Can Mathematica prove inequalities involving complex numbers?












1












$begingroup$


I am given two complex numbers $z$ and $w$ that satisfy the following constraint
$$ |z - overline{z}w| + |w|^2 < 1. $$



I want to see if the following inequality is true
$$ z^2 overline{w} + overline{z}^2w + |w|^2(z^2 overline{w} + overline{z}^2w - 4|z|^2) geq 0. $$
Is it possible for Mathematica to prove or disprove the above inequality?






inequalities







share|improve this question









$endgroup$








  • 1




    $begingroup$
    Could you please provide code for those inequalities?
    $endgroup$
    – Andrew
    Dec 29 '18 at 19:32












  • $begingroup$
    |z - conjugate[z]w| + |w|^2 < 1
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:39










  • $begingroup$
    z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:47










  • $begingroup$
    That's not what he was looking for
    $endgroup$
    – user6014
    Dec 30 '18 at 17:33










  • $begingroup$
    People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful
    $endgroup$
    – Michael E2
    Jan 1 at 16:28
















1












$begingroup$


I am given two complex numbers $z$ and $w$ that satisfy the following constraint
$$ |z - overline{z}w| + |w|^2 < 1. $$



I want to see if the following inequality is true
$$ z^2 overline{w} + overline{z}^2w + |w|^2(z^2 overline{w} + overline{z}^2w - 4|z|^2) geq 0. $$
Is it possible for Mathematica to prove or disprove the above inequality?






inequalities







share|improve this question









$endgroup$








  • 1




    $begingroup$
    Could you please provide code for those inequalities?
    $endgroup$
    – Andrew
    Dec 29 '18 at 19:32












  • $begingroup$
    |z - conjugate[z]w| + |w|^2 < 1
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:39










  • $begingroup$
    z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:47










  • $begingroup$
    That's not what he was looking for
    $endgroup$
    – user6014
    Dec 30 '18 at 17:33










  • $begingroup$
    People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful
    $endgroup$
    – Michael E2
    Jan 1 at 16:28














1












1








1


2



$begingroup$


I am given two complex numbers $z$ and $w$ that satisfy the following constraint
$$ |z - overline{z}w| + |w|^2 < 1. $$



I want to see if the following inequality is true
$$ z^2 overline{w} + overline{z}^2w + |w|^2(z^2 overline{w} + overline{z}^2w - 4|z|^2) geq 0. $$
Is it possible for Mathematica to prove or disprove the above inequality?






inequalities







share|improve this question









$endgroup$




I am given two complex numbers $z$ and $w$ that satisfy the following constraint
$$ |z - overline{z}w| + |w|^2 < 1. $$



I want to see if the following inequality is true
$$ z^2 overline{w} + overline{z}^2w + |w|^2(z^2 overline{w} + overline{z}^2w - 4|z|^2) geq 0. $$
Is it possible for Mathematica to prove or disprove the above inequality?






inequalities




inequalities






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Dec 29 '18 at 19:17









JaikrishnanJaikrishnan

1062




1062








  • 1




    $begingroup$
    Could you please provide code for those inequalities?
    $endgroup$
    – Andrew
    Dec 29 '18 at 19:32












  • $begingroup$
    |z - conjugate[z]w| + |w|^2 < 1
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:39










  • $begingroup$
    z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:47










  • $begingroup$
    That's not what he was looking for
    $endgroup$
    – user6014
    Dec 30 '18 at 17:33










  • $begingroup$
    People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful
    $endgroup$
    – Michael E2
    Jan 1 at 16:28














  • 1




    $begingroup$
    Could you please provide code for those inequalities?
    $endgroup$
    – Andrew
    Dec 29 '18 at 19:32












  • $begingroup$
    |z - conjugate[z]w| + |w|^2 < 1
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:39










  • $begingroup$
    z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    $endgroup$
    – Jaikrishnan
    Dec 29 '18 at 19:47










  • $begingroup$
    That's not what he was looking for
    $endgroup$
    – user6014
    Dec 30 '18 at 17:33










  • $begingroup$
    People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful
    $endgroup$
    – Michael E2
    Jan 1 at 16:28








1




1




$begingroup$
Could you please provide code for those inequalities?
$endgroup$
– Andrew
Dec 29 '18 at 19:32






$begingroup$
Could you please provide code for those inequalities?
$endgroup$
– Andrew
Dec 29 '18 at 19:32














$begingroup$
|z - conjugate[z]w| + |w|^2 < 1
$endgroup$
– Jaikrishnan
Dec 29 '18 at 19:39




$begingroup$
|z - conjugate[z]w| + |w|^2 < 1
$endgroup$
– Jaikrishnan
Dec 29 '18 at 19:39












$begingroup$
z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
$endgroup$
– Jaikrishnan
Dec 29 '18 at 19:47




$begingroup$
z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
$endgroup$
– Jaikrishnan
Dec 29 '18 at 19:47












$begingroup$
That's not what he was looking for
$endgroup$
– user6014
Dec 30 '18 at 17:33




$begingroup$
That's not what he was looking for
$endgroup$
– user6014
Dec 30 '18 at 17:33












$begingroup$
People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful
$endgroup$
– Michael E2
Jan 1 at 16:28




$begingroup$
People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful
$endgroup$
– Michael E2
Jan 1 at 16:28










3 Answers
3






active

oldest

votes


















3












$begingroup$

Your inequalities:



z = x + I y;

w = u + I v;

ineq1 = Abs[z - Conjugate[z] w] + Abs[w]^2 < 1 // ComplexExpand;

ineq2 = z^2*Conjugate[w] + Conjugate[z]^2*w + 

    Abs[w]^2*(z^2*Conjugate[w] + Conjugate[z]^2*w - 4 Abs[z]^2) >= 0 //

   ComplexExpand;


This gives a counterexample:



res = FindInstance[{ineq1, ! ineq2}, {x, y, u, v}, Reals]


$
left{left{xto -frac{11}{32},yto frac{29}{32},uto -frac{3}{4},vto
-frac{7}{16}right}right}
$



Check:



{ineq1, ineq2} /. res


$left(
begin{array}{cc}
text{True} & text{False} \
end{array}
right)$






share|improve this answer









$endgroup$





















    3












    $begingroup$

    Resolve[
    
  ForAll[{z, w},
    
    Abs[z - Conjugate[z] w] + Abs[w]^2 < 1, 
    
    z^2 Conjugate[w] + Conjugate[z]^2 w +
    
    Abs[w]^2 (z^2 Conjugate[w] + Conjugate[z]^2 w - 4 Abs[z]^2) >= 0
    
  ],
    
  Complexes
    
]



    False







    share|improve this answer









    $endgroup$













    • $begingroup$
      Or simply inequality /. {z -> I, w -> 1}.
      $endgroup$
      – Michael E2
      Dec 29 '18 at 20:53










    • $begingroup$
      @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
      $endgroup$
      – Thies Heidecke
      Dec 29 '18 at 20:56












    • $begingroup$
      My first try on complicated expressions is usually to plug in a bunch of Random* stuff and check, which is pretty easy in M. This inequality came out mostly false, so I tried something simple. Already upvoted cuz I like quantifiers too. :)
      $endgroup$
      – Michael E2
      Dec 29 '18 at 21:00



















    1












    $begingroup$

    With your inequality you should be careful about what you mean by "greater" or "less". Since the complex plane is principally vectors it is natural to compare magnitudes or 'radus' from the origin. Hence people normally use the absolute value.



    You can naievely compare the real parts however this can result in absurd inequalities since moving along the complex axis retains the equality. 2 + i = 2 + 100 i



    I'm afraid in the way you constructed your second inequality this is how Mathematica will handle the problem.



    https://math.stackexchange.com/questions/1116022/can-a-complex-number-ever-be-considered-bigger-or-smaller-than-a-real-number






    share|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Your inequalities:



      z = x + I y;
      
w = u + I v;
      
ineq1 = Abs[z - Conjugate[z] w] + Abs[w]^2 < 1 // ComplexExpand;
      
ineq2 = z^2*Conjugate[w] + Conjugate[z]^2*w + 
      
    Abs[w]^2*(z^2*Conjugate[w] + Conjugate[z]^2*w - 4 Abs[z]^2) >= 0 //
      
   ComplexExpand;


      This gives a counterexample:



      res = FindInstance[{ineq1, ! ineq2}, {x, y, u, v}, Reals]


      $
      left{left{xto -frac{11}{32},yto frac{29}{32},uto -frac{3}{4},vto
      -frac{7}{16}right}right}
      $



      Check:



      {ineq1, ineq2} /. res


      $left(
      begin{array}{cc}
      text{True} & text{False} \
      end{array}
      right)$






      share|improve this answer









      $endgroup$


















        3












        $begingroup$

        Your inequalities:



        z = x + I y;
        
w = u + I v;
        
ineq1 = Abs[z - Conjugate[z] w] + Abs[w]^2 < 1 // ComplexExpand;
        
ineq2 = z^2*Conjugate[w] + Conjugate[z]^2*w + 
        
    Abs[w]^2*(z^2*Conjugate[w] + Conjugate[z]^2*w - 4 Abs[z]^2) >= 0 //
        
   ComplexExpand;


        This gives a counterexample:



        res = FindInstance[{ineq1, ! ineq2}, {x, y, u, v}, Reals]


        $
        left{left{xto -frac{11}{32},yto frac{29}{32},uto -frac{3}{4},vto
        -frac{7}{16}right}right}
        $



        Check:



        {ineq1, ineq2} /. res


        $left(
        begin{array}{cc}
        text{True} & text{False} \
        end{array}
        right)$






        share|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Your inequalities:



          z = x + I y;
          
w = u + I v;
          
ineq1 = Abs[z - Conjugate[z] w] + Abs[w]^2 < 1 // ComplexExpand;
          
ineq2 = z^2*Conjugate[w] + Conjugate[z]^2*w + 
          
    Abs[w]^2*(z^2*Conjugate[w] + Conjugate[z]^2*w - 4 Abs[z]^2) >= 0 //
          
   ComplexExpand;


          This gives a counterexample:



          res = FindInstance[{ineq1, ! ineq2}, {x, y, u, v}, Reals]


          $
          left{left{xto -frac{11}{32},yto frac{29}{32},uto -frac{3}{4},vto
          -frac{7}{16}right}right}
          $



          Check:



          {ineq1, ineq2} /. res


          $left(
          begin{array}{cc}
          text{True} & text{False} \
          end{array}
          right)$






          share|improve this answer









          $endgroup$



          Your inequalities:



          z = x + I y;
          
w = u + I v;
          
ineq1 = Abs[z - Conjugate[z] w] + Abs[w]^2 < 1 // ComplexExpand;
          
ineq2 = z^2*Conjugate[w] + Conjugate[z]^2*w + 
          
    Abs[w]^2*(z^2*Conjugate[w] + Conjugate[z]^2*w - 4 Abs[z]^2) >= 0 //
          
   ComplexExpand;


          This gives a counterexample:



          res = FindInstance[{ineq1, ! ineq2}, {x, y, u, v}, Reals]


          $
          left{left{xto -frac{11}{32},yto frac{29}{32},uto -frac{3}{4},vto
          -frac{7}{16}right}right}
          $



          Check:



          {ineq1, ineq2} /. res


          $left(
          begin{array}{cc}
          text{True} & text{False} \
          end{array}
          right)$







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 29 '18 at 19:55









          AndrewAndrew

          1,9511215




          1,9511215























              3












              $begingroup$

              Resolve[
              
  ForAll[{z, w},
              
    Abs[z - Conjugate[z] w] + Abs[w]^2 < 1, 
              
    z^2 Conjugate[w] + Conjugate[z]^2 w +
              
    Abs[w]^2 (z^2 Conjugate[w] + Conjugate[z]^2 w - 4 Abs[z]^2) >= 0
              
  ],
              
  Complexes
              
]



              False







              share|improve this answer









              $endgroup$













              • $begingroup$
                Or simply inequality /. {z -> I, w -> 1}.
                $endgroup$
                – Michael E2
                Dec 29 '18 at 20:53










              • $begingroup$
                @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
                $endgroup$
                – Thies Heidecke
                Dec 29 '18 at 20:56












              • $begingroup$
                My first try on complicated expressions is usually to plug in a bunch of Random* stuff and check, which is pretty easy in M. This inequality came out mostly false, so I tried something simple. Already upvoted cuz I like quantifiers too. :)
                $endgroup$
                – Michael E2
                Dec 29 '18 at 21:00
















              3












              $begingroup$

              Resolve[
              
  ForAll[{z, w},
              
    Abs[z - Conjugate[z] w] + Abs[w]^2 < 1, 
              
    z^2 Conjugate[w] + Conjugate[z]^2 w +
              
    Abs[w]^2 (z^2 Conjugate[w] + Conjugate[z]^2 w - 4 Abs[z]^2) >= 0
              
  ],
              
  Complexes
              
]



              False







              share|improve this answer









              $endgroup$













              • $begingroup$
                Or simply inequality /. {z -> I, w -> 1}.
                $endgroup$
                – Michael E2
                Dec 29 '18 at 20:53










              • $begingroup$
                @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
                $endgroup$
                – Thies Heidecke
                Dec 29 '18 at 20:56












              • $begingroup$
                My first try on complicated expressions is usually to plug in a bunch of Random* stuff and check, which is pretty easy in M. This inequality came out mostly false, so I tried something simple. Already upvoted cuz I like quantifiers too. :)
                $endgroup$
                – Michael E2
                Dec 29 '18 at 21:00














              3












              3








              3





              $begingroup$

              Resolve[
              
  ForAll[{z, w},
              
    Abs[z - Conjugate[z] w] + Abs[w]^2 < 1, 
              
    z^2 Conjugate[w] + Conjugate[z]^2 w +
              
    Abs[w]^2 (z^2 Conjugate[w] + Conjugate[z]^2 w - 4 Abs[z]^2) >= 0
              
  ],
              
  Complexes
              
]



              False







              share|improve this answer









              $endgroup$



              Resolve[
              
  ForAll[{z, w},
              
    Abs[z - Conjugate[z] w] + Abs[w]^2 < 1, 
              
    z^2 Conjugate[w] + Conjugate[z]^2 w +
              
    Abs[w]^2 (z^2 Conjugate[w] + Conjugate[z]^2 w - 4 Abs[z]^2) >= 0
              
  ],
              
  Complexes
              
]



              False








              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Dec 29 '18 at 20:50









              Thies HeideckeThies Heidecke

              6,9462638




              6,9462638












              • $begingroup$
                Or simply inequality /. {z -> I, w -> 1}.
                $endgroup$
                – Michael E2
                Dec 29 '18 at 20:53










              • $begingroup$
                @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
                $endgroup$
                – Thies Heidecke
                Dec 29 '18 at 20:56












              • $begingroup$
                My first try on complicated expressions is usually to plug in a bunch of Random* stuff and check, which is pretty easy in M. This inequality came out mostly false, so I tried something simple. Already upvoted cuz I like quantifiers too. :)
                $endgroup$
                – Michael E2
                Dec 29 '18 at 21:00


















              • $begingroup$
                Or simply inequality /. {z -> I, w -> 1}.
                $endgroup$
                – Michael E2
                Dec 29 '18 at 20:53










              • $begingroup$
                @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
                $endgroup$
                – Thies Heidecke
                Dec 29 '18 at 20:56












              • $begingroup$
                My first try on complicated expressions is usually to plug in a bunch of Random* stuff and check, which is pretty easy in M. This inequality came out mostly false, so I tried something simple. Already upvoted cuz I like quantifiers too. :)
                $endgroup$
                – Michael E2
                Dec 29 '18 at 21:00
















              $begingroup$
              Or simply inequality /. {z -> I, w -> 1}.
              $endgroup$
              – Michael E2
              Dec 29 '18 at 20:53




              $begingroup$
              Or simply inequality /. {z -> I, w -> 1}.
              $endgroup$
              – Michael E2
              Dec 29 '18 at 20:53












              $begingroup$
              @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
              $endgroup$
              – Thies Heidecke
              Dec 29 '18 at 20:56






              $begingroup$
              @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
              $endgroup$
              – Thies Heidecke
              Dec 29 '18 at 20:56














              $begingroup$
              My first try on complicated expressions is usually to plug in a bunch of Random* stuff and check, which is pretty easy in M. This inequality came out mostly false, so I tried something simple. Already upvoted cuz I like quantifiers too. :)
              $endgroup$
              – Michael E2
              Dec 29 '18 at 21:00




              $begingroup$
              My first try on complicated expressions is usually to plug in a bunch of Random* stuff and check, which is pretty easy in M. This inequality came out mostly false, so I tried something simple. Already upvoted cuz I like quantifiers too. :)
              $endgroup$
              – Michael E2
              Dec 29 '18 at 21:00











              1












              $begingroup$

              With your inequality you should be careful about what you mean by "greater" or "less". Since the complex plane is principally vectors it is natural to compare magnitudes or 'radus' from the origin. Hence people normally use the absolute value.



              You can naievely compare the real parts however this can result in absurd inequalities since moving along the complex axis retains the equality. 2 + i = 2 + 100 i



              I'm afraid in the way you constructed your second inequality this is how Mathematica will handle the problem.



              https://math.stackexchange.com/questions/1116022/can-a-complex-number-ever-be-considered-bigger-or-smaller-than-a-real-number






              share|improve this answer









              $endgroup$


















                1












                $begingroup$

                With your inequality you should be careful about what you mean by "greater" or "less". Since the complex plane is principally vectors it is natural to compare magnitudes or 'radus' from the origin. Hence people normally use the absolute value.



                You can naievely compare the real parts however this can result in absurd inequalities since moving along the complex axis retains the equality. 2 + i = 2 + 100 i



                I'm afraid in the way you constructed your second inequality this is how Mathematica will handle the problem.



                https://math.stackexchange.com/questions/1116022/can-a-complex-number-ever-be-considered-bigger-or-smaller-than-a-real-number






                share|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  With your inequality you should be careful about what you mean by "greater" or "less". Since the complex plane is principally vectors it is natural to compare magnitudes or 'radus' from the origin. Hence people normally use the absolute value.



                  You can naievely compare the real parts however this can result in absurd inequalities since moving along the complex axis retains the equality. 2 + i = 2 + 100 i



                  I'm afraid in the way you constructed your second inequality this is how Mathematica will handle the problem.



                  https://math.stackexchange.com/questions/1116022/can-a-complex-number-ever-be-considered-bigger-or-smaller-than-a-real-number






                  share|improve this answer









                  $endgroup$



                  With your inequality you should be careful about what you mean by "greater" or "less". Since the complex plane is principally vectors it is natural to compare magnitudes or 'radus' from the origin. Hence people normally use the absolute value.



                  You can naievely compare the real parts however this can result in absurd inequalities since moving along the complex axis retains the equality. 2 + i = 2 + 100 i



                  I'm afraid in the way you constructed your second inequality this is how Mathematica will handle the problem.



                  https://math.stackexchange.com/questions/1116022/can-a-complex-number-ever-be-considered-bigger-or-smaller-than-a-real-number







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 30 '18 at 16:16









                  DrewDrew

                  111




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