0
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I have the following function of ω

f[ω_] := (
2 Sqrt[Γ] (4*
 g2^2 + (κ1 - 2*I*ω) (κ2 - 2*I*ω)))/(
4*g2^2 (Γ - 
 2*I*ω) + (4*
  g1^2 + (Γ - 2*I*ω) (κ1 - 
    2*I*ω)) (κ2 - 2*I*ω))

And I intend to find the roots of the denominator. I do the following:

wroots1 = x /. Solve[(Denominator[f[ω]] /. {ω -> x}) == 0, x] // FullSimplify

And I get a really long, ugly solution. However, the problem lies with my attempt to obtain the conjugate of wroots1:

wroots2 = FullSimplify[Conjugate[wroots1], Assumptions -> {Γ ∈ 
 Reals, κ1 ∈ Reals, κ2 ∈ Reals, 
g1 ∈ Reals, g2 ∈ Reals, 
4 (-16 (Γ + κ1 + κ2)^2 + 
     48 (4 g1^2 + 
        4 g2^2 + Γ κ1 + (Γ + \
κ1) κ2))^3 + 
  4096 (-36 g1^2 (Γ + κ1 - 
        2 κ2) + (2 Γ - κ1 - \
κ2) (36 g2^2 + (Γ + κ1 - 
           2 κ2) (Γ - 
           2 κ1 + κ2)))^2 > 0}]

Rather than getting the conjugate of the 3 roots from wroots1, I obtain the same solutions for wroots1 with the exception of the word "conjugate" being tacked in front of certain terms. What is the issue here? I could use any help I can get.

Thank you in advance for the help.

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  • $\begingroup$ It doesn't work because expressions depend not only on parameters being reals, but also on their values. You cannot conjugate square root without knowing if expression inside positive or negative. $\endgroup$ – Vsevolod A. Mar 18 '18 at 20:23
  • $\begingroup$ @VsevolodA. Thanks for your response. If you were to compile and print wroots1, you'd see that the terms in the square root are precisely part of the conditions that I set to be bigger than 0. Namely: 4(-16([Gamma]....) > 0. Those are the only terms that are in the square root of wroots1. $\endgroup$ – kowalski Mar 18 '18 at 20:31
  • $\begingroup$ If all roots are bigger than 0 why don't you just /. I -> -I $\endgroup$ – Vsevolod A. Mar 18 '18 at 20:34
  • $\begingroup$ @VsevolodA. I'm not saying that all roots are bigger than 0, I don't know that. If I understand what you said earlier correctly, conjugate doesn't work because it doesn't know if the terms inside the square root is positive or negative. I've circumvented that by setting said condition (the quantities in the square root) to be more than 0 (positive) but it still didn't work $\endgroup$ – kowalski Mar 18 '18 at 20:40
  • $\begingroup$ By roots I meant square roots. $\endgroup$ – Vsevolod A. Mar 18 '18 at 20:50

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