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Does there exist an analytic function on $\mathbb{C}$ such that $f(1/2n)=f(1/2n+1)=1/2n$?

Well, I considered a new function $g(z)=f(z)-z$. The zeroes of $g$ has limit point $0$ in $\mathbb{C}$ so $g(z)\equiv 0\Rightarrow f(z)=z$ but I checked that both the condition is not satisfied by $f$ i.e $f(1/2n+1)=1/2n\Rightarrow 1=0 \Leftrightarrow$ so such a non constant analytic function does not exists.

Am I right? Please help!

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As pointed out in the comments, you are correct, but you could make the argument a little bit clearer towards the end. In particular, once you’ve shown that $f(z) = z$, just note that $$f\left(\frac{1}{2n+1}\right) = \frac{1}{2n+1} \neq \frac{1}{2n}$$ so there is no analytic function $f$ satisfying the desired condition.

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