Does there exist $f:B(0,1) \to \Bbb{C}$, holomorphic such that $f(\frac{1}{n})=\frac{1}{3-2n}$ for all $n >2$? 
Does there exist $f:B(0,1) \to \Bbb{C}$, holomorphic such that $f(\frac{1}{n})=\frac{1}{3-2n}$ for all $n >2$?

I just did a very similar exercise in which I used the identity priciple. (If there exist $z_n \to z_0$ with $z_n,z_0 \in B(0,1)$ and $f(z_n)= 0 $ for all $n$ then $f \equiv 0$). But Im not being able to apply a similar argument to this case. Any hints?

$$f\left(\frac 1n\right)=\frac{1}{3-2n} = \frac{1}{3- 2\frac{1}{\frac 1n}}  = \frac{ \frac 1n}{3\frac 1n -2}$$
Together with $f(0) = 0$, we have that $f(z) = \frac{z}{3z-2}$ if $f$ is holomorphic by the identity theorem. However, $\frac{z}{3z - 2}$ is not holomorphic in $B(0,1)$ as it has a pole at $2/3$, so such a holomorphic $f$ does not exist. 
