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












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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 $|z|,|z_0|<1$?






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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


















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









$endgroup$



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
















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









$endgroup$





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






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asked Jan 22 at 20:59









campecampe

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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.









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




















  • $begingroup$
    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












1 Answer
1






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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.






share|cite|improve this answer









$endgroup$




















    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









    $endgroup$


















      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









      $endgroup$
















        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









        $endgroup$



        $$|g(z)|^2=frac{|z-z_0|^2}{|overline{z_0}z-1|^2}
        =frac{|z|^2+|z_0|^2-(zoverline{z_0}+overline{z}z_0)}
        {|zz_0|^2+1-(zoverline{z_0}+overline{z}z_0)}{}
        $$
        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












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 21:07









        Lord Shark the UnknownLord Shark the Unknown

        106k1161133




        106k1161133















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