Is there an holomorphic function? If that function exists, is it unique? I am solving a problem that asked me if exist an holomorphic function which satisfies only two condition. both equally:
$$ f\left(\frac{1}{\alpha n} \right)=0\ \ \ \ and\ \ \ f\left( \frac{1}{\alpha n+1}\right)=1$$
where $\alpha, n \in \mathbb{N}$ where $\alpha$ is unique
I think that I have to use Identity theorem for holomorphic function. Someone could help me with this? thanks for all.
Note: I do not know if $f$ is or not holomorphic. (namely, I have to prove that exist, first, at least holomorphic function)
 A: Depending on the domain of your function, either no such function exists or it does exist but is not unique.
If the domain is $\mathbb C$, then note that both $\lim_{n\to \infty}\frac{1}{\alpha n}=\lim_{n \to \infty}\frac{1}{\alpha n +1}=0$. So, since holomorphic functions are continuous, $f(0)=\lim_{n\to \infty} f(\frac{1}{\alpha n})=\lim_{n \to \infty}f(\frac{1}{\alpha n +1})$ must hold, but this statement is not compatible with the conditions given.
If you have the domain be $\mathbb C - \{0\}$, then the situation is different.
Let $\alpha \in \mathbb N$. If $\alpha = 1$ the condition is obviously contradicts itself.
Otherwise
$$e^{i 2\pi z/\alpha}=\begin{cases}1 &z= \alpha n \\ e^{i2\pi/\alpha}\neq 1 &z=\alpha n +1 \end{cases}$$
So with
$$f(z):=\frac{1-e^{i 2 \pi z /\alpha}}{1-e^{i2\pi/\alpha}}$$
You have $f(\alpha n)= 0$ and $f(\alpha n +1) = 1$ and $f$ is holomorphic on $\mathbb C$.
As a consequence $f(\frac{1}{z})$ is a holomorphic function on $\mathbb C - \{0\}$ that satisfies the condition. Now note that $e^{2\pi i z}$ evaluates to $1$ whenver $z \in \mathbb N$. For that reason, adding $a (1-e^{i2\pi /z})$ for any number $a \in \mathbb{C}$ will also satisfy the condtion and the solution is not unique.
