I am faced with the following problem:
Does there exist a function that is analytic on a neighborhood of $z=0$ and satisfies the following condition for every positive $n$:
(a) $f(1/n)=f(-1/n)=1/n^{2}$ (I was given the hint that the answer is yes)
(b) $f(1/n)=f(-1/n)=1/n^{3}$ (I was given the hint that the answer is no)
However, I have absolutely no idea how to even get started. I know I'm supposed to use the interior uniqueness theorem, but exactly how to use it is baffling me a bit. Perhaps if someone could tell me how to do part (a), I could use it to help me figure out part (b).
Thank you.
For the question $a$, it is clear that $$f(z) = z^2$$ is a solution to the problem. A more difficult problem is to show that this solution is unique but it is not asked. However I will give you the hint to prove that and it will be helpful for the second exercice.
Suppose, there is another function $g$ holomorphic on a neighborhood $V$ of $0$ such that $$g(1/n)=f(1/n)=1/n^2$$ then taking the limit $n \to +\infty$, we find by continuity that $g(0)=f(0).$ Hence $f=g$ on the set $$\left\{\frac{1}{n} : n \in \mathbb{N}\right\} \cup \{0\}.$$ In this set $0$ is a limit point so by the uniqueness theorem you find that $f(z)=g(z)$ for all $z\in V$.