@John Fredsted I obtained the same conclusion yesterday evening by setting k=r*exp(theta*I).
It seemed a little bit simpler to analyze because r is always positive and sin(theta) is bounded

I will send you the .mw file asap. For the moment the uploading doesn't work correctly (I'm in the office and it could be related to the firewall ?)

@Kitonum Thanks.

It is a little bit clever than what I coded myself.

I just hoped that the equivalent of `intersect`(set1, set2) for countable sets existed for no countable ones

(Maple is sometimes so "elegant" ... it could have been the case).


Thanks again

@John Fredsted I am really sorry : I realized my mistake as I was pushing the "submit" button !!!

In addition to your previous reply  "Whether this is the complete solution for k remains to be investigated" I obtained these results (I hope there is no mistake this time) :

phi2  := (x,k) -> expand(phi(x,k)*conjugate(phi(x,k))) :
PHI2 := phi2(x, a+b*I) assuming x::real, a::real b::real :

limit(PHI2, x=+infinity) assuming b::positive ;    # returns 0
limit(PHI2, x=+infinity) assuming b::negative;    # returns infinity
limit(subs(b=0, PHI2), x=+infinity) ;                  # returns 1

It seems that the boundedness only occur for pure imaginary k ?

(sorry again)

@John Fredsted  Please enlighten me

In your last reply you study phi(x,k) for k complex of the form k=a+b*I (a and b real).
You draw phi(x,(a,b)) for the particular case a=0, b=1 and find thie last bell-shaped curve.

But if a=0 and b=1 then k = I and k^2 = -1

and then phi(x,k) is undefined ...

Did I miss something ?