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Show that if f is analytic in $|z|\leq 1$, there must be some positive integer n such that $f(\frac{1}{n})\neq \frac{1}{n+1}$


If $f(\frac{1}{n})=\frac{1}{n+1}$, then for all points $z_{n}=\frac{1}{n}$ , $f(z_{n})=\frac{1}{\frac{1}{z_{n}}+1}$ or $f(z_{n})=\frac{z_{n}}{1 + z_{n}}$ Because ${z_{n}}$ has an accumulation point at 0, this implies that $f(z)=\frac{z}{1+z}$ throughout its domain of analyticity which yields a contradiction since f was assumed analytic at $z=-1$.

Can anyone help me improve it?

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Your solution looks fine to me. – Brett Frankel May 1 '12 at 16:29
thanks,But it would be understandable to deliver the task, do not you need to justify anything? – Breton May 1 '12 at 17:43
I'm not sure I understand your comment. But your solution as written is fine. If you want to be nitpicky, you should point out that $\frac{1}{\frac{1}{z_n}+1}$ has a singularity at $0$, but it's removable so there aren't any serious issues here. – Brett Frankel May 1 '12 at 17:47
@BrettFrankel Or not to mention $\frac 1{\frac1{z_n}+1}$ at all and only verify that $f(z)$ coincides with $\frac z{1+z}$ for all $z=z_n$ – Hagen von Eitzen Apr 24 at 14:01

The answer is no; your solution is as good as it could be. (Posting this CW to remove the question from "unanswered")

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