# If $f$ is analytic in $|z|\leq 1$ then $\exists n, f({1\over n})\ne{1\over 1+n}$

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.

• You can multiply by $z+1$ and then use the accumulation theorem to conclude that $(z+1)f(z)=z$. – Severin Schraven Apr 27 at 14:13

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.