Absolute value and argument of a complex expression

Question

I would like to find the absolute value and the argument of the expression: $ x + \sqrt { x^2 + y^2 } $ under the assumption that the absolute values of $ x $ and $ y $ are equal.

I have written:

Simplify[ComplexExpand[Abs[x + (x^2 + y^2)^(1/2)], {x, y}, TargetFunctions -> {Abs, Arg}], 
  Assumptions -> Abs[x] == Abs[y]] // FullSimplify

Why does Mathematica return the un-evaluated expression?

I would like to have a solution that will also find a simplified version of my expression when Arg is substituted for Abs.

Answer 1

Try this:

   expr1 = x + Sqrt[x^2 + y^2];

Let us use tha condition that the absolute values of x and y are equal:

expr2 = Simplify[
  expr1 /. {x -> \[Rho]*Exp[I \[CurlyPhi]1], 
    y -> \[Rho]*Exp[I \[CurlyPhi]2]}, {\[Rho] > 0, \[CurlyPhi]1 > 
    0, \[CurlyPhi]2 > 0}]



 (*  (E^(I \[CurlyPhi]1) + Sqrt[
       E^(2 I \[CurlyPhi]1) + E^(2 I \[CurlyPhi]2)]) \[Rho]  *) 

expr3 = ComplexExpand[expr2]

The result is rather long:

  (*     \[Rho] Cos[\[CurlyPhi]1] + \[Rho] Cos[
       1/2 Arg[E^(2 I \[CurlyPhi]1) + E^(
          2 I \[CurlyPhi]2)]] ((Cos[2 \[CurlyPhi]1] + 
          Cos[2 \[CurlyPhi]2])^2 + (Sin[2 \[CurlyPhi]1] + 
          Sin[2 \[CurlyPhi]2])^2)^(1/4) + 
     I (\[Rho] Sin[\[CurlyPhi]1] + \[Rho] ((Cos[2 \[CurlyPhi]1] + 
             Cos[2 \[CurlyPhi]2])^2 + (Sin[2 \[CurlyPhi]1] + 
             Sin[2 \[CurlyPhi]2])^2)^(1/4)
          Sin[1/2 Arg[E^(2 I \[CurlyPhi]1) + E^(2 I \[CurlyPhi]2)]])                               *)

Now this gives the answer to your task: the absolute value:

     expr4 = Simplify[
  Sqrt[Simplify[
     Re[expr3], {\[Rho] > 0, \[CurlyPhi]1 > 0, \[CurlyPhi]2 > 0}]^2 + 
    Simplify[
     Im[expr3], {\[Rho] > 0, \[CurlyPhi]1 > 0, \[CurlyPhi]2 > 
       0}]^2], {\[Rho] > 0, \[CurlyPhi]1 > 0, \[CurlyPhi]2 > 0}]

(* [Rho] [Sqrt](1 + 2 Abs[Cos[[CurlyPhi]1 - [CurlyPhi]2]] + 2 Sqrt[2] Sqrt[Abs[Cos[[CurlyPhi]1 - [CurlyPhi]2]]] Cos[[CurlyPhi]1 - 1/2 Arg[E^(2 I [CurlyPhi]1) + E^(2 I [CurlyPhi]2)]]) *)

and the phase:

  expr5 = ArcTan@
      Simplify[Simplify[
        Im[expr3], {\[Rho] > 0, \[CurlyPhi]1 > 0, \[CurlyPhi]2 > 0}]/
       Simplify[
        Re[expr3], {\[Rho] > 0, \[CurlyPhi]1 > 0, \[CurlyPhi]2 > 
          0}], {\[Rho] > 0, \[CurlyPhi]1 > 0, \[CurlyPhi]2 > 0}]

(*          ArcTan[(Sin[\[CurlyPhi]1] + 
  Sqrt[2] Sqrt[Abs[Cos[\[CurlyPhi]1 - \[CurlyPhi]2]]]
    Sin[1/2 Arg[E^(2 I \[CurlyPhi]1) + E^(2 I \[CurlyPhi]2)]])/(
 Cos[\[CurlyPhi]1] + 
  Sqrt[2] Sqrt[Abs[Cos[\[CurlyPhi]1 - \[CurlyPhi]2]]]
    Cos[1/2 Arg[E^(2 I \[CurlyPhi]1) + E^(2 I \[CurlyPhi]2)]])]          *)

Now, if you want it in terms of x and y, you might do something like this:

    expr4 /. {\[Rho] -> Abs[x], \[CurlyPhi]1 -> Arg[x], \[CurlyPhi]2 -> 
   Arg[y]}

yielding

(* Abs[x] [Sqrt](1 + 2 Abs[Cos[Arg[x] - Arg[y]]] + 2 Sqrt[2] Sqrt[Abs[Cos[Arg[x] - Arg[y]]]] Cos[1/2 Arg[E^(2 I Arg[x]) + E^(2 I Arg[y])] - Arg[x]]) *)

while this gives the phase:

      expr5 /. {\[Rho] -> Abs[x], \[CurlyPhi]1 -> Arg[x], \[CurlyPhi]2 -> 
   Arg[y]}



 (*         ArcTan[(Sqrt[2] Sqrt[Abs[Cos[Arg[x] - Arg[y]]]]
          Sin[1/2 Arg[E^(2 I Arg[x]) + E^(2 I Arg[y])]] + 
        Sin[Arg[x]])/(Sqrt[2] Sqrt[Abs[Cos[Arg[x] - Arg[y]]]]
          Cos[1/2 Arg[E^(2 I Arg[x]) + E^(2 I Arg[y])]] + Cos[Arg[x]])]      *)

These expressions are so long that I doubt that you will use them. It seems to be better in its original form.

Answer 2

Let's give Mathematica some help. The condition that Abs[x]=Abs[y] can be rewritten using the polar form of complex numbers (where m is the common magnitude and tx, ty are the angles). Then the expression can be simplified by making explicit assumptions about the domain of the parameters.

x = m*Exp[I tx];
y = m*Exp[I ty];
FullSimplify[x + Sqrt[x^2 + y^2], 
      Assumptions -> {m > 0, 0 <= tx <= 2 Pi, 0 <= ty <= 2 Pi}]

(E^(I tx) + Sqrt[E^(2 I tx) + E^(2 I ty)]) m

To keep the phases the same:

x = m1*Exp[I t];
y = m2*Exp[I t];
FullSimplify[x + Sqrt[x^2 + y^2], Assumptions -> {m1 > 0, m2 > 0, 0 <= t <= 2 Pi}]

E^(I t) m1 + Sqrt[E^(2 I t) (m1^2 + m2^2)]