complex analysis problems related to nalyticity...... I don't even know how to start for the following problem. It would be highlt appreciated if you could help me.
Let $z_n=\frac{1}{n},\forall n\in\Bbb N$. If possible find $f\in H(\bar{B}_1(0))$ [H means analyticity] such that $\{f(z_n)\}_{n=1}^\infty$ are given by each of the follwoing cases 
(a) {0,1,0,1,.....}
(b) {$1,\frac{1}{4},\frac{1}{9},\frac{1}{16}...$}
(c) {$0,\frac{1}{2},\frac{1}{3},\frac{1}{4}...$}
(d) {$\frac{1}{2},\frac{1}{2},\frac{1}{3},\frac{1}{3},\frac{1}{4},\frac{1}{4}......$}
If not possible, explain why such an $f$ may not exist.
 A: Is $\overline{B}_1(0)$ the closed unit disk?
In that case the first one is impossible because it has no limit as $z\to 0$, thus isn't continuous at $0$. The second one is solved by $f(z)=z^2$.
The third one is also impossible, because every point except the first would force the function to be $f(z)=z$ (remember, analytical functions are uniquely determined by their value on a converging sequence of points). But that doesn't match the first point.
The last one I is also impossible, by the same reasoning as the third one, only instead of ignoring the first term, you ignore every term of odd index.
A: The first two options can be handled without considering the analyticity of $f(z)$.
For the last one, consider $g \in H(B_1(0))$ defined as
$$g(z) = f(z/(1+z))-f(z)$$
so that $g(1/2n)=0$ (show).
This feels "wrong" for an analytic function. My complex analysis is rusty, but according to wikipedia if the zeros of an analytic function have an accumulation point in a connected domain, then the function is identically zero.
So $f(z/(1+z))=f(z)$ (show).
Use the substitution $z \to z/(z+1)$ repeatedly to conclude that $f(z)$ is constant (show). Hint: $f(z) = f(\frac{z}{nz+1})$ for all $n>0$.
