2.86295 + 3.22327 i

[tommy1729]

Hmmm, just to get more familiar to it....
\( z=z^z \)                           

\( 0=z^z-z \)                           

\( 0=z \cdot (z^{z-1} - 1) \)                           

\( 0= f^{\circ 2}(1) \)                           
where
.                      \( f(x)=z^x-1 \)                           


To look for zeros it might be helpful to list zeros of the real- and the imaginary parts separately first.

 
Interesting.

I prefer to look at it differently or at least used too.

Im not aware of any kind of closed form like lambert W but I have not really tried.

Here is what I did to get 2.86295 + 3.22327 i



Let z^z = z.

Take ln on both sides.

z ln(z) = ln(z) + 2 k pi I 

Now if we let k be 1 we get

2.86295 + 3.22327 i = z

And if k = -1 we get

2.86295 - 3.22327 i

And they appear the smallest ones.
That is probably easy to prove though I did not.

I have not considered other ln branches ( on the left side ) , my first guess is they are a invalid solutions ?

——

As for your solution you auto find z = 0 as a solution which is funny since it is invalid ; 0^0 = 1 !
Well at least that is standard. Limit cases and other opinions may differ.
As an explanation we could say you divided by z = 0. If that is satisfactory is another matter.

So as you say we end up with an expression that is a double iteration in disguise. 

Consider z^v - 1 = 0

So z^v = 1

This implies v = the period of the function z^t. 

Also v = z - 1.

——

We could use Newton iterations to find 2.86295 + 3.22327 i.

Or we have a Taylor series similar to lambert W.

I assume the real part , Imaginary part , norm and argument are all transcendental.
A proof would be nice.

Recall that e^(1/e) and other typical tetration number are also not proven transcendental or even irrational !

I assume there is a way to transform your equations to mine and vice versa without first going back to z^z = z. 
Not sure.

I welcome other attempts to solve that equation.

Notice that the real part of ln ln 2.86295 + 3.22327 i is larger than 0 hence it is not on the shell-tron boundary !! 

That is remarkable considering its cyclic behavior.

I was thinking about “ its cousin base “

z^z = z

z^^( “ oo “)  = z

z^(1/z) = Q 
(z^z)^(z^-z) = S

Now

Q^z = z^(z/z) = z = z^z

So Q = z.
It easy to show S = Q = z.

Nothing special 

However we also have

T^(1/T) = z

Where T =\= z.

And T within the Shel-tron.

I have to think more about that. 

Regards

Tommy1729