I got this question in an exam today, and I was wondering whether my reasoning is correct or whether there exists a counterexample:
Let f, g be two holomorphic functions from $\Bbb{C}$ to $\Bbb{C}$. Then, if for all n $\in$ $\Bbb{N}$ we have f($\frac 1n$) = g($\frac 1n$), we must have that f=g everywhere.
My reasoning was that since z=0 is an accumulation point in $\Bbb{C}$, and f=g at this point, by the identity theorem for holomorphic functions we can conclude that f=g everywhere.
This seemed rather straightforward to me but I just wanted to verify that I didn't miss any steps or make false assumptions that would imply I can't use the identity theorem. I appreciate all the help you can give me :) thanks in advance!
No, that is not correct. You are deducing that $f=g$ simply from the fact that $f(0)=g(0)$ and the fact that $0$ is an accumulation point of $\mathbb C$. By the same argument, if $f(z)=z$ and $g(z)=z^2$, then $f=g$.
Note that, since $f(0)=g(0)$, you know that$$\{z\in\mathbb C\mid f(z)=g(z)\}\supset\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}\cup\{0\}$$and therefore the set $\{z\in\mathbb C\mid f(z)=g(z)\}$ has an accumulation point. So, by the identity theorem, $f=g$.