Multiple choice question about an entire function $f:\Bbb{C}\to\Bbb{C}$ and the function $g :\Bbb{C}\to\Bbb{C} $ defined by $ g(z)= f(z) - f(z+1)$

Question

Let $ f: \mathbb{C} \rightarrow \mathbb{C} $ be an entire function and let $g : \mathbb{C} \rightarrow \mathbb{C} $ be defined by $$g(z)= f(z) - f(z+1)$$ for all $ z\in \mathbb{C}$. Which of the options are correct :

1. if $ f(\frac{1}{n}) = 0 $ for all positive integers n, then $f$ is a constant function.

2. if $ f(n) = 0 $ for all positive integers n, then $f$ is a constant function.

3. if $ f(\frac{1}{n}) = f(\frac{1}{n}+1)$ for all positive integers n, then $g$ is a constant function.

4. $ f(n) = f(n+1) $ for all positive integers $n$, then $g$ is a constant function

Please suggest which of the options are correct.

Using the Identity theorem, the options 1 and 3 seem to be correct as in both cases, the sequence of zeros for $\,f\,$ and $\,g\,$ is $ < \frac{1}{n} >$ that converges to zero which belongs to $\Bbb C$. Therefore, in both cases $\,f\,$ and $\,g\,$ are identically equal to zero. But in (2) and (4), we arrive for both $\,f\,$ and $\,g\,$, at the zeros sequence $ <{n}>$ diverges to infinity which does not ensure the required conclusion.

Answer 1

This answer is a compilation of the answers given in the comments above.

To answer these questions, you should be familiar with the identity theorem.

1. True. The set of zeros of $f(z)$ has an accumulation point at $0$, so $f$ is identically zero.

2. False. Consider $f(z)=\sin(\pi z)$.

3. True. We have $f(1/n)=f(1/n+1)$ for all $n\in\mathbb{N}$, so $g(1/n)=f(1/n)-f(1+1/n)=0$ for all $n\in\mathbb{N}$, and the set of zeros has an accumulation point at 0, so $g$ is identically zero.

4. False. Again, consider $f(z)=\sin(\pi z)$.