1
$\begingroup$

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 wish to obtain the poles for the denominator of the function:

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

The result is of the following:

{-(1/6) I (Γ + κ1 + κ2) + (I (-16 (\
Γ + κ1 + κ2)^2 + 
   48 (4 g1^2 + 
      4 g2^2 + Γ κ1 + Γ...}

Basically a really long ugly solution. My goal is to obtain the conjugate of wroots1, however, when I do it straightforwardly:

wroots2 = 
Simplify[Conjugate[wroots1], 
Assumptions -> {Γ ∈ 
  Reals, κ1 ∈ Reals, κ2 ∈ Reals, 
 g1 ∈ Reals, 
 g2 ∈ Reals, ω ∈ 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}];

I am returned with:

{1/48 I (8 (Γ + κ1 + κ2) + 
8 2^(1/3) (-12 g1^2 - 
   12 g2^2 + (Γ + κ1 + κ2)^2 - 
   3 (κ1 κ2 + Γ (κ1 + \
κ2))) Conjugate[
  1/(-36 g1^2 Γ + 72 g2^2 Γ + 
     2 Γ^3 - 36 g1^2 κ1 - 
     36 g2^2 κ1 - 3 Γ^2 κ1 - 
     3 Γ κ1^2 + 2 κ1^3 + 
     72 g1^2 κ2 - 36 g2^2 κ2 - 
     3 Γ^2 κ2 + 
     12 Γ κ1 κ2 - 
     3 κ1^2 κ2 - 3 Γ κ2^2 - 
     3 κ1 κ2^2 + 2 κ2^3...}

Clearly, Conjugate refuses to take the conjugate of said function. I tried doing

wroots2 = wroots1 /. {I -> -I}

But that doesn't work as well. I'm at lost at what to do here and I could use any help I can get. Thank you very much in advance.

$\endgroup$
1
  • $\begingroup$ You can get the Conjugate by wroots1 /. Complex[a_, b_] -> Complex[a, -b] , but I'm not clear, what you realy want. $\endgroup$
    – Akku14
    Commented May 2, 2018 at 7:22

2 Answers 2

1
$\begingroup$

If you Conjugate the equation before Solve, you'll get the complex-conjugated solutions (complex analysis is nice!):

wroots1 = w /. Solve[Denominator[f[w]] == 0, w];
wroots1C = w /. Solve[ComplexExpand[Conjugate[Denominator[f[w]]]] == 0, w];

This way you don't have to complex-conjugate the solutions. The downside is that the order of the solutions in wroots1 and wroots1C may not be the same.

$\endgroup$
0
$\begingroup$

You can use ComplexExpand, but the cube root term (what I define as cRoot below) interferes with its logic. The following could be a work-around (suppressing all output):

wroots2 = FullSimplify[Conjugate[wroots1], Assumptions -> {Γ ∈ Reals, κ1 ∈ Reals, κ2 ∈ Reals, g1 ∈ Reals, g2 ∈ Reals, ω ∈ 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}];

wroots2 = wroots2 /. (-36 g1^2 (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \[Kappa]2) \ (36 g2^2 + (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) (\[CapitalGamma] - 2 \[Kappa]1 + \[Kappa]2)) + 1/64 \[Sqrt](4 (-16 (\[CapitalGamma] + \[Kappa]1 + \[Kappa]2)^2 \ + 48 (4 g1^2 + 4 g2^2 + \[CapitalGamma] \[Kappa]1 + (\[CapitalGamma] \ + \[Kappa]1) \[Kappa]2))^3 + 4096 (-36 g1^2 (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \ \[Kappa]2) (36 g2^2 + (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) (\[CapitalGamma] - 2 \[Kappa]1 + \[Kappa]2)))^2))^(1/3) -> cRoot

wroots2 = wroots2 /. 1/(-36 g1^2 (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \[Kappa]2) \ (36 g2^2 + (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) (\[CapitalGamma] - 2 \[Kappa]1 + \[Kappa]2)) + 1/64 \[Sqrt](4 (-16 (\[CapitalGamma] + \[Kappa]1 + \[Kappa]2)^2 \ + 48 (4 g1^2 + 4 g2^2 + \[CapitalGamma] \[Kappa]1 + (\[CapitalGamma] \ + \[Kappa]1) \[Kappa]2))^3 + 4096 (-36 g1^2 (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) + (2 \[CapitalGamma] - \[Kappa]1 - \ \[Kappa]2) (36 g2^2 + (\[CapitalGamma] + \[Kappa]1 - 2 \[Kappa]2) (\[CapitalGamma] - 2 \[Kappa]1 + \[Kappa]2)))^2))^(1/3) -> (1/cRoot)

ComplexExpand[wroots2]

You can then substitute the expression for cRoot back into the answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.