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.
ComplexExpand
,Simplify
, andFullSimplify
all return your expression unevaluated? Are you aware thatFullSimplify
doesn't inherit your assumptions fromSimplify
? Do have any simplification of your expression in mind that you expect Mathematica to find? $\endgroup$