Removing Abs from Abs[a + Exp[I*c]b]^2

Question

Title says it all. Mathematica produces sometimes expressions of the form

Abs[a + Exp[I*c]b]^2

where all three quantities a, b and c are positive real numbers. That expression of course reduces simply to the law of cosines

a^2 + b^2 + 2a b Cos[c]

however, I cannot get Mathematica to display it in this form. Any ideas?

Answer 1

Update

Not sure why all responses here are not simple and direct, maybe in older versions this did not work, but now it does:

Abs[a+Exp[I*c] b]^2//ComplexExpand//FullSimplify

a^2+b^2+2 a b Cos[c]

(Abs[a+Exp[I*c] b]^2+Abs[a-Exp[I*c] b]^2)/(2 Abs[a+I b]^2)//ComplexExpand//FullSimplify

1

$Version

14.0.0 for Mac OS X x86 (64-bit) (December 1, 2023)

Original answer

Until someone figures out how to do it in a better way, here is a work around. Define a function (which is true only for your specific case of parameters):

RemoveAbs[x_] := FullSimplify[ComplexExpand[Sqrt[x Conjugate[x]]]]

Then

Abs[a + Exp[I*c] b]^2 /. Abs -> RemoveAbs

a^2 + b^2 + 2 a b Cos[c]

There is a simpler version inspired by @Nasser answer. Define a function

RemoveAbs[x_] := ComplexExpand[Abs[x]]

and now, for the sake of variety, apply simplification at the end:

Abs[a + Exp[I*c] b]^2 /. Abs -> RemoveAbs // FullSimplify

a^2 + b^2 + 2 a b Cos[c]

Point is, all the above will work for more complicated cases because /. is a ReplaceAll. Even for cases that are convoluted via things like, for example, TraditionalForm - the only thing you need is that InputForm still has Abs in it. For instance, imagine a beast of expression, here rather a simpler one but in reality they could go for pages:

(Abs[a + Exp[I*c] b]^2 + Abs[a - Exp[I*c] b]^2)/(2 Abs[a + I b]^2) // TraditionalForm

and now, voila, you get your simplification:

% /. Abs -> RemoveAbs // FullSimplify

1

Answer 2

ClearAll[a, c, b];
expr = a + Exp[I*c] b;
Simplify@Expand@(ComplexExpand[Abs@expr]^2)

Or in non @ friendly way

Simplify [ Expand [ ComplexExpand[Abs[expr]]^2 ] ]

Answer 3

I like very much the solution of Nasser, especially since it is direct, i.e. it does not use any workaround, but just the Mathematica technique. Not to leave such a basic exercise with only one solution I would like to offer a very simple one:

    Clear[exprA, exprB];
exprA = a + Exp[I*c] b
exprB = exprA /. c -> -c (* This produces the conjugated expression *)
   (* a + b E^(I c)    *)
   (*  a + b E^(-I c)  *)

Now let us recall that Abs[z]^2=z z* and just multiply

    ComplexExpand[exprA*exprB] // Simplify

     (*   a^2 + b^2 + 2 a b Cos[c]  *)

It should be noted that I used the replacement c->-c to obtain the conjugated value using that it is known all parameters to be positive, including c. It can be done in a more general way using Conjugate[]. In this case the final result is the same, but the intermediate results (if one chooses to look at them) are more cumbersome. Anyway,

    ComplexExpand[exprA*Conjugate[exprA]] // Simplify
    (* a^2 + b^2 + 2 a b Cos[c]  *)