The super of exp(z)(z^2 + 1) + z.
#1
Im very intrested in The super of exp(z)(z^2 + 1) + z.
Notice it has only 2 fixpoints.

Also The super of exp(z)(z^2 + 1) + z = L.
Does it have solutions z for every L ? Why ?
They determine the branch structure / singularities.

Regards

Tommy1729
Reply
#2
So we need to show f = exp(z) (1+x^2) + z is surjective to the complex plane.

F maps [- oo , oo ] to [- oo , oo ].

Also conj(f(z)) = f(conj(z)).

Hence f is surjective on the reals.

By picard if f(z) =\= L then this L is unique.

But by the above f(conj(z)) =\= conj L.
Hence contradicting picard.
Therefore f is surjective to the complex plane.

So the singularities of its super are bounded.

Q.E.D.

Regards

Tommy1729
Reply


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