If functions $f$ and $g$ are entire such as: $f(1/n) = g(1/n)$ $\forall n\in \Bbb{N}$, then $f=g$ in $\Bbb{C}$
Is it true or not?
In order to answer this question I want use a statement which says:
If $f$ and $g$ are holomorphic functions in $D$ and a set $\{z \in D : f(z)=g(z) \}$ has a cluster point in $D$ then $f \equiv g$ in $D$
My functions are holomorphic in $\Bbb{C}$ so I have to show that a set $S=\{z \in \Bbb{C} : f(z)=g(z) \}$ has a limit point in $\Bbb{C}$. Point $p$ will be a cluster point of $S$ if I find a sequence with terms from $S$ which converges to $p$. I don't know how to use a information that $f(1/n)=g(1/n)$. I know that $(1/n)_{n=1}^{\infty} \to 0$
The set $\{z \in \mathbb{C}\,|\,f(z)=g(z) \}$ contains all points of the form $\frac1n$ ($n\in\mathbb N$) and therefore it has a cluster point: $0$. So, $f=g$.