Can Mathematica prove inequalities involving complex numbers?












1














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







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  • Could you please provide code for those inequalities?
    – Andrew
    6 hours ago












  • |z - conjugate[z]w| + |w|^2 < 1
    – Jaikrishnan
    6 hours ago










  • z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    – Jaikrishnan
    6 hours ago
















1














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







New contributor




Jaikrishnan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Could you please provide code for those inequalities?
    – Andrew
    6 hours ago












  • |z - conjugate[z]w| + |w|^2 < 1
    – Jaikrishnan
    6 hours ago










  • z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    – Jaikrishnan
    6 hours ago














1












1








1


2





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







New contributor




Jaikrishnan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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







New contributor




Jaikrishnan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Jaikrishnan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






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Jaikrishnan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 6 hours ago









Jaikrishnan

1062




1062




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Jaikrishnan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Jaikrishnan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • Could you please provide code for those inequalities?
    – Andrew
    6 hours ago












  • |z - conjugate[z]w| + |w|^2 < 1
    – Jaikrishnan
    6 hours ago










  • z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    – Jaikrishnan
    6 hours ago


















  • Could you please provide code for those inequalities?
    – Andrew
    6 hours ago












  • |z - conjugate[z]w| + |w|^2 < 1
    – Jaikrishnan
    6 hours ago










  • z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
    – Jaikrishnan
    6 hours ago
















Could you please provide code for those inequalities?
– Andrew
6 hours ago






Could you please provide code for those inequalities?
– Andrew
6 hours ago














|z - conjugate[z]w| + |w|^2 < 1
– Jaikrishnan
6 hours ago




|z - conjugate[z]w| + |w|^2 < 1
– Jaikrishnan
6 hours ago












z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
– Jaikrishnan
6 hours ago




z^2 * conjugate[w] + conjugate[z]^2 * w + |w|^2 * (z^2 * conjugate[w] + conjugate[z]^2 * w - 4|z|^2) >= 0
– Jaikrishnan
6 hours ago










2 Answers
2






active

oldest

votes


















3














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





























    2














    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





















    • Or simply inequality /. {z -> I, w -> 1}.
      – Michael E2
      5 hours ago










    • @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
      – Thies Heidecke
      5 hours ago












    • 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. :)
      – Michael E2
      5 hours ago











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    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


























      3














      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
























        3












        3








        3






        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












        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 6 hours ago









        Andrew

        1,8161115




        1,8161115























            2














            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





















            • Or simply inequality /. {z -> I, w -> 1}.
              – Michael E2
              5 hours ago










            • @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
              – Thies Heidecke
              5 hours ago












            • 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. :)
              – Michael E2
              5 hours ago
















            2














            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





















            • Or simply inequality /. {z -> I, w -> 1}.
              – Michael E2
              5 hours ago










            • @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
              – Thies Heidecke
              5 hours ago












            • 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. :)
              – Michael E2
              5 hours ago














            2












            2








            2






            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












            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 5 hours ago









            Thies Heidecke

            6,8662438




            6,8662438












            • Or simply inequality /. {z -> I, w -> 1}.
              – Michael E2
              5 hours ago










            • @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
              – Thies Heidecke
              5 hours ago












            • 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. :)
              – Michael E2
              5 hours ago


















            • Or simply inequality /. {z -> I, w -> 1}.
              – Michael E2
              5 hours ago










            • @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
              – Thies Heidecke
              5 hours ago












            • 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. :)
              – Michael E2
              5 hours ago
















            Or simply inequality /. {z -> I, w -> 1}.
            – Michael E2
            5 hours ago




            Or simply inequality /. {z -> I, w -> 1}.
            – Michael E2
            5 hours ago












            @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
            – Thies Heidecke
            5 hours ago






            @MichaelE2 It's always easier if you already have a counterexample ;) Nice catch though! The truth is i just love existential quantifiers ^_^
            – Thies Heidecke
            5 hours ago














            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. :)
            – Michael E2
            5 hours ago




            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. :)
            – Michael E2
            5 hours ago










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