This question was asked in my complex analysis quiz and I was unable to solve it.
If $\Omega= \{ z: Re(z) >0 \}$ and f is a bounded holomorphic function on $\Omega$ with $f(n) = 0$ for all $n \in \mathbb{N} $ then show that $f(z) =0$ for all z $\in \Omega$ .
I am sorry but I will not be able to provide anything as attempt as I am absolutely clueless on how to approach this problem in both exam time and while trying again . Kindly believe me and tell me which result should I use.
Thank you!!
Consider the function
$$\varphi(z)=\frac{z-1}{1+z}; \varphi:\Omega\leftrightarrow \mathbb{D}$$
Defining $g(z):=f(\varphi^{-1}(z))$, we have translated the problem to a bounded holomorphic function on the unit disk.
It is easy to see that the zeros of $g$ are of the form $\frac{n-1}{1+n}$. Thus, applying the Blaschke condition (see here) , we obtain that $g\equiv0$. It easily follows that $f\equiv 0$