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If you consider $f=\frac{P}{Q}$ the quotient of two polynomial function then $\frac{f'}{1+\vert f\vert^2}$ decrease like $\frac{1}{z}$. My question is, is the converse true? is an meromorphic function(define on the whole plane) which satisfies
$$\frac{f'}{1+\vert f\vert^2}=O\left(\frac{1}{z}\right)$$ as $z$ goes to infinity, is the quotient of two polynomial function?

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    Welcome to math.stackexchange! My first instinct would be no - it certainly doesn't "feel" like it should be true. Being a "polynomial" is a different kind of property than having a certain type of growth condition - You can make all kinds of weird functions which grow similarly to a given polynomial just be changing it in some little way, such as adding $ + \sin x $ to the end. So perhaps we should look for a counterexample. But before we proceed any further, you should specify: Holomorphic *where* ? And the asymptotics are as z goes where?2011-10-18
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    @Ragib, being polynomial is not quite as special a condition in the complex case as it is for reals. In particular, if a function is meromorphic on all of $\mathbb C$ _and_ at infinity, then it has to be a quotient of polynomials; and having a pole (or a removable singularity) at infinity certainly feels like a kind of growth condition to me.2011-10-18
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    If you cross-post a question to MathOverflow, it would be good if you edited your question to include a link to there as well, and vice versa. I notice that someone gave a reference to the Lehto-Virtanen paper there the day before I posted my answer below. I do enjoy thinking about questions, but it is still somewhat annoying to spend time on one and then realize that the OP is no longer interested because they have already received a reference elsewhere ...2011-10-26
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    sorry for this, but i start to ask on mathstackexchange. Since there is no answer, i ask to mathoverflow...i will close the mathoverflow post.2011-11-01
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    Hi Paul, that's quite a common and sensible thing to do, it would just be good to edit your answer to include a link to the mathoverflow question, and vice versa. That way people can just check on the other thread before posting an answer that might be duplicated elsewhere.2011-11-01

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