1
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ 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
1
$\begingroup$

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.