Does there exist a holomorphic function $f : B(0; 1) \rightarrow \mathbb C$ such that $f(1/n) = z_n$ where
(a) $z_n = (-1)^n$
(b) $z_n = \frac{n+1}{n}$
(c) $z_n = 0$ if $n$ is even and $z_n = \frac1{n}$ if $n$ is odd.
For (a), I showed that function is not continuous at $0$ as $\lim_{n \rightarrow \infty}(-1)^n$ does not exist, so function not holomorphic.
For (b), $f = g $ for infinitely many points of the sequence $1/n$, where $g(z)=1+z$ is holomorphic on $B$. So, by identity theorem, $f=g$ $\forall z$.
For (c), I am confused as if I apply Identity theorem for the sequence of even integers I get $f=z$, but that contradicts at, say $z=1/3$. Does that say there is no such function?
Did I give correct proofs of these questions? Thanks in advance.
Reasoning for part a) as well as part b) is correct as discussed in comments. For part c) use identity theorem for even $n$ and odd $n$ to get a contradiction(Like you observed, one way you will get it is equal to identity function and another way you will get that it is identically zero function.)