Otdfbt

Secure offsite backup, even in the case of hacker root access

Building a road to escape Earth's gravity by making a pyramid on Antartica

What are the words for people who cause trouble believing they know better?

Incremental Ranges!

What are they doing to this rocket following its test fire?

Meaning of constructor with multiple pairs of parentheses

In this example, which path would a monster affected by the Dissonant Whispers spell take?

Company did not petition for visa in a timely manner. Is asking me to work from overseas, but wants me to take a paycut

Removing applications from Show Applications without uninstalling

What happened to all the nuclear material being smuggled after the fall of the USSR?

Is it possible to trip with natural weapon?

Java guess the number

How bad would a partial hash leak be, realistically?

Pronoun introduced before its antecedent

Importance sampling estimation of power function

Pros and cons of writing a book review?

How could a government be implemented in a virtual reality?

Traffic law UK, pedestrians

Should I "tell" my exposition or give it through dialogue?

Bent spoke design wheels — feasible?

Is it a problem that pull requests are approved without any comments

If Boris Johnson were prosecuted and convicted of lying about Brexit, can that be used to cancel Brexit?

How do I write "Show, Don't Tell" as an Asperger

What is the purpose of building foundations?

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 $(1)$
$$fracw-2 =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3viota|=2|u-2+viota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
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

52.7k1365132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1077

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $(1)$
$$fracw-2 =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3viota|=2|u-2+viota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
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

52.7k1365132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1077

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $(1)$
$$fracw-2 =2$$
$$|4w+3|=2|w-2|$$
$$|3u+2+3viota|=2|u-2+viota|$$
$$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$
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

52.7k1365132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1077

$endgroup$

share|cite|improve this question

edited May 19 at 19:48

Maria Mazur

52.7k1365132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1077

share|cite|improve this question

edited May 19 at 19:48

Maria Mazur

52.7k1365132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1077

edited May 19 at 19:48

Maria Mazur

52.7k1365132

edited May 19 at 19:48

Maria Mazur

52.7k1365132

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1077

asked May 19 at 15:43

Vedant ChoureyVedant Chourey

1077

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

52.7k1365132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

15.9k3828

$endgroup$

add a comment | 

Your Answer

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3231935%2ftrouble-in-mapping-of-m%25c3%25b6bius-transformation%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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

52.7k1365132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

52.7k1365132

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

52.7k1365132

$endgroup$

share|cite|improve this answer

edited May 19 at 16:23

answered May 19 at 16:09

Maria MazurMaria Mazur

52.7k1365132

answered May 19 at 16:09

Maria MazurMaria Mazur

52.7k1365132

answered May 19 at 16:09

Maria MazurMaria Mazur

52.7k1365132

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

15.9k3828

$endgroup$

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

15.9k3828

$endgroup$

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

15.9k3828

$endgroup$

share|cite|improve this answer

edited May 19 at 17:03

answered May 19 at 16:50

Chris CusterChris Custer

15.9k3828

answered May 19 at 16:50

Chris CusterChris Custer

15.9k3828

answered May 19 at 16:50

Chris CusterChris Custer

15.9k3828