0
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    FindInstance[(Abs[(
     2 β)/(β + 
      Sqrt[-4 α + (β - γ)^2] + γ)] + 
    Abs[(4 α - 
      2 β (β + 
         Sqrt[-4 α + (β - γ)^2] - γ))/(\
β + 
       Sqrt[-4 α + (β - γ)^2] + γ)^2]) < 
  1, {α, β, γ}, 10]

It was running for a long time. Unfortunately no output came. Can anyone help to get it done? Also I was trying to reduce the inequality in terms of $\alpha$, $\beta$ and $\gamma$ but again same problem happened. Help to get it done, please. Note that $\alpha$, $\beta$ and $\gamma$ are complex numbers.

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  • $\begingroup$ Integers are complex numbers with the coefficient of the imaginary part being 0, so both answers below are finding valid solutions. $\endgroup$ – bobbym Jun 10 '16 at 4:28
  • $\begingroup$ True. When I mean complex numbers, they are truly complex number, meaning is imaginary part is non-zero. $\endgroup$ – Sk Sarif Hassan Jun 10 '16 at 4:40
2
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The problem seems to be with the Abs[ ] function. Since you have the sum of two Abs[ ] terms, and since you are only looking for specific examples, you can rewrite your FindInstance as:

FindInstance[{-1/2 < (2 \[Beta])/(\[Beta] + 
       Sqrt[-4 \[Alpha] + (\[Beta] - \[Gamma])^2] + \[Gamma]) < 1/2, 
       -1/2 < (4 \[Alpha] - 2 \[Beta] (\[Beta] + 
     Sqrt[-4 \[Alpha] + (\[Beta] - \[Gamma])^2] - \[Gamma]))/
       (\[Beta] + Sqrt[-4 \[Alpha] + (\[Beta] - \[Gamma])^2] + \[Gamma])^2 < 1/2}, 
     {\[Alpha], \[Beta], \[Gamma]}]

This returns {[Alpha] -> -7, [Beta] -> -1, [Gamma] -> 2}, which can be verified to fulfill the original equation as well.

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  • $\begingroup$ Can we reduce the inequality? Actually the $\alpha$, $\beta$ and $\gamma$ are all complex numbers. $\endgroup$ – Sk Sarif Hassan Jun 10 '16 at 3:59
0
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This gets some answers:

FindInstance[{10 > \[Alpha] > 0, 10 > \[Beta] > 0, 
10 > \[Gamma] > 
 0, (Abs[(2 \[Beta])/(\[Beta] + 
 Sqrt[-4 \[Alpha] + (\[Beta] - \[Gamma])^2] + \[Gamma])] + 
 Abs[(4 \[Alpha] - 
 2 \[Beta] (\[Beta] + 
 Sqrt[-4 \[Alpha] + (\[Beta] - \[Gamma])^2] - \[Gamma]))/(\
 \[Beta] + Sqrt[-4 \[Alpha] + (\[Beta] - \[Gamma])^2]  +\[Gamma])^2]) <1}, {\[Alpha], \[Beta], \[Gamma]}, Integers, 10]

I upvoted bill s' solution but some answers are better than nothing.

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  • $\begingroup$ Actually the α, β and γ are all complex numbers. What is more important to get complex instances rather than real numbers. Also if we could make the reduced inequality, that would be sufficient. $\endgroup$ – Sk Sarif Hassan Jun 10 '16 at 4:04
  • $\begingroup$ Hi; please add that to the original problem. $\endgroup$ – bobbym Jun 10 '16 at 4:06

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