Hello,

I tried to find h(z) = h((I/2)^(2/I)) = h( exp(pi)/exp(-I*2*ln2)

ln ( exp*pi/exp(-I*2*ln2)) = pi+ I*2*ln2

-ln(z) = - pi- I*2*ln2

so as W(-pi-I*2*ln2) = ln2 - I*(pi/2)

Than h(z) = (ln2-I*(pi/2))/ (-pi - i*2*ln2)

This can be braught to :

h(z) = 0.5*I ( i guess with plus sign, not sure)

now, since I/2 = sin (I*ln(phi)) than

h((i/2)^(2/i)) = sin(I*ln(phi))

so Arcsin(h(I/2^(2/I)) = I*ln(phi)

phi= -I*e^Arcsin(h(I/2)^(2/I))

so taking a self root of I/2 can be related to phi in this way. Without an additional partitioning, 2 in any side of the formula involving h(i^(1/i)), it is not possible to get from I to phi.

I think it is possible to continue with I/3, I/4 etc in the same way to find interesting things.

is this well known, if there is no mistake?

I tried to find h(z) = h((I/2)^(2/I)) = h( exp(pi)/exp(-I*2*ln2)

ln ( exp*pi/exp(-I*2*ln2)) = pi+ I*2*ln2

-ln(z) = - pi- I*2*ln2

so as W(-pi-I*2*ln2) = ln2 - I*(pi/2)

Than h(z) = (ln2-I*(pi/2))/ (-pi - i*2*ln2)

This can be braught to :

h(z) = 0.5*I ( i guess with plus sign, not sure)

now, since I/2 = sin (I*ln(phi)) than

h((i/2)^(2/i)) = sin(I*ln(phi))

so Arcsin(h(I/2^(2/I)) = I*ln(phi)

phi= -I*e^Arcsin(h(I/2)^(2/I))

so taking a self root of I/2 can be related to phi in this way. Without an additional partitioning, 2 in any side of the formula involving h(i^(1/i)), it is not possible to get from I to phi.

I think it is possible to continue with I/3, I/4 etc in the same way to find interesting things.

is this well known, if there is no mistake?