# Existence of an analytic function under some given conditions

Which of the followings is(/are) correct?

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

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

• For (1), Where $f(0)=1$ & $f(1)=1$ are necessary? – Empty Jan 17 '15 at 16:56
• @Panja.S. To show that $g$ is non-constant. – N. S. Jan 17 '15 at 16:57
• (3) is not clear to me. Please clear it. – Empty Jan 17 '15 at 17:00
• @Panja.S. If two analytic functions in a simply connected domain are equal on a set which has an accumulation point, then the functions..... – N. S. Jan 17 '15 at 17:06
• So, $f(z)=0$ identically. And so (3) is true. Am I right? – Empty Jan 17 '15 at 17:12

For 1, do you know the open mapping theorem?

For 2, do you know the Mittag-Leffler theorem (or maybe Weierstrass factorization)?

For 3, do you know the identity theorem?

1. False as range(f) cannot have dimension one { Consequence of Open mapping theorem} unless f is a constant.
2. 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.
3. 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.
• How $\sin (\pi z)$ is an example for 2 ? – Empty Aug 22 '15 at 1:53
• S.Panja..thanks for your concern.I have edited it. – Nitin Uniyal Aug 23 '15 at 6:58