Existence of an analytic function under some given conditions Which of the followings is(/are) correct?



There exists an entire function $f:\mathbb C \to \mathbb C$ which takes only real values &amp; is such that $f(0)=0$ &amp; $f(1)=1$.
There exists an entire function $f:\mathbb C \to \mathbb C$ such that $f\bigl(n+\dfrac{1}{n}\bigr)=0$ , $\forall n\in \mathbb N$.
There exists an entire function $f:\mathbb C \to \mathbb C$ which is onto &amp; which is such that, $f(\frac{1}{n})=0$ , $\forall n\in \mathbb N.$


I tried through the Uniqueness Theorem for options (2) &amp; (3) &amp; apply consequence of Schwarz-Pick lemma &amp; I get (1) &amp; (3) are correct.

Am I right? I think I am wrong, since it contradicts the answer of the book?

Where my fallacy ?
1. If $f$ would be such a function, then $g(z)=e^{if(z)}$ would be entire, bounded, and non-constant.

2. $f(z) =0$...

3. The functions $f$ and $0$ are entire and equal on a Set which has an accumulation point....
False as range(f) cannot have dimension one { Consequence of Open mapping theorem} unless f is a constant.
True as f(n+1/n)=0 for all n implies f assumes value 0 in the nbd. of each n€N. By Identity theorem f must be identically zero.
False as f(1/n)=0 for all n implies f assumes value 0 in the nbd. of zero. Use Identity theorem to get f is identically zero hence it cannot be onto.