Maybe the solution at z=0 to f(f(z))=-z+z^2
#2
maybe a silly comment but

why not consider h(z) = - z + z^2 , g(z) = h(h(z))

and then consider the super and abel of g(z) ?

regards

tommy1729
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RE: Maybe the solution at z=0 to f(f(z))=-z+z^2 - by tommy1729 - 01/19/2023, 11:20 PM

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