I ran into the problem that Mathematica's (full-) simplification does not replace
t Conjugate[t]
by
Abs[t]^2
In my case, I get pretty ugly expressions which greatly simplify when I replace the expressions manually. As an example:
FullSimplify[( 2 (-((Abs[a - b + Sqrt[(a - b)^2 + 4 t Conjugate[t]]]^2 +
4 t Conjugate[t])/(
2 (a + b - 2 ω + Sqrt[(a - b)^2 + 4 t Conjugate[t]]))) + (
Abs[-a + b + Sqrt[(a - b)^2 + 4 t Conjugate[t]]]^2 +
4 t Conjugate[t])/(
2 (-a - b + 2 ω +
Sqrt[(a - b)^2 + 4 t Conjugate[t]]))))/((a - b) (Conjugate[a] -
Conjugate[b]) + 4 t Conjugate[t] + Sqrt[(a - b)^2 + 4 t Conjugate[t]]
Conjugate[Sqrt[(a - b)^2 + 4 t Conjugate[t]]]), {{a,
b, ω} ∈ Reals, t ∈ Complexes}]
Does not yield any simplification. However, if I replace
t Conjugate[t] -> Abs[t]^2
or, if I change $t$ to be real, I find the desired result $\frac{b-\omega}{ |t|^2-(a-\omega)(b-\omega)}$
PS: I just realized this is appears to be the same problem as posted here: FullSimplify on complex numbers seems inconsistent. However I'd much appreciate a less "crazy" solution…
ComplexExpand[t Conjugate[t]]
. Lookup upComplexExpand
in the help. $\endgroup$Arg[(a - b)^2 + 4 t Conjugate[t]]
which are zero as the argument is positive. $\endgroup$Simplify
that both parts are real, then I find the final simplified result. But… that's not a great solution? $\endgroup$TransformationFunctions -> {Automatic, # /. x_ Conjugate[x_] :> Abs[x]^2 &}
does give the result you want, for what it's worth. $\endgroup$