Existence of a holomorphic function on a open unit disc

Question

Let $f$ be a holomorphic function on the disc $|z|<1$.Which are true?

• If $f(\dfrac{1}{n})=\dfrac{1}{n^2}\forall n$ then $f(z)=z^2$ on the unit disc.
• If $f(1-\dfrac{1}{n})=(1-\dfrac{1}{n})^2$ then $f(z)=z^2$ on the unit disc.
• $f$ can't satisfy $f(\dfrac{1}{n})=\dfrac{(-1)^n}{n}\forall n$.
• $f$ can't satisfy $f(\dfrac{1}{n})=\dfrac{1}{n+1}\forall n$

1.Define $g(z)=f(z)-z^2$ .Now $g(\dfrac{1}{n})=0$ and $g$ is holomorphic and set of zeros of $g$ has a limit point $0$ so $g\equiv 0\implies f(z)=z^2$.

2. .Define $g(z)=f(z)-z^2$ .Now $g(1-\dfrac{1}{n})=0$ and $g$ is holomorphic and set of zeros of $g$ has a limit point $0$ so $g\equiv 0\implies f(z)=z^2$.

3. I am unable to do this problem

4..Define $g(z)=f(z)-\dfrac{z}{z+1}$ .Then $g(\dfrac{1}{n})=0$ $g$ is holomorphic and set of zeros of $g$ has a limit point $0$ so $g\equiv 0\implies f(z)=\dfrac{z}{z+1}$.

The answer is given to be 1,3

Please help whether the arguements are correct/not .Please help me in case of 3.

Answer 1

Your answer to 2 is wrong. The function $z^2+\sin (\pi/(1-z))$ is holomorphic in the open unit disc and satisfies the requirement. Hint for 3: Consider separately $n$ even, $n$ odd.