I understand that there are many solutions that will produce the same arctangent, nevertheless I really need a simplification:

$Assumptions = {w0 > 0, a > 0, G \[Element] Reals}
FullSimplify[ArcTan[E^(-I G w0)]]

I want to get out G w0

Does anybody know something that will let Mathematica simplify this expression?

Attempted solution:

FullSimplify[Log[E^(-I G w0)]]
  • 1
    $\begingroup$ If I Plot[{Re[ArcTan[E^(-I x)]], Im[ArcTan[E^(-I x)]]}, {x, -Pi, Pi}] I don't think I see that being just x. And I think your G Real times w0>0 means G w0 is equivalent to just x Real. So I think the reason it might be resisting your attempts to simplify to G w0 is that it just isn't equal. Have you mangled your question? $\endgroup$ – Bill Apr 25 '15 at 2:59

Simplifications work only if you avoid branch cuts of the inverse function. Here I do that for the ArcTan function, but using the relation that I think you meant to prove (I may have guessed wrong, of course):

 ArcTan[-I (Exp[2 I x] - 1)/(Exp[2 I x] + 1)], -Pi/2 < x < Pi/2]

(* ==> x *)

For the second part of the question, how to simplify Log[E^(-I G w0)], the simplest would be a replacement rule, but it may be of interest to try a different approach based on integral transforms (because they automatically make assumptions that we want here, too):

In particular, the Laplace transform (if it has an inverse) can be used to transform an expression f[x] back into itself, or into an equivalent, simplified version of itself:

InverseLaplaceTransform[LaplaceTransform[f[ x], x, t], t, x]

(* ==> f[x] *)

 LaplaceTransform[Log[E^(-2 I  x)], x, t], t, x]

(* ==> -2 I x *)

The first line shows that the composition with the inverse in general leaves the original expression unchanged. Then I apply that to Log[E^(-2 I x)] and get the desired simplification.

  • $\begingroup$ Thanks, do you happen to know what is conceptually different between the following two expressions:FullSimplify[Log[Exp[-I x]], -Pi < x < Pi] FullSimplify[ArcTan[Exp[-I x]], -Pi < x < Pi] $\endgroup$ – Mikhail Apr 26 '15 at 23:55
  • $\begingroup$ They both have branch cuts. The difference is that ArcTan has two branch points at $\pm i$, while Log only has one at 0. In both cases, the unit circle generated by E^(-2 I x) hits these cuts, so Mathematica doesn't want to perform the inverse. $\endgroup$ – Jens Apr 27 '15 at 0:40

I think you may have entered the wrong equation, but for yours one gets:

Simplify@TrigToExp[ArcTan[E^(-I G w0)]]

$\frac{1}{2} i \left(\log \left(1-i e^{-i\ G\ \text{w0}}\right)-\log \left(1+i e^{-i\ G\ \text{w0}}\right)\right)$


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