Interesting value for W, h involving phi,Omega?

[Ivars]

This is obvious- imaginary selfroot of Omega, but might prove useful, linking I and Omega and e:

Omega^(I/Omega)=e^-I=1/(e^I)=0.540302306-I*0.841470985=cos1-I*sin1

(Omega^(I/Omega))^pi/2=(e^(-I*pi/2))=-i
(Omega^(I/Omega))^pi=(e^(-I*pi))=-1
(Omega^(I/Omega))^2*pi=e^(I*pi) = 1

Basically, if we have Omega, there is no need for e as it can be always defined and found as 1/self root of Omega. Opposite is not so easy.

That means base 1/e is Omega^(1/Omega), base 1/e^I is Omega^(I/Omega), similarly formed as base i^1/i = e^(pi/2) (and all inverses).

This begs a question is Omega itself an interesting base, or 1/Omega,or I*Omega, or I/Omega.

For exapmple, log omega (I) = -(I/Omega)*pi/2

To get rid of e more, i will use formula i derived earlier in this thread, putting x=1/Omega.

W(-(1/Omega)*(pi/2) - I*(1/Omega)*ln((1/Omega)) ) = ln(1/Omega) -I*(pi/2) for x>1;

(W(-pi/(2*Omega)-I) = Omega-(I*pi/2)

Then:

h(((e^(pi/2))^(1/Omega))*(e^I)) = (-Omega+(I*(pi/2))/(pi/(2*Omega)+I) and, since

e= Omega^(-1/Omega) , e^(I) =( Omega^(-I/Omega))

h(Omega^(pi/2)*Omega^ (-I/Omega)) = h((Omega^((Omega*pi/2-I)/Omega))= (-Omega+(I*(pi/2))/(pi/(2*Omega)+I)

The same without pi/2 in h:

h(Omega^(pi/2)*Omega^ (-I/Omega)) = h((Omega^((1-I)/Omega))*(I^(1/I))=(-Omega+(I*(pi/2))/(pi/(2*Omega)+I)

This value is:

h(((e^(pi/2))^(1/Omega))*(e^I)) =(-Omega+(I*(pi/2))/(pi/(2*Omega)+I)= Omega*I=I*0,56714329..

The other branch of W should give -I*Omega?

This was changed from previous mistaken result, due to wrong signs.