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Could you solve this system of equations, please?

Posted 10 years ago

I want to solve this:

Solve[{-k - 
    I Sqrt[1 - 
       k^2] == (1 - (2 (e + I Sqrt[1 - e^2]) (f + 
           I Sqrt[1 - f^2]))/(e + I Sqrt[1 - e^2] + f + 
         I Sqrt[1 - f^2]))/(1 - (2 (e - I Sqrt[1 - e^2]) (f - 
           I Sqrt[1 - f^2]))/(e - I Sqrt[1 - e^2] + f - 
         I Sqrt[1 - f^2])), -m - 
    I Sqrt[1 - 
       m^2] == (1 - (2 (d + I Sqrt[1 - d^2]) (f + 
           I Sqrt[1 - f^2]))/(d + I Sqrt[1 - d^2] + f + 
         I Sqrt[1 - f^2]))/(1 - (2 (d - I Sqrt[1 - d^2]) (f - 
           I Sqrt[1 - f^2]))/(d - I Sqrt[1 - d^2] + f - 
         I Sqrt[1 - f^2])), -(k + I Sqrt[1 - k^2]) (m + 
      I Sqrt[1 - 
         m^2]) == (1 - (2 (d + I Sqrt[1 - d^2]) (e + 
           I Sqrt[1 - e^2]))/(d + I Sqrt[1 - d^2] + e + 
         I Sqrt[1 - e^2]))/(1 - (2 (d - I Sqrt[1 - d^2]) (e - 
           I Sqrt[1 - e^2]))/(d - I Sqrt[1 - d^2] + e - 
         I Sqrt[1 - e^2]))}, {d, e, f}]

But my computer probably isn't good enough for advanced computation, so maybe someone will help me. Could you paste it to Mathematica in your computer and tell me solution?

POSTED BY: Jan Redwi?ski
2 Replies

Also, the way you have Solve written here, it will look for Complex solutions for d, e, and f. I'm guessing you don't that?. If that's the case, you would begin by using ComplexExpand to separate the equations in their real and complex components and set them equal to each other. That way you'd only be working with Real equations. Unfortunately, you don't want to do this for this equation, because you'd loose the important insight the complex numbers are giving you.

POSTED BY: Sean Clarke

Could you paste it to Mathematica in your computer and tell me solution?

That doesn't work most of the time. If doing math were just about more powerful computers life would be too easy. You have to do some work yourself to solve the equation.

The first step in working with an equation like this is to make it easier to understand. You can make a substitution that would make this simpler.

newd = d + I Sqrt[1 - d^2]
Conjugate[newd] = d - I Sqrt[1 - d^2]

What does your equation look like when you rewrite it with these substitutions?

POSTED BY: Sean Clarke
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