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I'm trying to get the norm of a complex function with symbolic notation. But really I'm very inexperienced at this.

FullSimplify[Abs[ExpToTrig[Exp[I*x*t]]], Assumptions -> {t ∈ Reals, x ∈ Reals}]

With this code, I got 1. That's right!

So, now I'm trying to use this in the following problem: $\frac{e^{it(w_{21}+w)}-1}{w+w_{21}}+\frac{e^{it(w_{21}-w)}-1}{w_{21}-w}$

F[w_, t_] = Exp[I*w*t];
FullSimplify[
  Abs[
    ExpToTrig[
      (F[w + Subscript[w, 21], t] - 1)/(w + Subscript[w, 21]) + 
      (F[Subscript[w, 21] - w, t] - 1)/(Subscript[w, 21] - w)]],
  Assumptions -> {t ∈ Reals,Subscript[w, 21] ∈ Reals, w ∈ Reals}]

But in this case the function Abs doesn't work. Can you tell me where I made my mistakes?

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  • $\begingroup$ How do you expect Mathematica to do anything when it doesn't know anything about the function F[]? $\endgroup$ – Feyre May 1 '17 at 16:16
  • $\begingroup$ I had forgotten to put that part. $\endgroup$ – 7919 May 1 '17 at 16:30
  • $\begingroup$ Your first expression can be simplified to Abs[Exp[I*x*t]] // ComplexExpand which as expected evaluates to 1 $\endgroup$ – Bob Hanlon May 1 '17 at 16:30
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Use ComplexExpand

F[w_, t_] = Exp[I*w*t];

expr = Abs[(F[w + Subscript[w, 21], t] - 1)/(w + 
       Subscript[w, 21]) + (F[Subscript[w, 21] - w, t] - 
       1)/(Subscript[w, 21] - w)] // ComplexExpand

enter image description here

An alternate representation is

expr // ExpandAll // Simplify

enter image description here

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