existence of analytic function related to identity theorm Does there exist an analytic function $f$ in closed unit disk such that $f(z)\neq 0$
identically and $f\left(\frac{ni^n}{1+n}\right)=0$ for $n\in\Bbb N$?
 A: I'm referring to the corrected version given by Martín-Blas Pérez Pinilla, 
$$f\left(\frac{n}{n+1} i^n\right)=0,\quad n=1,2,\dots \tag{1}$$
Being analytic in closed unit disk is an important assumption here: it means that $f$ extends to an analytic function on an open set $\Omega$ containing the closed unit disk. If such $f$ is not zero identically, then (by the identity theorem) every compact subset of $\Omega$ contains only finitely many zeros of $f$. This rules out (1), and the form of the sequence does not really matter. 

In fact, the same conclusion can be reached merely by knowing that $f$ is continuous in the closed unit disk, or even that it has integrable boundary values (in the sense of nontangential limits). However, this is a deeper result, due to Szegő: the zeros of an $H^1$ function satisfy the Blaschke condition $\sum (1-|a_n|)<\infty$.  The sequence in (1) fails the Blaschke condition: 
$$\sum (1-|a_n|) = \sum\frac{1}{n+1} =\infty$$

If we only asked $f$ to be analytic in the open unit disk, then it is possible for $f$ to be not identically zero and satisfy (1). This is a generalized form of the Weierstrass theorem on the existence of holomorphic function with specified zeroes.
