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Let $f$ be an analytic function in $|z|\leq 1$. Prove that there must be a positive number $n$ such that $f({1\over n})\ne {1\over 1+n}$.

Attampt:

$f$ is analytic in $|z|\leq 1$. Thus, by defenition there's some open set $\{|z|\leq1\}\subseteq U$ that $f$ is analytic there. Let $g:U\to\mathbb{C}, g(z)={z\over z+1}$. For all $z_n:={1\over n}, f(z_n)=g(z_n)$. The point is that because $(z_n)$ has an accumalation point $0$ then $f=g$ in $U$, but $g$ is not analytic in $-1\in U$ so this is a contradiction.

My problem is that if I want to use the accumalation theorem it is needed that $f,g$ will be analytic in $U$ in a first place, which doesn't happan here.

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    $\begingroup$ You can multiply by $z+1$ and then use the accumulation theorem to conclude that $(z+1)f(z)=z$. $\endgroup$ – Severin Schraven Apr 27 at 14:13
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What you did is fine. It proves that there is no analytic function from $D(0,1)$ into $\mathbb C$ such that$$(\forall n\in\mathbb N):f\left(\frac1n\right)=\frac1{n+1}.\tag1$$But then of course that there is no analytic function from $\overline{D(0,1)}$ into $\mathbb C$ such that $(1)$ holds.

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