Does there exist an analytic function $f$ on $U_2(0)$ such that $f(\frac{1}{n}) = \frac{n}{1-2n}$?

Question

I found, by a little bit of trial and error and algebra, that the function \begin{equation} f(z) = \frac{1}{z - 2} \end{equation} has this property. However, how do we prove that this is the only function, or find other functions with this property?

Answer 1

Since$$(\forall n\in\mathbb N):f\left(\frac1n\right)=\frac n{1-2n}=\frac1{\frac1n-2},$$the function $f(z)=\frac1{z-2}$ will do. And it is the only one, by the identity theorem. In fact, if $g\colon U_2(0)\longrightarrow\mathbb C$ is such that$$(\forall n\in\mathbb N):g\left(\frac1n\right)=\frac n{1-2n},$$then$$\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}\subset\{z\in U_2(0)\mid f(z)=g(z)\}$$and therefore the set $\{z\in U_2(0)\mid f(z)=g(z)\}$ has an accumulation point, which is $0$. So, $f=g$.

Answer 2

I would substitute $$x=\frac{1}{n}$$ so $$f(x)=\frac{\frac{1}{x}}{1-\frac{2}{x}}=\frac{1}{x-2}$$