Trouble in mapping of möbius transformationFind Möbius transformation that send Re(z)=Im(z) to a circle and the real axis to itselfMöbius transformation mapping problemMapping the upper half plane to unit discFind Möbius transformation fixing $i$,mapping im-axis to im-axis and $vert z-ivert < 2$ to the upper half plane,Show that the image of the extended real line under any Möbius transformation is either a circle or a straight line.Constructing Möbius transformationMobius transformation produces either a circle or a line…Is this a simpler proof that all transformations mapping D(1:0) onto D(1;0) are of the form $;;e^ilambdafracz-alphabaralphaz-1$Mobius Transformation viewed as a mapping on $bar mathbb C$Finding the image of the unit disc under a Mobius transform
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Trouble in mapping of möbius transformation
Find Möbius transformation that send Re(z)=Im(z) to a circle and the real axis to itselfMöbius transformation mapping problemMapping the upper half plane to unit discFind Möbius transformation fixing $i$,mapping im-axis to im-axis and $vert z-ivert < 2$ to the upper half plane,Show that the image of the extended real line under any Möbius transformation is either a circle or a straight line.Constructing Möbius transformationMobius transformation produces either a circle or a line…Is this a simpler proof that all transformations mapping D(1:0) onto D(1;0) are of the form $;;e^ilambdafracz-alphabaralphaz-1$Mobius Transformation viewed as a mapping on $bar mathbb C$Finding the image of the unit disc under a Mobius transform
$begingroup$
Question:-
Show that the transformation $$ w = frac2z+3z-4$$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$
My attempt:-
The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$
So the inverse mapping of the given bilinear transformation is:-
$$z= frac4w+3w-2 $$
Now substituting the value of $z$ in $(1)$
$$frac3w+1 =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3viota|=2|u-2+viota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
On solving these it appears as
$$5u^2+28u-12=0$$
I can not come at the conclusion as stated in question, is my method correct ?
Suggestions are highly appreciated
Thankyou
transformation mobius-transformation geometric-transformation
edited 53 mins ago
Maria Mazur
52.2k1363131
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
$endgroup$
|
show 1 more comment
$begingroup$
Question:-
Show that the transformation $$ w = frac2z+3z-4$$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$
My attempt:-
The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$
So the inverse mapping of the given bilinear transformation is:-
$$z= frac4w+3w-2 $$
Now substituting the value of $z$ in $(1)$
$$frac3w+1 =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3viota|=2|u-2+viota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
On solving these it appears as
$$5u^2+28u-12=0$$
I can not come at the conclusion as stated in question, is my method correct ?
Suggestions are highly appreciated
Thankyou
transformation mobius-transformation geometric-transformation
edited 53 mins ago
Maria Mazur
52.2k1363131
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
$endgroup$
$begingroup$
The circle is centered at $(2,0)$ though.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I apologize for that mistake. Although i have corrected that but still the problem is same
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Nice points, like $(0,0),(4,0)$ and $(2,2)$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I will try this method
$endgroup$
– Vedant Chourey
4 hours ago
|
show 1 more comment
$begingroup$
Question:-
Show that the transformation $$ w = frac2z+3z-4$$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$
My attempt:-
The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$
So the inverse mapping of the given bilinear transformation is:-
$$z= frac4w+3w-2 $$
Now substituting the value of $z$ in $(1)$
$$frac3w+1 =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3viota|=2|u-2+viota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
On solving these it appears as
$$5u^2+28u-12=0$$
I can not come at the conclusion as stated in question, is my method correct ?
Suggestions are highly appreciated
Thankyou
transformation mobius-transformation geometric-transformation
edited 53 mins ago
Maria Mazur
52.2k1363131
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
$endgroup$
Question:-
Show that the transformation $$ w = frac2z+3z-4$$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$
My attempt:-
The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$
So the inverse mapping of the given bilinear transformation is:-
$$z= frac4w+3w-2 $$
Now substituting the value of $z$ in $(1)$
$$frac3w+1 =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3viota|=2|u-2+viota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
On solving these it appears as
$$5u^2+28u-12=0$$
I can not come at the conclusion as stated in question, is my method correct ?
Suggestions are highly appreciated
Thankyou
transformation mobius-transformation geometric-transformation
transformation mobius-transformation geometric-transformation
edited 53 mins ago
Maria Mazur
52.2k1363131
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
edited 53 mins ago
Maria Mazur
52.2k1363131
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
edited 53 mins ago
Maria Mazur
52.2k1363131
edited 53 mins ago
Maria Mazur
52.2k1363131
edited 53 mins ago
Maria Mazur
52.2k1363131
52.2k1363131
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
asked 4 hours ago
Vedant ChoureyVedant Chourey
1096
1096
$begingroup$
The circle is centered at $(2,0)$ though.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I apologize for that mistake. Although i have corrected that but still the problem is same
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Nice points, like $(0,0),(4,0)$ and $(2,2)$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I will try this method
$endgroup$
– Vedant Chourey
4 hours ago
|
show 1 more comment
$begingroup$
The circle is centered at $(2,0)$ though.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I apologize for that mistake. Although i have corrected that but still the problem is same
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Nice points, like $(0,0),(4,0)$ and $(2,2)$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I will try this method
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
The circle is centered at $(2,0)$ though.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
The circle is centered at $(2,0)$ though.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I apologize for that mistake. Although i have corrected that but still the problem is same
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
I apologize for that mistake. Although i have corrected that but still the problem is same
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Nice points, like $(0,0),(4,0)$ and $(2,2)$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Nice points, like $(0,0),(4,0)$ and $(2,2)$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I will try this method
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
I will try this method
$endgroup$
– Vedant Chourey
4 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$Big|Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-3over 4$ or $$4w+3=0$$
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
$endgroup$
$begingroup$
Thankyou. For explanation
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Fell free to upvote and accept the answer if you think it was usefull answer to you.
$endgroup$
– Maria Mazur
4 hours ago
$begingroup$
Ok. I'll take care of this next time
$endgroup$
– Vedant Chourey
4 hours ago
add a comment |
$begingroup$
$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.
The result follows.
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$Big|Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-3over 4$ or $$4w+3=0$$
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
$endgroup$
$begingroup$
Thankyou. For explanation
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Fell free to upvote and accept the answer if you think it was usefull answer to you.
$endgroup$
– Maria Mazur
4 hours ago
$begingroup$
Ok. I'll take care of this next time
$endgroup$
– Vedant Chourey
4 hours ago
add a comment |
$begingroup$
If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$Big|Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-3over 4$ or $$4w+3=0$$
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
$endgroup$
$begingroup$
Thankyou. For explanation
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Fell free to upvote and accept the answer if you think it was usefull answer to you.
$endgroup$
– Maria Mazur
4 hours ago
$begingroup$
Ok. I'll take care of this next time
$endgroup$
– Vedant Chourey
4 hours ago
add a comment |
$begingroup$
If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$Big|Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-3over 4$ or $$4w+3=0$$
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
$endgroup$
If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$Big|Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-3over 4$ or $$4w+3=0$$
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
edited 4 hours ago
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
answered 4 hours ago
Maria MazurMaria Mazur
52.2k1363131
52.2k1363131
$begingroup$
Thankyou. For explanation
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Fell free to upvote and accept the answer if you think it was usefull answer to you.
$endgroup$
– Maria Mazur
4 hours ago
$begingroup$
Ok. I'll take care of this next time
$endgroup$
– Vedant Chourey
4 hours ago
add a comment |
$begingroup$
Thankyou. For explanation
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Fell free to upvote and accept the answer if you think it was usefull answer to you.
$endgroup$
– Maria Mazur
4 hours ago
$begingroup$
Ok. I'll take care of this next time
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Thankyou. For explanation
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Thankyou. For explanation
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Fell free to upvote and accept the answer if you think it was usefull answer to you.
$endgroup$
– Maria Mazur
4 hours ago
$begingroup$
Fell free to upvote and accept the answer if you think it was usefull answer to you.
$endgroup$
– Maria Mazur
4 hours ago
$begingroup$
Ok. I'll take care of this next time
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Ok. I'll take care of this next time
$endgroup$
– Vedant Chourey
4 hours ago
add a comment |
$begingroup$
$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.
The result follows.
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
$endgroup$
add a comment |
$begingroup$
$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.
The result follows.
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
$endgroup$
add a comment |
$begingroup$
$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.
The result follows.
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
$endgroup$
$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.
The result follows.
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
edited 3 hours ago
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
answered 3 hours ago
Chris CusterChris Custer
15.5k3827
15.5k3827
add a comment |
add a comment |
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$begingroup$
The circle is centered at $(2,0)$ though.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I apologize for that mistake. Although i have corrected that but still the problem is same
$endgroup$
– Vedant Chourey
4 hours ago
$begingroup$
Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
Nice points, like $(0,0),(4,0)$ and $(2,2)$.
$endgroup$
– Chris Custer
4 hours ago
$begingroup$
I will try this method
$endgroup$
– Vedant Chourey
4 hours ago