Holomorphic function on the open unit disc Question: Let f be a holomorphic function on the unit disc $\{|z|<1\}$, which of the following is/are necessarily true?


*

*If for each positive integer n we have $f(1/n)=1/n^2$ then $f(z)=z^2$ on the unit disc.

*If for each positive integer n we have $f(1-1/n)=(1-1/n)^2$ then $f(z)=z^2$ on the unit disc.

*f cannot satisfy $f(1/n)=(-1)^n/n$ for each positive integer n.

*f cannot satisfy $f(1/n)=1/(n+1)$ for each positive integer n.
I have an intuitive feeling that (1) and (4) are true, but that's not helping! 
Thanks in advance!
 A: Statements 1 and 3 are true, but Statements 2 and 4 are false. 
Statement 1 follows from the identity theorem for holomorphic functions. For the conditions imply that $f(z)$ agrees with $z^2$ on the infinite set $\{1/n : n\in \Bbb N\}$, which has $0$ as an accumulation point in the disc. So $f(z) = z^2$ for all $z$ in the disc.
For Statement 2, consider the function $f(z) = z^2 e^{2\pi i/(1 - z)}$ is holomorphic in the open unit disc and satisifies $f(1 - 1/n) = (1 - 1/n)^2 e^{2\pi i n} = (1 - 1/n)^2$ for all $n\in \Bbb N$, but $f(z)\neq z^2$ for all $z$ in the disc. Therefore, Statement 2 is false.
To see why Statement 3 is true, note that if $f$ satisfies the hypotheses, then $f(z)$ agrees with $g(z) := z$ on the infinite set $\{\frac{1}{2n} : n\in \Bbb N\}$, which has $0$ as a point of accumulation lying in the disc.  So $f \equiv g$ by the identity theorem. However, $f(1) = -1 \neq 1 = g(1)$.
Since $f(z) = z/(z + 1)$ is holomorphic in the open unit disc and satisfies $f(1/n) = 1/(n + 1)$ for all $n\in \Bbb N$, Statement 4 is false.
