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 The super of exp(z)(z^2 + 1) + z. tommy1729 Ultimate Fellow Posts: 1,614 Threads: 364 Joined: Feb 2009 03/14/2016, 01:26 PM 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 tommy1729 Ultimate Fellow Posts: 1,614 Threads: 364 Joined: Feb 2009 03/15/2016, 01:02 PM (This post was last modified: 03/15/2016, 01:06 PM by tommy1729.) 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 « Next Oldest | Next Newest »

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