What I feel there can't be any such function..but what if we assume a piecewise function? Will that be ok if we assume it like that..?
What's the proper approach to this question. And furthermore, I don't know what is meant by $A(z=0)$ I just assumed it may be analytic at $z=0$ The boundary condition shows discontinuity..so it may not be analytic. So I thought there can't be such function
If I understand the question correctly, I think that the point is that since $f(z)=z$ for $z={1\over2n}$, the identity is the only possible analytic function satisfying the condition in a neighborhood of $0$. (Two analytic functions that agree on a set with a limit point are identical.) But $f$ is clearly not the identity, so there is no such functions.