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The function

    fun[x_, y_, 
   z_] = (4 y^2 - x^2 Cos[1/2 Sqrt[-x^2 + 4 y^2] z] - 
     x Sqrt[-x^2 + 4 y^2]
       Sin[1/2 Sqrt[-x^2 + 4 y^2] z])/(\[Sqrt](2 x^2 y^2 + 16 y^4 - 
       8 x^2 y^2 Cos[1/2 Sqrt[-x^2 + 4 y^2] z] + 
       x^2 (x^2 - 2 y^2) Cos[Sqrt[-x^2 + 4 y^2] z] - 
       8 x y^2 Sqrt[-x^2 + 4 y^2] Sin[1/2 Sqrt[-x^2 + 4 y^2] z] + 
       x^3 Sqrt[-x^2 + 4 y^2] Sin[Sqrt[-x^2 + 4 y^2] z]));

when checked for different values of x,y,z always comes out to be 1, say fun[0.1, 0.2, 0.3]=1. However, Mathematica is not able to show this in the first place!

     Simplify[fun[x, y, z]]

Out[25]= (4 y^2 - x^2 Cos[1/2 Sqrt[-x^2 + 4 y^2] z] - 
   x Sqrt[-x^2 + 4 y^2]
     Sin[1/2 Sqrt[-x^2 + 4 y^2] z])/(\[Sqrt](2 x^2 y^2 + 16 y^4 - 
     8 x^2 y^2 Cos[1/2 Sqrt[-x^2 + 4 y^2] z] + 
     x^2 (x^2 - 2 y^2) Cos[Sqrt[-x^2 + 4 y^2] z] - 
     8 x y^2 Sqrt[-x^2 + 4 y^2] Sin[1/2 Sqrt[-x^2 + 4 y^2] z] + 
     x^3 Sqrt[-x^2 + 4 y^2] Sin[Sqrt[-x^2 + 4 y^2] z]))
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2
  • $\begingroup$ Are you making assumptions on x, y, and z? Mathematica won't know about those unless you specify them. $\endgroup$ Feb 10, 2021 at 10:35
  • $\begingroup$ I even tried with assumptions that $x\ge0, y\ge0,z\ge0$, but it doesn't work! $\endgroup$
    – Mike
    Feb 10, 2021 at 10:46

2 Answers 2

2
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Look for instance at

z = 0.7; Plot3D[fun[x, y, z], {x, -3, 3}, {y, -3, 3}]

then you see why.

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0
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With

fun[x_, y_, z_] = (4 y^2 - x^2 Cos[1/2 Sqrt[-x^2 + 4 y^2] z] - 
 x Sqrt[-x^2 + 4 y^2]
   Sin[1/2 Sqrt[-x^2 + 4 y^2] z])/(\[Sqrt](2 x^2 y^2 + 16 y^4 - 
   8 x^2 y^2 Cos[1/2 Sqrt[-x^2 + 4 y^2] z] + 
   x^2 (x^2 - 2 y^2) Cos[Sqrt[-x^2 + 4 y^2] z] - 
   8 x y^2 Sqrt[-x^2 + 4 y^2] Sin[1/2 Sqrt[-x^2 + 4 y^2] z] + 
   x^3 Sqrt[-x^2 + 4 y^2] Sin[Sqrt[-x^2 + 4 y^2] z]));

Try instead

fun[x, y, z]^2 // FullSimplify

The problem is in the denominator's square root.

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