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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

2

$begingroup$

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

share|cite|improve this question

edited May 19 at 19:48

Maria Mazur

52.4k1363132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1097

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The circle is centered at $(2,0)$ though.
  $endgroup$
  – Chris Custer
  May 19 at 15:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I apologize for that mistake. Although i have corrected that but still the problem is same
  $endgroup$
  – Vedant Chourey
  May 19 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
  $endgroup$
  – Chris Custer
  May 19 at 16:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice points, like $(0,0),(4,0)$ and $(2,2)$.
  $endgroup$
  – Chris Custer
  May 19 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will try this method
  $endgroup$
  – Vedant Chourey
  May 19 at 16:12
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

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

share|cite|improve this question

edited May 19 at 19:48

Maria Mazur

52.4k1363132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1097

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The circle is centered at $(2,0)$ though.
  $endgroup$
  – Chris Custer
  May 19 at 15:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I apologize for that mistake. Although i have corrected that but still the problem is same
  $endgroup$
  – Vedant Chourey
  May 19 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points).
  $endgroup$
  – Chris Custer
  May 19 at 16:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice points, like $(0,0),(4,0)$ and $(2,2)$.
  $endgroup$
  – Chris Custer
  May 19 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will try this method
  $endgroup$
  – Vedant Chourey
  May 19 at 16:12
  

  
  
  
  

 | 
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

share|cite|improve this question

edited May 19 at 19:48

Maria Mazur

52.4k1363132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1097

$endgroup$

share|cite|improve this question

edited May 19 at 19:48

Maria Mazur

52.4k1363132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1097

share|cite|improve this question

edited May 19 at 19:48

Maria Mazur

52.4k1363132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1097

edited May 19 at 19:48

Maria Mazur

52.4k1363132

edited May 19 at 19:48

Maria Mazur

52.4k1363132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1097

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1097

2 Answers
2

active

oldest

votes

3

$begingroup$

If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get

$$Big|4w+3 -2w+4over w-2Big| = 2$$ 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 $$4w+3=0$$

share|cite|improve this answer

edited May 19 at 16:23

answered May 19 at 16:09

Maria MazurMaria Mazur

52.4k1363132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thankyou. For explanation
  $endgroup$
  – Vedant Chourey
  May 19 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fell free to upvote and accept the answer if you think it was usefull answer to you.
  $endgroup$
  – Maria Mazur
  May 19 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. I'll take care of this next time
  $endgroup$
  – Vedant Chourey
  May 19 at 16:26
  

  
  
  
  

add a comment | 

2

$begingroup$

$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.

The result follows.

share|cite|improve this answer

edited May 19 at 17:03

answered May 19 at 16:50

Chris CusterChris Custer

15.6k3827

$endgroup$

add a comment | 

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3

$begingroup$

If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get

$$Big|4w+3 -2w+4over w-2Big| = 2$$ 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 $$4w+3=0$$

share|cite|improve this answer

edited May 19 at 16:23

answered May 19 at 16:09

Maria MazurMaria Mazur

52.4k1363132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thankyou. For explanation
  $endgroup$
  – Vedant Chourey
  May 19 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fell free to upvote and accept the answer if you think it was usefull answer to you.
  $endgroup$
  – Maria Mazur
  May 19 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. I'll take care of this next time
  $endgroup$
  – Vedant Chourey
  May 19 at 16:26
  

  
  
  
  

add a comment | 

3

$begingroup$

If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get

$$Big|4w+3 -2w+4over w-2Big| = 2$$ 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 $$4w+3=0$$

share|cite|improve this answer

edited May 19 at 16:23

answered May 19 at 16:09

Maria MazurMaria Mazur

52.4k1363132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thankyou. For explanation
  $endgroup$
  – Vedant Chourey
  May 19 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fell free to upvote and accept the answer if you think it was usefull answer to you.
  $endgroup$
  – Maria Mazur
  May 19 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. I'll take care of this next time
  $endgroup$
  – Vedant Chourey
  May 19 at 16:26
  

  
  
  
  

add a comment | 

$begingroup$

If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get

$$Big|4w+3 -2w+4over w-2Big| = 2$$ 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 $$4w+3=0$$

share|cite|improve this answer

edited May 19 at 16:23

answered May 19 at 16:09

Maria MazurMaria Mazur

52.4k1363132

$endgroup$

share|cite|improve this answer

edited May 19 at 16:23

answered May 19 at 16:09

Maria MazurMaria Mazur

52.4k1363132

answered May 19 at 16:09

Maria MazurMaria Mazur

52.4k1363132

answered May 19 at 16:09

Maria MazurMaria Mazur

52.4k1363132

2

$begingroup$

$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.

The result follows.

share|cite|improve this answer

edited May 19 at 17:03

answered May 19 at 16:50

Chris CusterChris Custer

15.6k3827

$endgroup$

add a comment | 

2

$begingroup$

$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.

The result follows.

share|cite|improve this answer

edited May 19 at 17:03

answered May 19 at 16:50

Chris CusterChris Custer

15.6k3827

$endgroup$

add a comment | 

$begingroup$

$0to-dfrac 34, 4toinfty $ and $2+2ito -dfrac 34-dfrac114i$.

The result follows.

share|cite|improve this answer

edited May 19 at 17:03

answered May 19 at 16:50

Chris CusterChris Custer

15.6k3827

$endgroup$

share|cite|improve this answer

edited May 19 at 17:03

answered May 19 at 16:50

Chris CusterChris Custer

15.6k3827

answered May 19 at 16:50

Chris CusterChris Custer

15.6k3827

answered May 19 at 16:50

Chris CusterChris Custer

15.6k3827