Complex numbers proof with modulus argument question

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I'm having trouble with this complex numbers proof.



Prove $|(z-a)/(overline az-1)|=1$ if $a,z$ are any complex numbers, where $zne a$ and $|z|=1$.



I tried substituing $z,a$ for general Cartesian and polar forms, but I couldn't get past the algebra or required simplifications.






complex-numbers







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    up vote
    1
    down vote

    favorite












    I'm having trouble with this complex numbers proof.



    Prove $|(z-a)/(overline az-1)|=1$ if $a,z$ are any complex numbers, where $zne a$ and $|z|=1$.



    I tried substituing $z,a$ for general Cartesian and polar forms, but I couldn't get past the algebra or required simplifications.






    complex-numbers







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm having trouble with this complex numbers proof.



      Prove $|(z-a)/(overline az-1)|=1$ if $a,z$ are any complex numbers, where $zne a$ and $|z|=1$.



      I tried substituing $z,a$ for general Cartesian and polar forms, but I couldn't get past the algebra or required simplifications.






      complex-numbers







      share|cite|improve this question















      I'm having trouble with this complex numbers proof.



      Prove $|(z-a)/(overline az-1)|=1$ if $a,z$ are any complex numbers, where $zne a$ and $|z|=1$.



      I tried substituing $z,a$ for general Cartesian and polar forms, but I couldn't get past the algebra or required simplifications.






      complex-numbers




      complex-numbers






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 27 mins ago









      Parcly Taxel

      36.3k136994




      36.3k136994










      asked 29 mins ago









      anonymous

      1,5901036




      1,5901036




















          3 Answers
          3






          active

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          up vote
          2
          down vote



          accepted










          $$|(z-a)/(overline az-1)|=1iff|z-a|=|overline az-1|$$
          Now, since $|z|=1$,
          $$|overline az-1|=|overline a-1/z|=|overline a-overline z|=|a-z|=|z-a|$$






          share|cite|improve this answer



























            up vote
            2
            down vote













            That's equivalent to $|z-a|=|overline az-1|$. As $|z|=1$, $overline z=z^-1$ and
            so
            $$|overline az-1|=|z||overline a -z^-1|=|overline a -overline z|
            =|overlinea-z|=|a-z|.$$






            share|cite|improve this answer



























              up vote
              2
              down vote













              Another approach is using $z=e^it$ then
              $$left|dfracz-aoverlineaz-1right|=left|dfrace^it-aoverlineae^-it-1right|=dfrac=dfrac=1$$






              share|cite|improve this answer




















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                2
                down vote



                accepted










                $$|(z-a)/(overline az-1)|=1iff|z-a|=|overline az-1|$$
                Now, since $|z|=1$,
                $$|overline az-1|=|overline a-1/z|=|overline a-overline z|=|a-z|=|z-a|$$






                share|cite|improve this answer
























                  up vote
                  2
                  down vote



                  accepted










                  $$|(z-a)/(overline az-1)|=1iff|z-a|=|overline az-1|$$
                  Now, since $|z|=1$,
                  $$|overline az-1|=|overline a-1/z|=|overline a-overline z|=|a-z|=|z-a|$$






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote



                    accepted







                    up vote
                    2
                    down vote



                    accepted






                    $$|(z-a)/(overline az-1)|=1iff|z-a|=|overline az-1|$$
                    Now, since $|z|=1$,
                    $$|overline az-1|=|overline a-1/z|=|overline a-overline z|=|a-z|=|z-a|$$






                    share|cite|improve this answer












                    $$|(z-a)/(overline az-1)|=1iff|z-a|=|overline az-1|$$
                    Now, since $|z|=1$,
                    $$|overline az-1|=|overline a-1/z|=|overline a-overline z|=|a-z|=|z-a|$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 19 mins ago









                    Parcly Taxel

                    36.3k136994




                    36.3k136994




















                        up vote
                        2
                        down vote













                        That's equivalent to $|z-a|=|overline az-1|$. As $|z|=1$, $overline z=z^-1$ and
                        so
                        $$|overline az-1|=|z||overline a -z^-1|=|overline a -overline z|
                        =|overlinea-z|=|a-z|.$$






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          That's equivalent to $|z-a|=|overline az-1|$. As $|z|=1$, $overline z=z^-1$ and
                          so
                          $$|overline az-1|=|z||overline a -z^-1|=|overline a -overline z|
                          =|overlinea-z|=|a-z|.$$






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            That's equivalent to $|z-a|=|overline az-1|$. As $|z|=1$, $overline z=z^-1$ and
                            so
                            $$|overline az-1|=|z||overline a -z^-1|=|overline a -overline z|
                            =|overlinea-z|=|a-z|.$$






                            share|cite|improve this answer












                            That's equivalent to $|z-a|=|overline az-1|$. As $|z|=1$, $overline z=z^-1$ and
                            so
                            $$|overline az-1|=|z||overline a -z^-1|=|overline a -overline z|
                            =|overlinea-z|=|a-z|.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 22 mins ago









                            Lord Shark the Unknown

                            92.9k956122




                            92.9k956122




















                                up vote
                                2
                                down vote













                                Another approach is using $z=e^it$ then
                                $$left|dfracz-aoverlineaz-1right|=left|dfrace^it-aoverlineae^-it-1right|=dfrac=dfrac=1$$






                                share|cite|improve this answer
























                                  up vote
                                  2
                                  down vote













                                  Another approach is using $z=e^it$ then
                                  $$left|dfracz-aoverlineaz-1right|=left|dfrace^it-aoverlineae^-it-1right|=dfrac=dfrac=1$$






                                  share|cite|improve this answer






















                                    up vote
                                    2
                                    down vote










                                    up vote
                                    2
                                    down vote









                                    Another approach is using $z=e^it$ then
                                    $$left|dfracz-aoverlineaz-1right|=left|dfrace^it-aoverlineae^-it-1right|=dfrac=dfrac=1$$






                                    share|cite|improve this answer












                                    Another approach is using $z=e^it$ then
                                    $$left|dfracz-aoverlineaz-1right|=left|dfrace^it-aoverlineae^-it-1right|=dfrac=dfrac=1$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 12 mins ago









                                    Nosrati

                                    24.1k61952




                                    24.1k61952



























                                         

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