find entire functions such that $f( e^\frac{1}{n})=\frac{n+1}{n}$ for all $n\in \mathbb{N}$

Question

I've got to find out if there are entire functions such that $f( e^\frac{1}{n})=\frac{n+1}{n}$ for all $n\in \mathbb{N}$.

We can write down $\frac{n+1}{n}=1+\frac{1}{n}$

I'm trying to see that $g(z):=f(e^z)-z-1$ gives $f(e^z)=1+z$ by identity theorem as $g(1/n)=0$ and has a limit point $0\in \mathbb{C}$. So $1+z$ would be the entire function. Is this right?

Answer 1

What you did is correct. And now you can deduce that there is no such entire function $f$, since $1+1\neq1+2\pi i$, but $f(e^0)=f\left(e^{2\pi i}\right)$. Therefore, you don't have $(\forall z\in\mathbb C):f(e^z)=1+z$.

Answer 2

Alternative ending once $f(e^z)=1+z$ is proved: $\frac{d}{dz}f(e^z) = e^z f'(e^z) = 1$, so $f'(e^z)=\frac{1}{e^z}$ and $f'(w)=\frac{1}{w}$ for any $w\neq 0$. This gives that $f'$ is not bounded in a neighbourhood of the origin, contradicting the entire-ness of $f$.