Sequence of uniformly converging entire function is equivalent to its limit for sufficiently large $n$ given $f_n(z)=f(z)=0$

Question

A question from my assignment. Suppose $\{f_n\}$ is a sequence of entire function converge uniformly to $f$ and satisfy for any $n$, $f_n(0)=f(0)=0$. Prove that there exist $N$ such that $n>N$ implies $f_n=f$.

Since it is a complex analysis problem, I suppose it should be a bit related to the Cauchy's integral formula?

Answer 1

Let $g_n := f_n - f$, so that $(g_n)$ converges uniformly to $0$ and $g_n(0) = 0$. Since $g_n \to 0$ uniformly, $g_n$ is bounded for every $n$. Being an entire function, $g_n$ is constant, i.e. $g_n(z) = g_n(0) = 0$ for every $z\in \mathbb{C}$.

Answer 2

Find $N$ such that $\sup_{\mathbb{C}} | f_n(z) -f(z) |<1,~\forall n\geq N$ .

Then define $g_n=f-f_n$, and notice $ |g_n| <1$ hence $g_n$ is constant as a bounded entire function. Also $g_n(0)=0$ therefore $g_n\equiv 0$. Therefore $f=f_n$ for all $n\geq N$.