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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e))
#29
Ivars Wrote:Logarithm itself has 2 branches such that log (-1) = +i *pi or - i*pi.

The logarithm has a branch for each integer , that is for you get . This is because the logarithm is the inverse of the exponential and .

So its not really about chosing the sign of but about chosing the .

Quote:h(e^pi/2) = +-i +- 2pik.

Again, this is not true. If we have a fixed point of then (or ) is usually not again a fixed point (and hence not a branch of ).
For example then but also .

PS: the name is sqrt and not sgrt Wink
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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 bo198214 - 11/04/2007, 10:53 PM
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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