$f: \mathbb C \to \mathbb C$ is analytic function and $z_0 \in \mathbb C$,$f(r)=z_0 \forall r \in \mathbb Q\cap[1,2]$

Question

$f: \mathbb C \to \mathbb C$ is analytic function and $z_0 \in \mathbb C$, I am investigating the character of $f$, when

$f(\frac{1}{n})=z_0 \forall n \in \mathbb N$

and if,

$f(r)=z_0 \forall r \in \mathbb Q\cap[1,2]$

and if,

$f(n)=z_0 \forall n \in \mathbb N$,

For theirst two cases what I am doing is that,

define $g(z):=f(z)-z_0$, the the set of zeros of $g$ has limit point $0$ (as $\frac{1}{n} \to 0 ,\text as. n \to \infty$)which belongs to domain of $g$,similarly for $r_n$ be rational sequence in $[1,2]$ will have limit point =$1,2$ as $[1,2]$closed. Then applying Identity mapping theorem $g=0$ identically in $\mathbb C$,

i.e. $f(z)=z_0$

Now for this case $f(n)=z_0 \forall n \in \mathbb N$, can we conclude that $f$ is constant $z_0?$

Can $f$ take the value $z_0$ for finitely many points in the set {$\frac{1}{n}|n \in \mathbb N$}?

Answer 1

Your answer for the first two cases is correct. There are lots of entire functions $f$ with $f(n)=z_0$ for all $n \in \mathbb N$ . One example is $z_0+\sin (\pi z)$.

A general fact: if $S$ is any set of complex numbers with no limit points and $z_s$ is a complex number for each $s \in S$ then there is an entire function $f$ such that $f(s)=z_s$ for all $s \in S$. In particular this holds for a finite set $S$.