$z in mathbb{C}^-$ where $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$ identity












0












$begingroup$


Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,



where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.



Why is this true?






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is $C^-$?....
    $endgroup$
    – greedoid
    Dec 6 '18 at 21:54










  • $begingroup$
    the set of complex numbers z, where Re(z)<0
    $endgroup$
    – pablo_mathscobar
    Dec 6 '18 at 21:56






  • 4




    $begingroup$
    But your equality holds always, not just when $operatorname{Re}z<0$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 21:58










  • $begingroup$
    Do not remove your question and replace it with a completely different one after it has been answered.
    $endgroup$
    – T. Bongers
    Dec 6 '18 at 22:27
















0












$begingroup$


Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,



where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.



Why is this true?






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is $C^-$?....
    $endgroup$
    – greedoid
    Dec 6 '18 at 21:54










  • $begingroup$
    the set of complex numbers z, where Re(z)<0
    $endgroup$
    – pablo_mathscobar
    Dec 6 '18 at 21:56






  • 4




    $begingroup$
    But your equality holds always, not just when $operatorname{Re}z<0$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 21:58










  • $begingroup$
    Do not remove your question and replace it with a completely different one after it has been answered.
    $endgroup$
    – T. Bongers
    Dec 6 '18 at 22:27














0












0








0





$begingroup$


Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,



where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.



Why is this true?






complex-numbers







share|cite|improve this question











$endgroup$




Why is $|2-z|^2 = 4 + |z|^2 - 4Re(z)$,



where $z in mathbb{C}^-$ and $mathbb{C}^- := {z in mathbb{C} : Re(z) < 0}$.



Why is this true?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 22:36







pablo_mathscobar

















asked Dec 6 '18 at 21:53









pablo_mathscobarpablo_mathscobar

996




996








  • 1




    $begingroup$
    What is $C^-$?....
    $endgroup$
    – greedoid
    Dec 6 '18 at 21:54










  • $begingroup$
    the set of complex numbers z, where Re(z)<0
    $endgroup$
    – pablo_mathscobar
    Dec 6 '18 at 21:56






  • 4




    $begingroup$
    But your equality holds always, not just when $operatorname{Re}z<0$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 21:58










  • $begingroup$
    Do not remove your question and replace it with a completely different one after it has been answered.
    $endgroup$
    – T. Bongers
    Dec 6 '18 at 22:27














  • 1




    $begingroup$
    What is $C^-$?....
    $endgroup$
    – greedoid
    Dec 6 '18 at 21:54










  • $begingroup$
    the set of complex numbers z, where Re(z)<0
    $endgroup$
    – pablo_mathscobar
    Dec 6 '18 at 21:56






  • 4




    $begingroup$
    But your equality holds always, not just when $operatorname{Re}z<0$.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 21:58










  • $begingroup$
    Do not remove your question and replace it with a completely different one after it has been answered.
    $endgroup$
    – T. Bongers
    Dec 6 '18 at 22:27








1




1




$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54




$begingroup$
What is $C^-$?....
$endgroup$
– greedoid
Dec 6 '18 at 21:54












$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56




$begingroup$
the set of complex numbers z, where Re(z)<0
$endgroup$
– pablo_mathscobar
Dec 6 '18 at 21:56




4




4




$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58




$begingroup$
But your equality holds always, not just when $operatorname{Re}z<0$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 21:58












$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27




$begingroup$
Do not remove your question and replace it with a completely different one after it has been answered.
$endgroup$
– T. Bongers
Dec 6 '18 at 22:27










2 Answers
2






active

oldest

votes


















1












$begingroup$

hint



$$|2-z|^2=(2-z)(2-bar{z})$$



$$=4+zbar{z}-2(z+bar{z})$$
$$=4+|z|^2-2(x+iy+x-iy)$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    This identity is true for any $zinmathbb C$



    $|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      hint



      $$|2-z|^2=(2-z)(2-bar{z})$$



      $$=4+zbar{z}-2(z+bar{z})$$
      $$=4+|z|^2-2(x+iy+x-iy)$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        hint



        $$|2-z|^2=(2-z)(2-bar{z})$$



        $$=4+zbar{z}-2(z+bar{z})$$
        $$=4+|z|^2-2(x+iy+x-iy)$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          hint



          $$|2-z|^2=(2-z)(2-bar{z})$$



          $$=4+zbar{z}-2(z+bar{z})$$
          $$=4+|z|^2-2(x+iy+x-iy)$$






          share|cite|improve this answer









          $endgroup$



          hint



          $$|2-z|^2=(2-z)(2-bar{z})$$



          $$=4+zbar{z}-2(z+bar{z})$$
          $$=4+|z|^2-2(x+iy+x-iy)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 21:58









          hamam_Abdallahhamam_Abdallah

          38.1k21634




          38.1k21634























              1












              $begingroup$

              This identity is true for any $zinmathbb C$



              $|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                This identity is true for any $zinmathbb C$



                $|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  This identity is true for any $zinmathbb C$



                  $|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$






                  share|cite|improve this answer









                  $endgroup$



                  This identity is true for any $zinmathbb C$



                  $|2-z|^2=(2-z)overline{(2-z)}=(2-z)(2-overline z)=4 + zoverline z - 2(z+overline z)=4+|z|^2-4Re(z)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 22:00









                  Shubham JohriShubham Johri

                  5,172717




                  5,172717






























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