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I have a complex function and I want to take a list of derivatives of it. My code reads as follows:

$Assumptions[w>0] 
f[r, t] = Exp[-I wt] (It Exp[-I wr]/r + R Exp[I w r]/r)
ComplexExpand[Conjugate[f[r, t]]]

In my function, I want the coefficients It and R to be complex numbers.
However, this gives me the following output:

(R Cos[r w] Cos[wt])/r + (It Cos[wr] Cos[wt])/r + (R Sin[r w] Sin[wt])/r - (It Sin[wr] Sin[wt])/r + I (-((R Cos[wt] Sin[r w])/r) + (It Cos[wt] Sin[wr])/r + (R Cos[r w] Sin[wt])/r + (It Cos[wr] Sin[wt])/r

This treats It and R as real numbers and it expands the exponentials using Euler's identity. How do I get the conjugate of the functions in a simpler form, and taking It and R as complex.

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  • $\begingroup$ REad the manual about "ComplexExpand" and you will see that you need to indicate which variables are complex. $\endgroup$ Oct 25, 2021 at 8:11
  • $\begingroup$ That still returns everything in terms of Sin and Cos even though the input is in terms of exponentials. The complex numbers themselves are also expressed as Re[] and Im[]. Is there some kind of z z^* style notation available? $\endgroup$
    – newtothis
    Oct 25, 2021 at 8:20
  • $\begingroup$ Look at "TrigToExp" $\endgroup$ Oct 25, 2021 at 10:14
  • $\begingroup$ Hard to tell if you mean wr or w*r in your code since you have it both ways. It puts what you mean bywt in doubt also. $\endgroup$
    – Bill Watts
    Jul 22 at 21:46

1 Answer 1

1
$\begingroup$
    $Assumptions[w > 0]
f[r, t] = Exp[-I w t] (It Exp[-I w r]/r + R Exp[I w r]/r)
ce = ComplexExpand[Conjugate[f[r, t]], {It, R}, 
TargetFunctions -> Conjugate] // Simplify
ce // TraditionalForm
ce // TrigToExp
ce // TrigToExp // TraditionalForm
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