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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#58
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.

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
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Messages In This Thread
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/15/2007, 09:40 AM
RE: Tetration below 1 - by Gottfried - 09/09/2007, 07:04 AM
RE: The Complex Lambert-W - by Gottfried - 09/09/2007, 04:54 PM
RE: The Complex Lambert-W - by andydude - 09/10/2007, 06:58 AM

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