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Let $U \subset \mathbb{C}$ be an open set such that {$ z\in\mathbb{C};|z|\leq2$}$ \subset U$ and $f:U\rightarrow\mathbb{C}$ a holomorphic function.

Show that there exists infinitely many natural numbers $n \in \mathbb{N}$ such that $f(\frac{1}{n})\neq(\frac{1}{n+1})$

So i have been told to start by comparing with the function $g(z)=(\frac{z}{z+1})$

however i am still not getting very far with this. Would i have to prove by a contradiction?

Completely stuck and unsure, some help would be fantastic, many thanks!

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Well, yes, by contradiction. Suppose the contrary, then $f(\frac1n)=\frac1{n+1}$ holds with finitely many exception, in particular there is a sequence $z_n\to 0$ (in fact, a subsequence of $(\frac1n)_{n\ge 1}$), such that $$f(z_n)=\frac1{\frac1{z_n}+1}=\frac{z_n}{1+z_n}=g(z_n)\,.$$ But, as both $f$ and $g$ are holomorphic around $0$, this implies that $f=g$. However, $g$ has a pole at $z=-1$, so it is not defined and not holomorphic on all the disk of radius $2$.

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  • $\begingroup$ thank you for this! what does a pole at z=-1 mean? sorry im new to this topic! $\endgroup$ Mar 27, 2013 at 11:27
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    $\begingroup$ Basically that it's not defined there, and can't be made continuous there, as $\lim_{z\to 1} |g(z)|=\infty$. So, $g$ can't be defined on that disk. $\endgroup$
    – Berci
    Mar 27, 2013 at 11:39

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