Maybe the solution at z=0 to f(f(z))=-z+z^2

[Leo.W]

Im not sure how these relate but 3 ideas cross my mind.
...
regards

tommy1729

What's more strange is, we assume (A) \(f(f(z))=-z+z^2\)
and (B) \(\log(az)=\log(a)+\log(z)\) for all \(a\) near \(z=0\) where log here represents the main value
and (D) a series for \(f\): \(f(z)=a_1z+a_2z^2+(a_{3,0}+a_{3,1}\log(z))z^3+(a_{4,0}+a_{4,1}\log(z)+a_{4,2}\log(z)^2)z^4+...\)
This gives a series dependent on \(a_{3,0}\)
Denote \(a_{3,0} = c, \log(z)=L\)
Choose the branch (E1): \(a_1=i\)
We have [A,B,D,E1]: 
\[f(z)=iz-\frac{1+i}{2}z^2+(c+\frac{2}{\pi}L)z^3+((-\frac{3}{2}+i)c-\frac{4-(5-4i)\pi}{4\pi}+\frac{2i-3}{\pi}L)z^4+...\]
\(c\) appears infinitely many times, for example in coefficients of \(z^5, z^5\log(z), z^6, z^6\log(z)\)
And choosing the branch (E2): \(a_1=-i\) the same thing happens. The posted version was with the condition (F) \(a_{3,0}=0.5\).
This is what I meant about how iteration by \(f(-z+z^2)=-f(z)+f(z)^2\) moves our branch to another, the function \(-z+z^2\) will affect the \(a_{3,0}\) term and so on.

Moreover, it's asymp, and will diverge.
Under [A,B,D,E2,F]:
\(f(z)=-iz-(0.5-0.5i)z^2+z^3(0.5+0.637L)+z^4(0.182+0.5i-(0.955+0.637i)L)+z^5(-0.352-0.307i\)
\(+(0.477+1.476i)L+0.608iL^2)+z^6(-2.339+1.779i-(0.832+0.478i)L+(0.811-1.512i)L^2)\)
\(+z^7((8.544-0.751i)L-(2.883-0.963i)L^2-0.645iL^3)+z^8(24.546+7.179i-(29.944+16.379i)L\)
\(+(1.991+0.251i)L^2+(2.2258+1.032i)L^3)+O(z^9L^4)\)
and \(a_{9,0}\approx -55.048-45.058i\)
The norm (abs) of the coefficients grows as \(\|a_{k,0}\|>\sim O(k^2)\)
My computer would explode to compute z^10 terms
SOOOOOOOOOOOOO MANY ISSUES! Literally just keel mi(lol)