7
$\begingroup$

How do I complex conjugate a vector?

E^(-((I β)/2)) p (Cos[α/2] (Cos[θ]^2 + Sin[θ]^2 Sin[ϕ] (-I Cos[ϕ] + Sin[ϕ]))
+ E^(I β)Sin[α/2] (Cos[θ]^2 + Sin[θ]^2 Sin[ϕ] (I Cos[ϕ] + Sin[ϕ])))

I tried doing Assuming[β ∈ Reals, c[[1]]^*, but that didn't work :\ ... I just want i's flipped, but I'm getting this:

enter image description here

Attempt at using ComplexExpand as suggested my LLiaMnYP. Still doesn't look as compact as c[[1]] if I just change all the i's by hand :(

enter image description here

Improve this question
$\endgroup$
4
  • $\begingroup$ You are getting this as there is no way for Mathematica to know, that all other variables are real. Check out the documentation for ComplexExpand IIRC. $\endgroup$
    – LLlAMnYP
    Commented Mar 17, 2015 at 23:30
  • $\begingroup$ tried it... see pic above... naturally, Mathematica decided to expand exponentials and rearrange everything %\ and cc* product has changed to a really long string compared to what it was when I just changed i's manually $\endgroup$
    – Raksha
    Commented Mar 17, 2015 at 23:39
  • $\begingroup$ Is p a factor or a function? $\endgroup$
    – Mahdi
    Commented Mar 17, 2015 at 23:43
  • 1
    $\begingroup$ p is a factor here $\endgroup$
    – Raksha
    Commented Mar 17, 2015 at 23:44

3 Answers 3

Reset to default
14
$\begingroup$

Is this what you wanted?

expr = E^(-((I β)/2)) p (Cos[α/2] (Cos[θ]^2 + Sin[θ]^2 Sin[ϕ] (-I Cos[ϕ] + Sin[ϕ])) + 
    E^(I β) Sin[α/2] (Cos[θ]^2 + Sin[θ]^2 Sin[ϕ] (I Cos[ϕ] + Sin[ϕ])))

Mathematica graphics

expr /. Complex[x_, y_] :> Complex[x, -y]

Mathematica graphics

Improve this answer
$\endgroup$
7
  • $\begingroup$ YES! Thank you! Tho it also returned "Complex::argr: Complex called with 1 argument; 2 arguments are expected." for some reason. $\endgroup$
    – Raksha
    Commented Mar 17, 2015 at 23:56
  • $\begingroup$ @Solarmew please try it now. $\endgroup$
    – Nasser
    Commented Mar 18, 2015 at 0:00
  • $\begingroup$ Perfect :] Thanks so much! $\endgroup$
    – Raksha
    Commented Mar 18, 2015 at 0:25
  • 1
    $\begingroup$ Cool trick. This does literally "flip the i's", as OP requested. Just be warned, if you have I represented as Sqrt[-1] anywhere, this will probably not work. $\endgroup$
    – LLlAMnYP
    Commented Mar 18, 2015 at 0:44
  • 1
    $\begingroup$ Agreed, I was too lazy to extend to more elaborate constructs such as Sqrt[-a] (where a may be positive, but not have a value assigned to it and remain in symbolic form) and a large class of functions, although elementary, but for which the relation Conjufate[f[x]]==f[Conjugate[x]] does not hold. Of course, OP's expression does happen to be one of those where you just need to flip the Is, I just wanted to point out, that your trick may not work in some cases even if all variables are real. $\endgroup$
    – LLlAMnYP
    Commented Mar 18, 2015 at 1:31
4
$\begingroup$

You might use:

  FullSimplify@ComplexExpand@Conjugate[(* expression *)]

in your case, it returns:

$$ e^{\frac{i \beta }{2}} p \left(\sin \left(\frac{\alpha }{2}\right) e^{-i (\beta -\phi )} \left(\cos ^2(\theta ) \cos (\phi )-i \sin (\phi )\right)+\cos \left(\frac{\alpha }{2}\right) \left(\cos ^2(\theta )+\sin ^2(\theta ) \sin (\phi ) (\sin (\phi )+i \cos (\phi ))\right)\right) $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Mine still returned the same expression as in the first picture, but without Conjugate[ ] around beta. (how did you input such pretty text so fast?) I also typed {beta, p, alpha, theta, phi} instead of just beta. That got p outside of Conjugate[ ] but left all the trig functions in. Tried including trig function, but that also didn't do anything. $\endgroup$
    – Raksha
    Commented Mar 17, 2015 at 23:43
  • 1
    $\begingroup$ @Solarmew such "pretty text" is input using $\LaTeX$. See a basic help page and an extended tutorial about what MathJax supports. $\endgroup$
    – Ruslan
    Commented Mar 18, 2015 at 4:45
3
$\begingroup$

You can also do

Refine[Conjugate@c[[1]],
       Assumptions->(\[Alpha] | \[Beta] | \[Phi]) \[Element] Reals]

This also works in cases when you have additional manifestly complex variables in your expression (ComplexExpand assumes that all variables are real). Of course in that case you would not add those in your Assumptions. For example:

Refine[Conjugate[a b], Assumptions -> a \[Element] Reals]
(*
==> a Conjugate[b]
*)
Improve this 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.