11/15/2007, 09:40 AM

Hej Gottfried,

I am not trying to show You have done something wrong , I think Your graph and approach is great.

I only want to make sure that reducing complex numbers to one root of sqrt(-1) = i so that a+ib work also in tetration and h(z) without loss of information. There fore I am trying to ask You since I can not do it- I understand You may feel annoyed - > if You feel any interest , of course, to carry the ambiquity of sign of sqrt(-1) through all derivation before getting rid of it.

But exp(alpha+I*beta) is not equal to exp (alpha-I*beta)?

so if we consistently keep u=alpha+-i beta as definition of u as complex number, than t has 2 values = e^alpha*e^(+- I*beta).

If You continue from that and find that this other value t=e^alpha*e^-I*beta does not add any new information about the whole thing, than let it be.

Ivars

I am not trying to show You have done something wrong , I think Your graph and approach is great.

I only want to make sure that reducing complex numbers to one root of sqrt(-1) = i so that a+ib work also in tetration and h(z) without loss of information. There fore I am trying to ask You since I can not do it- I understand You may feel annoyed - > if You feel any interest , of course, to carry the ambiquity of sign of sqrt(-1) through all derivation before getting rid of it.

Gottfried Wrote:t = exp(u) (whatever u) and this is unique, since exp is multiple->unique function (no two values/branches for function-result)

Gottfried

But exp(alpha+I*beta) is not equal to exp (alpha-I*beta)?

so if we consistently keep u=alpha+-i beta as definition of u as complex number, than t has 2 values = e^alpha*e^(+- I*beta).

If You continue from that and find that this other value t=e^alpha*e^-I*beta does not add any new information about the whole thing, than let it be.

Ivars