Finding the norm of $w + frac{1 - |w|^2}{|w - z|^2}(w - z)$, where $w$ and $z$ are in $mathbb{R}^n$












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$begingroup$


I'm trying to show a result from Manfred Stoll's book on Hyperbolic geometry. Consider the function
$$M_w(z) = w + frac{1 - |w|^2}{|w - z|^2}(w - z)$$ where $w,z in mathbb{R}^n$ and $|w| < 1$. (I am using different variable names than Stoll). This function generalizes the Möbius transformations on the complex numbers.



Stoll writes that
$$|M_w(z)|^2 = frac{|w - z|^2 + (1 - |w|^2)(1-|z|^2)}{|w - z|^2}$$
How do you show this? I tried computing $langle M_w(z), M_w(z) rangle$, but I'm not getting any closer.



This result ultimately shows that $M_w(z)$ maps into the unit ball.






hyperbolic-geometry mobius-transformation







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    3












    $begingroup$


    I'm trying to show a result from Manfred Stoll's book on Hyperbolic geometry. Consider the function
    $$M_w(z) = w + frac{1 - |w|^2}{|w - z|^2}(w - z)$$ where $w,z in mathbb{R}^n$ and $|w| < 1$. (I am using different variable names than Stoll). This function generalizes the Möbius transformations on the complex numbers.



    Stoll writes that
    $$|M_w(z)|^2 = frac{|w - z|^2 + (1 - |w|^2)(1-|z|^2)}{|w - z|^2}$$
    How do you show this? I tried computing $langle M_w(z), M_w(z) rangle$, but I'm not getting any closer.



    This result ultimately shows that $M_w(z)$ maps into the unit ball.






    hyperbolic-geometry mobius-transformation







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I'm trying to show a result from Manfred Stoll's book on Hyperbolic geometry. Consider the function
      $$M_w(z) = w + frac{1 - |w|^2}{|w - z|^2}(w - z)$$ where $w,z in mathbb{R}^n$ and $|w| < 1$. (I am using different variable names than Stoll). This function generalizes the Möbius transformations on the complex numbers.



      Stoll writes that
      $$|M_w(z)|^2 = frac{|w - z|^2 + (1 - |w|^2)(1-|z|^2)}{|w - z|^2}$$
      How do you show this? I tried computing $langle M_w(z), M_w(z) rangle$, but I'm not getting any closer.



      This result ultimately shows that $M_w(z)$ maps into the unit ball.






      hyperbolic-geometry mobius-transformation







      share|cite|improve this question











      $endgroup$




      I'm trying to show a result from Manfred Stoll's book on Hyperbolic geometry. Consider the function
      $$M_w(z) = w + frac{1 - |w|^2}{|w - z|^2}(w - z)$$ where $w,z in mathbb{R}^n$ and $|w| < 1$. (I am using different variable names than Stoll). This function generalizes the Möbius transformations on the complex numbers.



      Stoll writes that
      $$|M_w(z)|^2 = frac{|w - z|^2 + (1 - |w|^2)(1-|z|^2)}{|w - z|^2}$$
      How do you show this? I tried computing $langle M_w(z), M_w(z) rangle$, but I'm not getting any closer.



      This result ultimately shows that $M_w(z)$ maps into the unit ball.






      hyperbolic-geometry mobius-transformation




      hyperbolic-geometry mobius-transformation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 2:36









      Blue

      48k870153




      48k870153










      asked Dec 2 '18 at 21:47









      dinstructiondinstruction

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      554423






















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          $begingroup$

          $$begin{align}
          left|M_w(z)right|^2 &=
          |w|^2+frac{left(1-|w|^2right)^2}{|w-z|^4}|w-z|^2+2frac{1-|w|^2}{|w-z|^2};wcdot(w-z) \[8pt]
          |w-z|^2;left|M_w(z)right|^2 &=
          |w|^2|w-z|^2+left(1-|w|^2right)^2+2left(1-|w|^2right);left(|w|^2-wcdot zright) \[8pt]
          &=
          phantom{+2}|w|^2left(|w|^2+|z|^2-2wcdot zright)\
          &phantom{=}+phantom{2}left(1-2|w|^2+|w|^4right)\
          &phantom{=}+2left(|w|^2-wcdot z-|w|^4+|w|^2wcdot z right) \[8pt]
          &=
          left(-2wcdot zright)+ left(1 + |w|^2|z|^2right)\[6pt]
          &=
          left(|w|^2+|z|^2-2wcdot zright)+ left(1 -|w|^2-|z|^2+ |w|^2|z|^2right)\[6pt]
          &=
          |w-z|^2+ left(1 -|w|^2right)left(1-|z|^2right)\
          end{align}$$






          share|cite|improve this answer









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            $begingroup$

            $$begin{align}
            left|M_w(z)right|^2 &=
            |w|^2+frac{left(1-|w|^2right)^2}{|w-z|^4}|w-z|^2+2frac{1-|w|^2}{|w-z|^2};wcdot(w-z) \[8pt]
            |w-z|^2;left|M_w(z)right|^2 &=
            |w|^2|w-z|^2+left(1-|w|^2right)^2+2left(1-|w|^2right);left(|w|^2-wcdot zright) \[8pt]
            &=
            phantom{+2}|w|^2left(|w|^2+|z|^2-2wcdot zright)\
            &phantom{=}+phantom{2}left(1-2|w|^2+|w|^4right)\
            &phantom{=}+2left(|w|^2-wcdot z-|w|^4+|w|^2wcdot z right) \[8pt]
            &=
            left(-2wcdot zright)+ left(1 + |w|^2|z|^2right)\[6pt]
            &=
            left(|w|^2+|z|^2-2wcdot zright)+ left(1 -|w|^2-|z|^2+ |w|^2|z|^2right)\[6pt]
            &=
            |w-z|^2+ left(1 -|w|^2right)left(1-|z|^2right)\
            end{align}$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              $$begin{align}
              left|M_w(z)right|^2 &=
              |w|^2+frac{left(1-|w|^2right)^2}{|w-z|^4}|w-z|^2+2frac{1-|w|^2}{|w-z|^2};wcdot(w-z) \[8pt]
              |w-z|^2;left|M_w(z)right|^2 &=
              |w|^2|w-z|^2+left(1-|w|^2right)^2+2left(1-|w|^2right);left(|w|^2-wcdot zright) \[8pt]
              &=
              phantom{+2}|w|^2left(|w|^2+|z|^2-2wcdot zright)\
              &phantom{=}+phantom{2}left(1-2|w|^2+|w|^4right)\
              &phantom{=}+2left(|w|^2-wcdot z-|w|^4+|w|^2wcdot z right) \[8pt]
              &=
              left(-2wcdot zright)+ left(1 + |w|^2|z|^2right)\[6pt]
              &=
              left(|w|^2+|z|^2-2wcdot zright)+ left(1 -|w|^2-|z|^2+ |w|^2|z|^2right)\[6pt]
              &=
              |w-z|^2+ left(1 -|w|^2right)left(1-|z|^2right)\
              end{align}$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                $$begin{align}
                left|M_w(z)right|^2 &=
                |w|^2+frac{left(1-|w|^2right)^2}{|w-z|^4}|w-z|^2+2frac{1-|w|^2}{|w-z|^2};wcdot(w-z) \[8pt]
                |w-z|^2;left|M_w(z)right|^2 &=
                |w|^2|w-z|^2+left(1-|w|^2right)^2+2left(1-|w|^2right);left(|w|^2-wcdot zright) \[8pt]
                &=
                phantom{+2}|w|^2left(|w|^2+|z|^2-2wcdot zright)\
                &phantom{=}+phantom{2}left(1-2|w|^2+|w|^4right)\
                &phantom{=}+2left(|w|^2-wcdot z-|w|^4+|w|^2wcdot z right) \[8pt]
                &=
                left(-2wcdot zright)+ left(1 + |w|^2|z|^2right)\[6pt]
                &=
                left(|w|^2+|z|^2-2wcdot zright)+ left(1 -|w|^2-|z|^2+ |w|^2|z|^2right)\[6pt]
                &=
                |w-z|^2+ left(1 -|w|^2right)left(1-|z|^2right)\
                end{align}$$






                share|cite|improve this answer









                $endgroup$



                $$begin{align}
                left|M_w(z)right|^2 &=
                |w|^2+frac{left(1-|w|^2right)^2}{|w-z|^4}|w-z|^2+2frac{1-|w|^2}{|w-z|^2};wcdot(w-z) \[8pt]
                |w-z|^2;left|M_w(z)right|^2 &=
                |w|^2|w-z|^2+left(1-|w|^2right)^2+2left(1-|w|^2right);left(|w|^2-wcdot zright) \[8pt]
                &=
                phantom{+2}|w|^2left(|w|^2+|z|^2-2wcdot zright)\
                &phantom{=}+phantom{2}left(1-2|w|^2+|w|^4right)\
                &phantom{=}+2left(|w|^2-wcdot z-|w|^4+|w|^2wcdot z right) \[8pt]
                &=
                left(-2wcdot zright)+ left(1 + |w|^2|z|^2right)\[6pt]
                &=
                left(|w|^2+|z|^2-2wcdot zright)+ left(1 -|w|^2-|z|^2+ |w|^2|z|^2right)\[6pt]
                &=
                |w-z|^2+ left(1 -|w|^2right)left(1-|z|^2right)\
                end{align}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 2:33









                BlueBlue

                48k870153




                48k870153






























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