Find a center and radius of circle that is an image of Mobius Transformation of real axis
I need to find a center and radius of a circle that is an image of real axis under homography
$$ h(z)= frac{z-z_1}{z-z_2} $$
I found out that homography preserves symetric points, therefore because
$$ infty quad and quad 0 $$ are symmetric with respect to generalised circle (real axis) then their images are also symmetric with respect to image of real axis
$$ h(infty)=1 quad and quad h(0)=frac{z_1}{z_2} $$
Any hints or solutions would be appreciated.
complex-analysis complex-numbers mobius-transformation
asked Nov 28 at 9:20
Pablo
996
add a comment |
I need to find a center and radius of a circle that is an image of real axis under homography
$$ h(z)= frac{z-z_1}{z-z_2} $$
I found out that homography preserves symetric points, therefore because
$$ infty quad and quad 0 $$ are symmetric with respect to generalised circle (real axis) then their images are also symmetric with respect to image of real axis
$$ h(infty)=1 quad and quad h(0)=frac{z_1}{z_2} $$
Any hints or solutions would be appreciated.
complex-analysis complex-numbers mobius-transformation
asked Nov 28 at 9:20
Pablo
996
add a comment |
I need to find a center and radius of a circle that is an image of real axis under homography
$$ h(z)= frac{z-z_1}{z-z_2} $$
I found out that homography preserves symetric points, therefore because
$$ infty quad and quad 0 $$ are symmetric with respect to generalised circle (real axis) then their images are also symmetric with respect to image of real axis
$$ h(infty)=1 quad and quad h(0)=frac{z_1}{z_2} $$
Any hints or solutions would be appreciated.
complex-analysis complex-numbers mobius-transformation
asked Nov 28 at 9:20
Pablo
996
I need to find a center and radius of a circle that is an image of real axis under homography
$$ h(z)= frac{z-z_1}{z-z_2} $$
I found out that homography preserves symetric points, therefore because
$$ infty quad and quad 0 $$ are symmetric with respect to generalised circle (real axis) then their images are also symmetric with respect to image of real axis
$$ h(infty)=1 quad and quad h(0)=frac{z_1}{z_2} $$
Any hints or solutions would be appreciated.
complex-analysis complex-numbers mobius-transformation
complex-analysis complex-numbers mobius-transformation
asked Nov 28 at 9:20
Pablo
996
asked Nov 28 at 9:20
Pablo
996
asked Nov 28 at 9:20
Pablo
996
asked Nov 28 at 9:20
Pablo
996
asked Nov 28 at 9:20
Pablo
996
996
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The inverse funtion to $f$ is equal $$g(z) =h^{-1} (z) =frac{z_2 -z_1}{z-1}$$ thus $$f({z: mbox{Im}z =0} ) =g^{-1} ({z: mbox{Im}z =0})$$
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
add a comment |
That's not quite right. $0$ is symmetric to itself wrt the real line, and ditto for $infty$. Take the point $z_2$ instead. First consider what happens when $z_2 in mathbb R$. If $z_2 notin mathbb R$, then you can exploit the symmetry to find the point which is mapped to the center of the circle. The distance between the center and the image of $infty$ is the radius.
answered Nov 30 at 15:28
Maxim
4,433219
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
The inverse funtion to $f$ is equal $$g(z) =h^{-1} (z) =frac{z_2 -z_1}{z-1}$$ thus $$f({z: mbox{Im}z =0} ) =g^{-1} ({z: mbox{Im}z =0})$$
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
add a comment |
The inverse funtion to $f$ is equal $$g(z) =h^{-1} (z) =frac{z_2 -z_1}{z-1}$$ thus $$f({z: mbox{Im}z =0} ) =g^{-1} ({z: mbox{Im}z =0})$$
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
add a comment |
The inverse funtion to $f$ is equal $$g(z) =h^{-1} (z) =frac{z_2 -z_1}{z-1}$$ thus $$f({z: mbox{Im}z =0} ) =g^{-1} ({z: mbox{Im}z =0})$$
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
The inverse funtion to $f$ is equal $$g(z) =h^{-1} (z) =frac{z_2 -z_1}{z-1}$$ thus $$f({z: mbox{Im}z =0} ) =g^{-1} ({z: mbox{Im}z =0})$$
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
answered Nov 28 at 9:49
MotylaNogaTomkaMazura
6,579917
6,579917
add a comment |
add a comment |
That's not quite right. $0$ is symmetric to itself wrt the real line, and ditto for $infty$. Take the point $z_2$ instead. First consider what happens when $z_2 in mathbb R$. If $z_2 notin mathbb R$, then you can exploit the symmetry to find the point which is mapped to the center of the circle. The distance between the center and the image of $infty$ is the radius.
answered Nov 30 at 15:28
Maxim
4,433219
add a comment |
That's not quite right. $0$ is symmetric to itself wrt the real line, and ditto for $infty$. Take the point $z_2$ instead. First consider what happens when $z_2 in mathbb R$. If $z_2 notin mathbb R$, then you can exploit the symmetry to find the point which is mapped to the center of the circle. The distance between the center and the image of $infty$ is the radius.
answered Nov 30 at 15:28
Maxim
4,433219
add a comment |
That's not quite right. $0$ is symmetric to itself wrt the real line, and ditto for $infty$. Take the point $z_2$ instead. First consider what happens when $z_2 in mathbb R$. If $z_2 notin mathbb R$, then you can exploit the symmetry to find the point which is mapped to the center of the circle. The distance between the center and the image of $infty$ is the radius.
answered Nov 30 at 15:28
Maxim
4,433219
That's not quite right. $0$ is symmetric to itself wrt the real line, and ditto for $infty$. Take the point $z_2$ instead. First consider what happens when $z_2 in mathbb R$. If $z_2 notin mathbb R$, then you can exploit the symmetry to find the point which is mapped to the center of the circle. The distance between the center and the image of $infty$ is the radius.
answered Nov 30 at 15:28
Maxim
4,433219
answered Nov 30 at 15:28
Maxim
4,433219
answered Nov 30 at 15:28
Maxim
4,433219
answered Nov 30 at 15:28
Maxim
4,433219
4,433219
add a comment |
add a comment |
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