup$
– Did
Jan 11 at 9:02
add a comment |
$begingroup$
Let see where this transformation takes $1,-1$ and $i$:
begin{eqnarray}
1&longmapsto &{5over 7}\-1&longmapsto &1\i&longmapsto &{18+iover 25}\
end{eqnarray}
Now calculate the center and radius of a triangle on $alpha ={5over 7}$, $beta =1$ and $gamma ={18+iover 25}$.
Since this triangle is right at $gamma$ we see that midpoint of segment $alpha beta$, that is $sigma = {6over 7}$ is a center of new circle with $r = {1over 7}$.
edited Jan 10 at 22:41
Botond
5,7282732
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
$endgroup$
add a comment |
$begingroup$
Let see where this transformation takes $1,-1$ and $i$:
begin{eqnarray}
1&longmapsto &{5over 7}\-1&longmapsto &1\i&longmapsto &{18+iover 25}\
end{eqnarray}
Now calculate the center and radius of a triangle on $alpha ={5over 7}$, $beta =1$ and $gamma ={18+iover 25}$.
Since this triangle is right at $gamma$ we see that midpoint of segment $alpha beta$, that is $sigma = {6over 7}$ is a center of new circle with $r = {1over 7}$.
edited Jan 10 at 22:41
Botond
5,7282732
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
$endgroup$
add a comment |
$begingroup$
Let see where this transformation takes $1,-1$ and $i$:
begin{eqnarray}
1&longmapsto &{5over 7}\-1&longmapsto &1\i&longmapsto &{18+iover 25}\
end{eqnarray}
Now calculate the center and radius of a triangle on $alpha ={5over 7}$, $beta =1$ and $gamma ={18+iover 25}$.
Since this triangle is right at $gamma$ we see that midpoint of segment $alpha beta$, that is $sigma = {6over 7}$ is a center of new circle with $r = {1over 7}$.
edited Jan 10 at 22:41
Botond
5,7282732
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
$endgroup$
Let see where this transformation takes $1,-1$ and $i$:
begin{eqnarray}
1&longmapsto &{5over 7}\-1&longmapsto &1\i&longmapsto &{18+iover 25}\
end{eqnarray}
Now calculate the center and radius of a triangle on $alpha ={5over 7}$, $beta =1$ and $gamma ={18+iover 25}$.
Since this triangle is right at $gamma$ we see that midpoint of segment $alpha beta$, that is $sigma = {6over 7}$ is a center of new circle with $r = {1over 7}$.
edited Jan 10 at 22:41
Botond
5,7282732
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
edited Jan 10 at 22:41
Botond
5,7282732
edited Jan 10 at 22:41
Botond
5,7282732
edited Jan 10 at 22:41
Botond
5,7282732
5,7282732
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
answered Jan 10 at 16:02
greedoidgreedoid
41.1k1149101
41.1k1149101
add a comment |
add a comment |
$begingroup$
Using Inversive Geometry
For a given LFT $frac{az+b}{cz+d}$ and circle of radius $r$ centered at $k$, the antipodal points of that source circle
$$
kpmfrac{k+d/c}{|k+d/c|}rtag1
$$
get mapped by the LFT to antipodal points of the image circle.
This is because these points are on the line containing the center of the circle, $k$, and the center of the inversion, $-d/c$. Any line through the center of the inversion is mapped to a line, and since that line is perpendicular to the source circle at the points of intersection, the image line is perpendicular to the image circle; that is, they intersect at antipodal points.
If $c=0$ (the LFT is simply affine) or $k+d/c=0$ (the center of the source circle is the center of the inversion), then any two antipodal points get mapped to antipodal points, so replace $frac{k+d/c}{|k+d/c|}$ with any point on the unit circle in $mathbb{C}$.
If one of the points computed in $(1)$ equals $-frac dc$ (that is, that point is mapped to $infty$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line.
Given a pair of antipodal points on a circle, ${p_1,p_2}$, the radius, r, and center, k, of that circle are given by
$$
r=frac{|p_1-p_2|}2qquad k=frac{p_1+p_2}2tag2
$$
Application
In this case, we have $frac{3z+2}{4z+3}$, $k=0$, and $r=1$. Therefore, $(1)$ says that the antipodal points of the source circle
$$
0pmfrac{0+3/4}{|0+3/4|}cdot1={-1,1}tag3
$$
get mapped by the LFT to the antipodal points of the image circle
$$
left{1,frac57right}tag4
$$
then $(2)$ says that the radius, $r$, and the center, $k$, of the image circle are
$$
bbox[5px,border:2px solid #C0A000]{r=frac17qquad k=frac67}tag5
$$
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
$endgroup$
$begingroup$
I've applied this method to this answer and this answer.
$endgroup$
– robjohn♦
Jan 12 at 4:32
add a comment |
$begingroup$
Using Inversive Geometry
For a given LFT $frac{az+b}{cz+d}$ and circle of radius $r$ centered at $k$, the antipodal points of that source circle
$$
kpmfrac{k+d/c}{|k+d/c|}rtag1
$$
get mapped by the LFT to antipodal points of the image circle.
This is because these points are on the line containing the center of the circle, $k$, and the center of the inversion, $-d/c$. Any line through the center of the inversion is mapped to a line, and since that line is perpendicular to the source circle at the points of intersection, the image line is perpendicular to the image circle; that is, they intersect at antipodal points.
If $c=0$ (the LFT is simply affine) or $k+d/c=0$ (the center of the source circle is the center of the inversion), then any two antipodal points get mapped to antipodal points, so replace $frac{k+d/c}{|k+d/c|}$ with any point on the unit circle in $mathbb{C}$.
If one of the points computed in $(1)$ equals $-frac dc$ (that is, that point is mapped to $infty$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line.
Given a pair of antipodal points on a circle, ${p_1,p_2}$, the radius, r, and center, k, of that circle are given by
$$
r=frac{|p_1-p_2|}2qquad k=frac{p_1+p_2}2tag2
$$
Application
In this case, we have $frac{3z+2}{4z+3}$, $k=0$, and $r=1$. Therefore, $(1)$ says that the antipodal points of the source circle
$$
0pmfrac{0+3/4}{|0+3/4|}cdot1={-1,1}tag3
$$
get mapped by the LFT to the antipodal points of the image circle
$$
left{1,frac57right}tag4
$$
then $(2)$ says that the radius, $r$, and the center, $k$, of the image circle are
$$
bbox[5px,border:2px solid #C0A000]{r=frac17qquad k=frac67}tag5
$$
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
$endgroup$
$begingroup$
I've applied this method to this answer and this answer.
$endgroup$
– robjohn♦
Jan 12 at 4:32
add a comment |
$begingroup$
Using Inversive Geometry
For a given LFT $frac{az+b}{cz+d}$ and circle of radius $r$ centered at $k$, the antipodal points of that source circle
$$
kpmfrac{k+d/c}{|k+d/c|}rtag1
$$
get mapped by the LFT to antipodal points of the image circle.
This is because these points are on the line containing the center of the circle, $k$, and the center of the inversion, $-d/c$. Any line through the center of the inversion is mapped to a line, and since that line is perpendicular to the source circle at the points of intersection, the image line is perpendicular to the image circle; that is, they intersect at antipodal points.
If $c=0$ (the LFT is simply affine) or $k+d/c=0$ (the center of the source circle is the center of the inversion), then any two antipodal points get mapped to antipodal points, so replace $frac{k+d/c}{|k+d/c|}$ with any point on the unit circle in $mathbb{C}$.
If one of the points computed in $(1)$ equals $-frac dc$ (that is, that point is mapped to $infty$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line.
Given a pair of antipodal points on a circle, ${p_1,p_2}$, the radius, r, and center, k, of that circle are given by
$$
r=frac{|p_1-p_2|}2qquad k=frac{p_1+p_2}2tag2
$$
Application
In this case, we have $frac{3z+2}{4z+3}$, $k=0$, and $r=1$. Therefore, $(1)$ says that the antipodal points of the source circle
$$
0pmfrac{0+3/4}{|0+3/4|}cdot1={-1,1}tag3
$$
get mapped by the LFT to the antipodal points of the image circle
$$
left{1,frac57right}tag4
$$
then $(2)$ says that the radius, $r$, and the center, $k$, of the image circle are
$$
bbox[5px,border:2px solid #C0A000]{r=frac17qquad k=frac67}tag5
$$
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
$endgroup$
Using Inversive Geometry
For a given LFT $frac{az+b}{cz+d}$ and circle of radius $r$ centered at $k$, the antipodal points of that source circle
$$
kpmfrac{k+d/c}{|k+d/c|}rtag1
$$
get mapped by the LFT to antipodal points of the image circle.
This is because these points are on the line containing the center of the circle, $k$, and the center of the inversion, $-d/c$. Any line through the center of the inversion is mapped to a line, and since that line is perpendicular to the source circle at the points of intersection, the image line is perpendicular to the image circle; that is, they intersect at antipodal points.
If $c=0$ (the LFT is simply affine) or $k+d/c=0$ (the center of the source circle is the center of the inversion), then any two antipodal points get mapped to antipodal points, so replace $frac{k+d/c}{|k+d/c|}$ with any point on the unit circle in $mathbb{C}$.
If one of the points computed in $(1)$ equals $-frac dc$ (that is, that point is mapped to $infty$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line.
Given a pair of antipodal points on a circle, ${p_1,p_2}$, the radius, r, and center, k, of that circle are given by
$$
r=frac{|p_1-p_2|}2qquad k=frac{p_1+p_2}2tag2
$$
Application
In this case, we have $frac{3z+2}{4z+3}$, $k=0$, and $r=1$. Therefore, $(1)$ says that the antipodal points of the source circle
$$
0pmfrac{0+3/4}{|0+3/4|}cdot1={-1,1}tag3
$$
get mapped by the LFT to the antipodal points of the image circle
$$
left{1,frac57right}tag4
$$
then $(2)$ says that the radius, $r$, and the center, $k$, of the image circle are
$$
bbox[5px,border:2px solid #C0A000]{r=frac17qquad k=frac67}tag5
$$
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
edited Jan 11 at 21:27
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
answered Jan 11 at 5:51
robjohn♦robjohn
267k27308631
267k27308631
$begingroup$
I've applied this method to this answer and this answer.
$endgroup$
– robjohn♦
Jan 12 at 4:32
add a comment |
$begingroup$
I've applied this method to this answer and this answer.
$endgroup$
– robjohn♦
Jan 12 at 4:32
$begingroup$
I've applied this method to this answer and this answer.
$endgroup$
– robjohn♦
Jan 12 at 4:32
$begingroup$
I've applied this method to this answer and this answer.
$endgroup$
– robjohn♦
Jan 12 at 4:32
add a comment |
$begingroup$
The image circle is symmetric wrt the real axis, therefore $[f(1), f(-1)] = [5/7, 1]$ is a diameter.
answered Jan 12 at 7:41
MaximMaxim
5,1981219
$endgroup$
add a comment |
$begingroup$
The image circle is symmetric wrt the real axis, therefore $[f(1), f(-1)] = [5/7, 1]$ is a diameter.
answered Jan 12 at 7:41
MaximMaxim
5,1981219
$endgroup$
add a comment |
$begingroup$
The image circle is symmetric wrt the real axis, therefore $[f(1), f(-1)] = [5/7, 1]$ is a diameter.
answered Jan 12 at 7:41
MaximMaxim
5,1981219
$endgroup$
The image circle is symmetric wrt the real axis, therefore $[f(1), f(-1)] = [5/7, 1]$ is a diameter.
answered Jan 12 at 7:41
MaximMaxim
5,1981219
answered Jan 12 at 7:41
MaximMaxim
5,1981219
answered Jan 12 at 7:41
MaximMaxim
5,1981219
answered Jan 12 at 7:41
MaximMaxim
5,1981219
5,1981219
add a comment |
add a comment |
$begingroup$
Here's another solution I was able to find. Notice the matrix factorization:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
=
left[ begin{array}{cc} frac{1}{5} & 0\ 0 & 5 end{array} right] times
left[ begin{array}{cc} 1 & 18 \ 0 & 1 end{array} right] times
left[ begin{array}{cr} frac{3}{5}& -frac{4}{5}\ frac{4}{5}& frac{3}{5} end{array} right] = A times B times C
$$
The geometry behind this is that we have a Möbius transformation that factors into three parts:
$$ text{Möbius} = rotation times translation times dilation $$
Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = pm 1$ and $ z = i$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:
$$ (-1,1) stackrel{C}{to} (7, - frac{1}{7}) stackrel{B}{to} (25,frac{125}{7})stackrel{A}{to} (1, frac{5}{7}) $$
This corresponds to a circle centered at $z = frac{6}{7}$ with radius $frac{1}{7}$.
One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, vec{u}) = (i, (1,0)) in T_1(mathbb{H}) $.
A Möbius transformation on $mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(mathbb{H})$ like this:
$$
left[ z mapsto frac{az+b}{cz+d} right] to left[
(z, vec{u}) mapsto left( frac{az+b}{cz+d} , frac{vec{u}}{(cz+d)^2} right) right] $$
Let's see what happens when I try the previous example here:
$$ big(i, (1,0)big) mapsto left( frac{3i+2}{4i+3}, frac{(1,0)}{(4i+3)^2}right) = left( frac{18+i}{25} , frac{1}{25}(24,-7) right) $$
The factor of $frac{1}{25}$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $vec{u}$ would be tangent to a semi-circle with radius in the direction $vec{u}_perp$ passing through the point $f(z)=(frac{18}{25}, frac{1}{25})$. Therefore the center would be:
$$ (frac{18}{25}, frac{1}{25}) + frac{1}{7 times 25}(24,-7) = (frac{1}{7},0) $$
agreeing with the previous answer.
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
$endgroup$
add a comment |
$begingroup$
Here's another solution I was able to find. Notice the matrix factorization:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
=
left[ begin{array}{cc} frac{1}{5} & 0\ 0 & 5 end{array} right] times
left[ begin{array}{cc} 1 & 18 \ 0 & 1 end{array} right] times
left[ begin{array}{cr} frac{3}{5}& -frac{4}{5}\ frac{4}{5}& frac{3}{5} end{array} right] = A times B times C
$$
The geometry behind this is that we have a Möbius transformation that factors into three parts:
$$ text{Möbius} = rotation times translation times dilation $$
Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = pm 1$ and $ z = i$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:
$$ (-1,1) stackrel{C}{to} (7, - frac{1}{7}) stackrel{B}{to} (25,frac{125}{7})stackrel{A}{to} (1, frac{5}{7}) $$
This corresponds to a circle centered at $z = frac{6}{7}$ with radius $frac{1}{7}$.
One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, vec{u}) = (i, (1,0)) in T_1(mathbb{H}) $.
A Möbius transformation on $mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(mathbb{H})$ like this:
$$
left[ z mapsto frac{az+b}{cz+d} right] to left[
(z, vec{u}) mapsto left( frac{az+b}{cz+d} , frac{vec{u}}{(cz+d)^2} right) right] $$
Let's see what happens when I try the previous example here:
$$ big(i, (1,0)big) mapsto left( frac{3i+2}{4i+3}, frac{(1,0)}{(4i+3)^2}right) = left( frac{18+i}{25} , frac{1}{25}(24,-7) right) $$
The factor of $frac{1}{25}$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $vec{u}$ would be tangent to a semi-circle with radius in the direction $vec{u}_perp$ passing through the point $f(z)=(frac{18}{25}, frac{1}{25})$. Therefore the center would be:
$$ (frac{18}{25}, frac{1}{25}) + frac{1}{7 times 25}(24,-7) = (frac{1}{7},0) $$
agreeing with the previous answer.
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
$endgroup$
add a comment |
$begingroup$
Here's another solution I was able to find. Notice the matrix factorization:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
=
left[ begin{array}{cc} frac{1}{5} & 0\ 0 & 5 end{array} right] times
left[ begin{array}{cc} 1 & 18 \ 0 & 1 end{array} right] times
left[ begin{array}{cr} frac{3}{5}& -frac{4}{5}\ frac{4}{5}& frac{3}{5} end{array} right] = A times B times C
$$
The geometry behind this is that we have a Möbius transformation that factors into three parts:
$$ text{Möbius} = rotation times translation times dilation $$
Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = pm 1$ and $ z = i$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:
$$ (-1,1) stackrel{C}{to} (7, - frac{1}{7}) stackrel{B}{to} (25,frac{125}{7})stackrel{A}{to} (1, frac{5}{7}) $$
This corresponds to a circle centered at $z = frac{6}{7}$ with radius $frac{1}{7}$.
One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, vec{u}) = (i, (1,0)) in T_1(mathbb{H}) $.
A Möbius transformation on $mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(mathbb{H})$ like this:
$$
left[ z mapsto frac{az+b}{cz+d} right] to left[
(z, vec{u}) mapsto left( frac{az+b}{cz+d} , frac{vec{u}}{(cz+d)^2} right) right] $$
Let's see what happens when I try the previous example here:
$$ big(i, (1,0)big) mapsto left( frac{3i+2}{4i+3}, frac{(1,0)}{(4i+3)^2}right) = left( frac{18+i}{25} , frac{1}{25}(24,-7) right) $$
The factor of $frac{1}{25}$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $vec{u}$ would be tangent to a semi-circle with radius in the direction $vec{u}_perp$ passing through the point $f(z)=(frac{18}{25}, frac{1}{25})$. Therefore the center would be:
$$ (frac{18}{25}, frac{1}{25}) + frac{1}{7 times 25}(24,-7) = (frac{1}{7},0) $$
agreeing with the previous answer.
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
$endgroup$
Here's another solution I was able to find. Notice the matrix factorization:
$$
left[ begin{array}{cc} 3 & 2 \ 4 & 3 end{array} right]
=
left[ begin{array}{cc} frac{1}{5} & 0\ 0 & 5 end{array} right] times
left[ begin{array}{cc} 1 & 18 \ 0 & 1 end{array} right] times
left[ begin{array}{cr} frac{3}{5}& -frac{4}{5}\ frac{4}{5}& frac{3}{5} end{array} right] = A times B times C
$$
The geometry behind this is that we have a Möbius transformation that factors into three parts:
$$ text{Möbius} = rotation times translation times dilation $$
Now we have that $|z|=1$ is a circle centered at the origin passing through the points $z = pm 1$ and $ z = i$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:
$$ (-1,1) stackrel{C}{to} (7, - frac{1}{7}) stackrel{B}{to} (25,frac{125}{7})stackrel{A}{to} (1, frac{5}{7}) $$
This corresponds to a circle centered at $z = frac{6}{7}$ with radius $frac{1}{7}$.
One possibility for computing this image circle is to notice the circle $|z|=1$ is a geodesic curve in the upper-half plane (with metric $ds^2 = frac{dx^2 +dy^2}{y^2}$) and passing through the point $(z, vec{u}) = (i, (1,0)) in T_1(mathbb{H}) $.
A Möbius transformation on $mathbb{H}$ can be "lifted" to a Möbius transformation on $T_1(mathbb{H})$ like this:
$$
left[ z mapsto frac{az+b}{cz+d} right] to left[
(z, vec{u}) mapsto left( frac{az+b}{cz+d} , frac{vec{u}}{(cz+d)^2} right) right] $$
Let's see what happens when I try the previous example here:
$$ big(i, (1,0)big) mapsto left( frac{3i+2}{4i+3}, frac{(1,0)}{(4i+3)^2}right) = left( frac{18+i}{25} , frac{1}{25}(24,-7) right) $$
The factor of $frac{1}{25}$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $vec{u}$ would be tangent to a semi-circle with radius in the direction $vec{u}_perp$ passing through the point $f(z)=(frac{18}{25}, frac{1}{25})$. Therefore the center would be:
$$ (frac{18}{25}, frac{1}{25}) + frac{1}{7 times 25}(24,-7) = (frac{1}{7},0) $$
agreeing with the previous answer.
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
edited Jan 14 at 22:30
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
answered Jan 12 at 3:32
cactus314cactus314
15.4k42269
15.4k42269
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$begingroup$
Why is accepted solution more usefull than mine?
$endgroup$
– greedoid
Jan 12 at 9:58