How to factorize $zz^*-4z-4z^*+12=0$ (where $z^*$ is the complex conjugate of $z$)












1












$begingroup$


I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.



I'm trying to factorise this using the completed square method but had no luck so far.



Could use some help.






complex-numbers quadratics







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$endgroup$












  • $begingroup$
    Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 2 '18 at 5:25










  • $begingroup$
    Alright, got it now. Thanks a lot.
    $endgroup$
    – Usama Abdul
    Dec 2 '18 at 5:33


















1












$begingroup$


I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.



I'm trying to factorise this using the completed square method but had no luck so far.



Could use some help.






complex-numbers quadratics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 2 '18 at 5:25










  • $begingroup$
    Alright, got it now. Thanks a lot.
    $endgroup$
    – Usama Abdul
    Dec 2 '18 at 5:33
















1












1








1





$begingroup$


I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.



I'm trying to factorise this using the completed square method but had no luck so far.



Could use some help.






complex-numbers quadratics







share|cite|improve this question











$endgroup$




I'm trying to factorize this:
$$zz^*-4z-4z^*+12=0$$
to get this:
$$|z-4|^2 - 4 = 0$$
where $z=x+yi$ is a complex number and $z^*=x-yi$ is the conjugate complex number of $z$.



I'm trying to factorise this using the completed square method but had no luck so far.



Could use some help.






complex-numbers quadratics




complex-numbers quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 5:29









Blue

47.8k870152




47.8k870152










asked Dec 2 '18 at 5:15









Usama AbdulUsama Abdul

82




82












  • $begingroup$
    Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 2 '18 at 5:25










  • $begingroup$
    Alright, got it now. Thanks a lot.
    $endgroup$
    – Usama Abdul
    Dec 2 '18 at 5:33




















  • $begingroup$
    Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 2 '18 at 5:25










  • $begingroup$
    Alright, got it now. Thanks a lot.
    $endgroup$
    – Usama Abdul
    Dec 2 '18 at 5:33


















$begingroup$
Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
$endgroup$
– Mohammad Zuhair Khan
Dec 2 '18 at 5:25




$begingroup$
Note that $zz^*-4z-4z^*+12=0implies zz^*-4z-4z^*+16=4 implies z(z^*-4)-4(z^*-4)-4=0 implies (z-4)(z^*-4)-4=0.$ Also note that $zz^*=|z|^2$ so $(z-4)(z^*-4)-4=0 implies |z-4|^2-4=0$
$endgroup$
– Mohammad Zuhair Khan
Dec 2 '18 at 5:25












$begingroup$
Alright, got it now. Thanks a lot.
$endgroup$
– Usama Abdul
Dec 2 '18 at 5:33






$begingroup$
Alright, got it now. Thanks a lot.
$endgroup$
– Usama Abdul
Dec 2 '18 at 5:33












2 Answers
2






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3












$begingroup$

begin{align}
zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
&= z(z^* -4) -4(z^* -4) -4\
&= (z-4)(z^* -4) -4\
&= (z-4)(z-4)^* -4\
&= |z-4|^2 -4.
end{align}






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write



    $zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$



    Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

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      3












      $begingroup$

      begin{align}
      zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
      &= z(z^* -4) -4(z^* -4) -4\
      &= (z-4)(z^* -4) -4\
      &= (z-4)(z-4)^* -4\
      &= |z-4|^2 -4.
      end{align}






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        begin{align}
        zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
        &= z(z^* -4) -4(z^* -4) -4\
        &= (z-4)(z^* -4) -4\
        &= (z-4)(z-4)^* -4\
        &= |z-4|^2 -4.
        end{align}






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          begin{align}
          zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
          &= z(z^* -4) -4(z^* -4) -4\
          &= (z-4)(z^* -4) -4\
          &= (z-4)(z-4)^* -4\
          &= |z-4|^2 -4.
          end{align}






          share|cite|improve this answer









          $endgroup$



          begin{align}
          zz^* -4z -4z^* +12 &= zz^* -4z -4z^* +16 -4\
          &= z(z^* -4) -4(z^* -4) -4\
          &= (z-4)(z^* -4) -4\
          &= (z-4)(z-4)^* -4\
          &= |z-4|^2 -4.
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 5:28









          RócherzRócherz

          2,7762721




          2,7762721























              2












              $begingroup$

              Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write



              $zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$



              Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write



                $zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$



                Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write



                  $zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$



                  Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.






                  share|cite|improve this answer









                  $endgroup$



                  Remark, since $zz^*=x^2+y^2$ and $z+z^*=2x$ you can as well write



                  $zz^*-4z-4z^*+12=x^2+y^2-8x+12=(x-4)^2+y^2-4=|z-4|^2-4=0$



                  Since both represent the equation of a circle of centre $(4,0)$ and radius $2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 5:52









                  zwimzwim

                  11.7k729




                  11.7k729






























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