Trying to obtain the result of Conjugate[a/b+c/d] only gives the same result. As such, I use Refine as:

  Conjugate[a/b + c/d], a ∈ Complexes && b ∈ Reals && c ∈ Reals && d ∈ Complexes]

with the output

Conjugate[a/b + c/d]

Funny thing is that if I try

Refine[Conjugate[a/b + c/d],a ∈ Reals && b ∈ Reals && c ∈ Reals && d ∈ Reals]

I get

a/b + c/d

or, if I try

Refine[Conjugate[a/b + c], a ∈ Complexes && b ∈ Reals && c ∈ Reals]

I get

c + Conjugate[a]/b

As soon as I have two fractions inside Conjugate], with any of the elements as Complex, I do not get the conjugate of each element, which is my obvious goal. Any reason why this happens?

I manage to get the desired result in this case for example:

In[252]:= Distribute@
 Conjugate[fc (2 E^((2 I fc \[Pi] (R1 + R2))/\[ConstantC]) mu1 c)]

Out[252]= 2 E^(-((2 I \[Pi] Conjugate[fc (R1 + R2)])/
  Conjugate[\[ConstantC]])) Conjugate[c fc mu1]

In[253]:= FullSimplify[%]

Out[253]= 2 E^(-((2 I fc \[Pi] (R1 + R2))/\[ConstantC])) fc Conjugate[
  c] Conjugate[mu1]

In[254]:= TraditionalForm[%]

Out[254]//TraditionalForm= 2 fc c^\[Conjugate] mu1^\[Conjugate] E^(-((2 I \[Pi] fc (R1+R2))/\[ConstantC]))

However as soon as I try to do it for

Distribute@ Conjugate[ fc (2 E^((2 I fc [Pi] (R1 + R2))/[ConstantC]) mu1 c + E^((4 I fc [Pi] R1)/[ConstantC]) mu2 d)]

the same series of commands produces only

fc (2 c E^((2 I fc \[Pi] (R1+R2))/\[ConstantC]) mu1+d E^((4 I fc \[Pi] R1)/\[ConstantC]) mu2)^\[Conjugate]

not what I would expect the result to be

2 fc c^\[Conjugate] mu1^\[Conjugate] E^(-((2 I \[Pi] fc (R1+R2))/\[ConstantC])) + fc d^\[Conjugate] mu2^\[Conjugate] E^(-((4 I \[Pi] fc R1)/\[ConstantC]))
  • $\begingroup$ Do avoid using C and D for this, as they are reserved symbols. $\endgroup$ Commented May 24, 2016 at 13:37
  • $\begingroup$ Yes, it was just an example. I actually have more complicated terms than A, B, C or D. Thanks for the quick input! $\endgroup$
    – user40410
    Commented May 24, 2016 at 13:52
  • $\begingroup$ Welcome! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$
    – user9660
    Commented May 24, 2016 at 13:59
  • $\begingroup$ You should edit the question to fix the caps, it is distracting from the issue. I suspect the problem is simply that the result you "expect" is regarded as more complicated than the result you get. $\endgroup$
    – george2079
    Commented May 24, 2016 at 14:30
  • 1
    $\begingroup$ Conjugate[a/b + c/d] returns Conjugate[a/b + c/d]. What else did you expect? Your error is in using capital C and D. They are reserved, and Mma could not understand you. $\endgroup$ Commented May 24, 2016 at 14:47

2 Answers 2


I find that getting the answer I want with this type of expression requires the use of a range of techniques. In this case, I would use

ComplexExpand[Conjugate[a/b + c/d], {a, d},  TargetFunctions -> Conjugate]


Conjugate[a]/b + c/Conjugate[d]

You probably need to distribute every product over sums in your expression in order for Conjugate to propagate properly:

distributeProducts[expr_] := 
 Replace[expr, t : Times[___] :> Distribute[t], {0, \[Infinity]}]

  fc (2 c E^((2 I fc \[Pi] (R1 + R2))/\[ConstantC]) mu1 + 
     d E^((4 I fc \[Pi] R1)/\[ConstantC]) mu2)]

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