Different output using `FullSimplify` and `TargetFunctions`

Question

Let's start with this function

expr = Abs[((x + a) (y + b)) c]^2

with this assumptions

$Assumptions={Element[x,Reals], Element[y,Reals] ,a>0,b>0,Element[c,Complexes]};

Now something strange happens when I use the TargetFunctions options: If I calculate

FullSimplify[(ComplexExpand[
    D[expr, y, x] /.{x -> 0, y -> 0}
    , c, TargetFunctions -> {Abs, Arg}]) - (ComplexExpand[
    D[expr, y, x] /.{x -> 0, y -> 0}
    , c])]

the output is 0, as I expect.

But if I calculate the same with an extra FullSimplyfy I have a different output:

FullSimplify[(ComplexExpand[
    D[expr, y, x] /.{x -> 0, y -> 0}
    , c, TargetFunctions -> {Abs, Arg}]) - (FullSimplify[ComplexExpand[
     D[expr, y, x] /.{x -> 0, y -> 0}
     , c]])]

here the output is

-2 a b (-Sqrt[c^2] Abs[c] + Sqrt[c^3 Conjugate[c]])

• Why is the last result not zero?
• Why is the output different in the two cases?

Thank you

Answer 1

This isn't really an answer to the two questions, but may yield some insights.

In SE post #24514 Stephen Kuhn reproduced function PowerContract, originally from the book Quantum Methods with Mathematica by James Feagin. Using that function one can get to the desired (zero) output.

$Assumptions = {Element[x, Reals], Element[y, Reals], a > 0, b > 0, 
   Element[c, Complexes]};
PowerContract[expr_] := 
 expr //. {m_^q_ n_^q_ :> (m n)^q /; ! IntegerQ[m] && ! IntegerQ[n], 
   m_^q_ n_^p_ :> (m/n)^q /; 
     q >= 0 && p == -q && ! IntegerQ[m] && ! IntegerQ[n]}
expr1 = -2 a b (-Sqrt[c^2] Abs[c] + Sqrt[c^3 Conjugate[c]]);
expr2 = expr1/(-2 a b) (* Get rid of constants *)
(* -Sqrt[c^2] Abs[c]+Sqrt[c^3 Conjugate[c]] *)
expr3 = PowerExpand[expr2]
(* -c Abs[c]+c^(3/2) Sqrt[Conjugate[c]] *)
expr4 = Factor[expr3]/c (* Dividing expr3 by c doesn't help *)
(* -Abs[c]+Sqrt[c] Sqrt[Conjugate[c]] *)
expr5 = PowerContract /@ expr4
(* -Abs[c]+Sqrt[c Conjugate[c]] *)
expr5 // FullSimplify
(* 0 *)