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I have some problem with the derivative of the 'conjugate' expression: Here I define the functions:

f[x_, y_] := x + I y; g[x_, y_] := x + 3 I y y; h[x_, y_] := Derivative[0, 1][f[#, #2]\[Conjugate]/g[#, #2] &][x, y]

When I try to evaluate one of the expression, for example

h[1, 1]

I got something like 

Derivative[1][Conjugate][1 + I]

which is not true. If I try the code suggested in Derivative of conjugate multivariate function 

 excluded = "ExcludedFunctions" /. ("DifferentiationOptions" /. SystemOptions["DifferentiationOptions"]) SetSystemOptions[ "DifferentiationOptions" -> "ExcludedFunctions" -> Union[excluded, {Conjugate}]]

I got the error

 General::ivar: 1 is not a valid variable.
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  • 1
    $\begingroup$ I usually DIY my own "conjugate" by an explicit replacement rule. $\endgroup$ Commented Dec 12, 2018 at 7:32
  • $\begingroup$ Could you please show this on the example written above $\endgroup$ Commented Dec 12, 2018 at 7:41
  • $\begingroup$ I decided so because I think it not good enough to be posted. $\endgroup$ Commented Dec 12, 2018 at 8:50
  • $\begingroup$ Well, I hope it helps. $\endgroup$ Commented Dec 12, 2018 at 8:55

3 Answers 3

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Update

I am sorry, the old answer does not work, but it seems that the rule cannot be applied to pure functions.

A working way is, provided that all symbols are deemed real, which is what ComplexExpand does:

f[x_, y_] = ComplexExpand[ (*Expressions containing Conjugate*) ]; h[x_, y_] := Derivative[0, 1][f[#, #2]/g[#, #2] &][x, y] h[x, y] h[1, 1] 
-((6 I (x - I y) y)/(x + 3 I y^2)^2) - I/(x + 3 I y^2) -(9/50) + (37 I)/50

Old

It is long since I made the observation that Mathematica seems not to know what to do when Conjugate meets D.

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  • $\begingroup$ thx. I have already noticed that your first suggested example gives the wrong answer. Will your second example decrease the calculation rate? I have a very large amount of functions and need to do rather complex calculations. Is it possible not to replace all the functions like 'f[x,y][Conjugate]' by 'Evaluate[ComplexExpand[f[x, y][Conjugate]]]'? $\endgroup$ Commented Dec 12, 2018 at 8:24
  • $\begingroup$ @Chipa-Chipa Then I suggest that you do not use a pure function in the middle of Derivative, but use the explicit form of the function you want to go further. $\endgroup$ Commented Dec 12, 2018 at 8:27
  • $\begingroup$ Do you mean that I should make the replacement only when I define the derivative? $\endgroup$ Commented Dec 12, 2018 at 8:31
  • $\begingroup$ @Chipa-Chipa I mean to obtain the explicit expression of the function, before you put it into its derivative calculation. $\endgroup$ Commented Dec 12, 2018 at 8:33
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    $\begingroup$ I think the problem with your original way was a misplaced Evaluate. Try myConjugate = # /. {I -> -I, -I -> I} &; h[x_, y_] := Derivative[0, 1][Evaluate[myConjugate[f[#, #2]]/g[#, #2]] &][x, y] $\endgroup$ Commented Jan 11, 2019 at 19:54
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Another workaround. Looks like the Derivative format doesn't like conjugate, but this seems to work:

f[x_, y_] := x + I y; g[x_, y_] := x + 3 I y y; h[x_, y_] := D[Conjugate[f[x, yy]]/g[x, yy], yy] /. yy -> y h[1, 1] // ComplexExpand (*-(9/50) + (37 I)/50*)
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  • $\begingroup$ Hi, this code failed in v13. $\endgroup$ Commented Jun 21, 2023 at 3:36
  • $\begingroup$ @lotus2019 Are you sure? $\endgroup$ Commented Jun 23, 2023 at 6:17
  • $\begingroup$ Yes. In v13.0, the result is 3/25 + 1/10 Im[Derivative[1][Conjugate][1]] + I (21/25 - 3/10 Im[Derivative[1][Conjugate][1]] - 1/10 Re[Derivative[1][Conjugate][1]]) - 3/10 Re[Derivative[1][Conjugate][1]] $\endgroup$ Commented Jun 23, 2023 at 11:50
  • $\begingroup$ @lotus2019 I don't have v13.0, but in v13.2.1, the output is as I have it. $\endgroup$ Commented Jun 23, 2023 at 20:16
  • $\begingroup$ OK, it seems that I need to upgrade the version. Thank you for your reply. $\endgroup$ Commented Jun 24, 2023 at 0:03
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Here are three similar ways. The first seems best (simplest) to me:

ClearAll[h]; Block[{x, y}, h[x_, y_] = D[ComplexExpand[f[x, y]\[Conjugate]/g[x, y]], y]; ] h[x, y] // Simplify (* -((I (x + 6 x y - 3 I y^2))/(x + 3 I y^2)^2) *)

The following one generates message every time h[x, y] is evaluated, but it seems to give the correct answer. The reason for the messages is interesting. It implies a potential pitfall in using ComplexExpand: It appears the internal code for ComplexExpand[] creates a function with & using its arguments. The Slot[] expressions in f[#, #2] and g[#, #2] get caught up in this and lead to error messages; but the internal code seems to recover in this case. (Perhaps it should be considered a bug?)

ClearAll[h]; h[x_, y_] := Derivative[0, 1][Evaluate@ComplexExpand[f[#, #2]\[Conjugate]/g[#, #2]] &][x, y]

One could also do something like this:

ClearAll[h]; h[x_, y_] := Derivative[0, 1][ Block[{xx, yy}, Function[{xx, yy}, Evaluate@ComplexExpand[f[xx, yy]\[Conjugate]/g[xx, yy]]]]][x, y];
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