Problem with a Moebius transformation from disk to itself [duplicate]

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This question already has an answer here:

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  Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  
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I need to show that a) every Möbius transformation that maps the disk onto itself can be written as $g(z)=e^{itheta}frac{z-z_0}{bar{z_0}z-1}$, and b) that this function maps the unit disk onto the unit disk. I'm currently trying to make sense of the last part. Taking modules I get the following result: $|z|^2+|z_0|^2<1+|z||z_0|$, and I have been stuck there since. How can I prove this inequality given that $|z|,|z_0|<1$?

complex-analysis

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asked Jan 22 at 20:59

campecampe

256

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marked as duplicate by Martin R, Lord Shark the Unknown, mrtaurho, A. Pongrácz, Chris Custer Jan 23 at 9:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Without any hypothesis concerning $z_0$?
  $endgroup$
  – José Carlos Santos
  Jan 22 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  $endgroup$
  – Martin R
  Jan 22 at 21:17
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  
  1 answer
  
  

I need to show that a) every Möbius transformation that maps the disk onto itself can be written as $g(z)=e^{itheta}frac{z-z_0}{bar{z_0}z-1}$, and b) that this function maps the unit disk onto the unit disk. I'm currently trying to make sense of the last part. Taking modules I get the following result: $|z|^2+|z_0|^2<1+|z||z_0|$, and I have been stuck there since. How can I prove this inequality given that $|z|,|z_0|<1$?

complex-analysis

share|cite|improve this question

asked Jan 22 at 20:59

campecampe

256

$endgroup$

marked as duplicate by Martin R, Lord Shark the Unknown, mrtaurho, A. Pongrácz, Chris Custer Jan 23 at 9:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Without any hypothesis concerning $z_0$?
  $endgroup$
  – José Carlos Santos
  Jan 22 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  $endgroup$
  – Martin R
  Jan 22 at 21:17
  
  
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  
  1 answer
  
  

I need to show that a) every Möbius transformation that maps the disk onto itself can be written as $g(z)=e^{itheta}frac{z-z_0}{bar{z_0}z-1}$, and b) that this function maps the unit disk onto the unit disk. I'm currently trying to make sense of the last part. Taking modules I get the following result: $|z|^2+|z_0|^2<1+|z||z_0|$, and I have been stuck there since. How can I prove this inequality given that $|z|,|z_0|<1$?

complex-analysis

share|cite|improve this question

asked Jan 22 at 20:59

campecampe

256

$endgroup$

This question already has an answer here:

• 
  Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  
  1 answer
  
  

I need to show that a) every Möbius transformation that maps the disk onto itself can be written as $g(z)=e^{itheta}frac{z-z_0}{bar{z_0}z-1}$, and b) that this function maps the unit disk onto the unit disk. I'm currently trying to make sense of the last part. Taking modules I get the following result: $|z|^2+|z_0|^2<1+|z||z_0|$, and I have been stuck there since. How can I prove this inequality given that $|z|,|z_0|<1$?

This question already has an answer here:

• 
  Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  
  1 answer
  
  

complex-analysis

complex-analysis

share|cite|improve this question

asked Jan 22 at 20:59

campecampe

256

share|cite|improve this question

asked Jan 22 at 20:59

campecampe

256

share|cite|improve this question

share|cite|improve this question

asked Jan 22 at 20:59

campecampe

256

asked Jan 22 at 20:59

campecampe

256

asked Jan 22 at 20:59

campecampe

256

256

marked as duplicate by Martin R, Lord Shark the Unknown, mrtaurho, A. Pongrácz, Chris Custer Jan 23 at 9:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Martin R, Lord Shark the Unknown, mrtaurho, A. Pongrácz, Chris Custer Jan 23 at 9:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Without any hypothesis concerning $z_0$?
  $endgroup$
  – José Carlos Santos
  Jan 22 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  $endgroup$
  – Martin R
  Jan 22 at 21:17
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Without any hypothesis concerning $z_0$?
  $endgroup$
  – José Carlos Santos
  Jan 22 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
  $endgroup$
  – Martin R
  Jan 22 at 21:17
  
  
  

  
  
  
  

$begingroup$
Without any hypothesis concerning $z_0$?
$endgroup$
– José Carlos Santos
Jan 22 at 21:05

$begingroup$
Without any hypothesis concerning $z_0$?
$endgroup$
– José Carlos Santos
Jan 22 at 21:05

4

4

$begingroup$
Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
$endgroup$
– Martin R
Jan 22 at 21:17

$begingroup$
Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
$endgroup$
– Martin R
Jan 22 at 21:17

add a comment | 

1 Answer
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$$|g(z)|^2=frac{|z-z_0|^2}{|overline{z_0}z-1|^2}
=frac{|z|^2+|z_0|^2-(zoverline{z_0}+overline{z}z_0)}
{|zz_0|^2+1-(zoverline{z_0}+overline{z}z_0)}{}
$$
Both numerator and denominator are positive, and denominator minus
numerator is
$$1+|zz_0|^2-|z|^2-|z_0|^2=(1-|z^2|)(1-|z_0^2|)>0$$
whenever $z$ and $z_0$ are in the open unit disc.

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answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

$$|g(z)|^2=frac{|z-z_0|^2}{|overline{z_0}z-1|^2}
=frac{|z|^2+|z_0|^2-(zoverline{z_0}+overline{z}z_0)}
{|zz_0|^2+1-(zoverline{z_0}+overline{z}z_0)}{}
$$
Both numerator and denominator are positive, and denominator minus
numerator is
$$1+|zz_0|^2-|z|^2-|z_0|^2=(1-|z^2|)(1-|z_0^2|)>0$$
whenever $z$ and $z_0$ are in the open unit disc.

share|cite|improve this answer

answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

add a comment | 

1

$begingroup$

$$|g(z)|^2=frac{|z-z_0|^2}{|overline{z_0}z-1|^2}
=frac{|z|^2+|z_0|^2-(zoverline{z_0}+overline{z}z_0)}
{|zz_0|^2+1-(zoverline{z_0}+overline{z}z_0)}{}
$$
Both numerator and denominator are positive, and denominator minus
numerator is
$$1+|zz_0|^2-|z|^2-|z_0|^2=(1-|z^2|)(1-|z_0^2|)>0$$
whenever $z$ and $z_0$ are in the open unit disc.

share|cite|improve this answer

answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

add a comment | 

1

1

1

$begingroup$

$$|g(z)|^2=frac{|z-z_0|^2}{|overline{z_0}z-1|^2}
=frac{|z|^2+|z_0|^2-(zoverline{z_0}+overline{z}z_0)}
{|zz_0|^2+1-(zoverline{z_0}+overline{z}z_0)}{}
$$
Both numerator and denominator are positive, and denominator minus
numerator is
$$1+|zz_0|^2-|z|^2-|z_0|^2=(1-|z^2|)(1-|z_0^2|)>0$$
whenever $z$ and $z_0$ are in the open unit disc.

share|cite|improve this answer

answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

106k1161133

$endgroup$

$$|g(z)|^2=frac{|z-z_0|^2}{|overline{z_0}z-1|^2}
=frac{|z|^2+|z_0|^2-(zoverline{z_0}+overline{z}z_0)}
{|zz_0|^2+1-(zoverline{z_0}+overline{z}z_0)}{}
$$
Both numerator and denominator are positive, and denominator minus
numerator is
$$1+|zz_0|^2-|z|^2-|z_0|^2=(1-|z^2|)(1-|z_0^2|)>0$$
whenever $z$ and $z_0$ are in the open unit disc.

share|cite|improve this answer

answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

106k1161133

share|cite|improve this answer

share|cite|improve this answer

answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

106k1161133

answered Jan 22 at 21:07

Lord Shark the UnknownLord Shark the Unknown

106k1161133

106k1161133

add a comment | 

add a comment | 

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