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Does there exist a function $f$ which is holomorphic on $B_0(2)$ (open disc of radius 2 in the complex plane) such that $f(1/n)=1/(n+1) \forall n \in \mathbb{N}$? At the moment I'm thinking not but a proof is seeming elusive. Any hints would be appreciated.

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up vote 9 down vote accepted

Let $a_n:=\frac 1n$. Such a function should satisfy $f(a_n)=\frac{a_n}{1+a_n}$. Let $g(z):=f(z)-\frac z{z+1}$ on $B(0,1)$, the open ball. What can you say about the zeros of $g$?

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Perfect, the first part of the question was to show that a holomorphic function has isolated zeros, so this must be the right approach. Obvious now I look at it. – user61496 Feb 22 '13 at 11:17
@DavideGiraudo I am dealing with an almost identical case , and I am at the point where we can see that as $\alpha_n\to0, g(\alpha_n)=f(\alpha_n) -\frac{\alpha_n}{\alpha_n+1}=0$ , hence $f(z)=\frac{z}{z+1}$ by the identity theorem . Shouldn't there be a contradiction that would prove such a fuction cannot exist? – helplessKirk Aug 24 '14 at 22:52
@helplessKirk Here the key point is that we consider the ball centered at the origin and of radius 2. If we had for example considered the radius $1/2$, there wouldn't be a contradiction. – Davide Giraudo Aug 25 '14 at 15:17
@DavideGiraudo So it has to do with where the zeros lie (and maybe if a maximum is achieved inside)? In my case we consider $D(0,1) $ and we need to prove that this function cannot exist except (maybe) for some finite cases. Thanks for answering! – helplessKirk Aug 25 '14 at 15:42

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