Does there exist an analytic function $f$ on $D(0,1)$ such that $f(z_n)=0$ for even $n$ and $f(z_n)=1$ for odd $n$?

Question

Given that $(z_n)$ is a sequence of distinct points in $D(0,1)=\{z \in \Bbb C : |z| \lt 1\}$ with $\lim_{n \to \infty} z_n=0$, Can we find an analytic function $f$ such that

$f(z_n)= \begin{cases} 0, & \text{if n is even} \\ 1, & \text{if n is odd} \end{cases}$ ?

I know the Uniqueness theorem and Identity theorem. How can I use them?

Also as $n \to \infty$, we have $f(z_n)$ to be divergent. Is this information of any help?

Answer 1

As already said in the comments, every analytic function is in particular continuous (since differentiability in a point implies continuity in that point).

If an analytic function $f$ would exist with the given properties, then both $$ \lim_{n \to \infty} z_{2n} = 0 \quad \Longrightarrow \quad f(0) = \lim_{n \to \infty} f(z_{2n}) = \lim_{n \to \infty} 0 = 0 $$ and $$ \lim_{n \to \infty} z_{2n+1} = 0 \quad \Longrightarrow \quad f(0) = \lim_{n \to \infty} f(z_{2n+1}) = \lim_{n \to \infty} 1 = 1 $$ which is apparently not possible.

Answer 2

The zeros of a nonzero analytic function are isolated and so there exists no function satisfying your hypotheses.