Proving the limit of a sequence of functions must be polynomial. Let $n \geq 1$ and let $\left(f_n (z)\right)_n$ be a sequence of polynomials whose coefficients are in $\mathbb{C}$. Suppose that $(f_n (z))_n$ converges uniformly to a function $f(z)$ on $\mathbb{C}$. Show that $f(z)$ is a polynomial.
I know that "$(f_n (z))_n$ converges uniformly to $f(z)$ on $\mathbb{C}$" is equivalent to 
$$\sup_{z\in \mathbb{C}}|f_n(z)-f(z)| \to 0 \, \, \text{as} \,\, n\to \infty$$
which I think is equivalent to 
$$ \forall \epsilon >0\,\,\, \exists N\,\, \text{s.t.}\,\, \forall z \in \mathbb{C} \,\,\,|f_n(z) - f(z)|<\epsilon \,\,\text{when} \,\, n>N \,.$$
Unfortunately I don't know where to go from here.
Thank you in advance.
 A: Find $N$ such that $|f_N(z)-f(z)|<1$ for al $z$. 
Then the function $f_N-f$ is entire(!) and bounded, hence a constant $c$, and therefore $f=f_N+c$.
A: A stronger result holds, and it doesn't have that much to do with complex analysis: Let $E\subset \mathbb C$ be unbounded. If $p_n$ is a sequence of polynomials with $p_n\to f$ uniformly on $E,$ then $f$ is a polynomial. (There can be at most one such polynomial, since $E$ is infinite.) 
Proof: First, note that if a polynomial $p$ is bounded on $E,$ then $p$ is a constant. That's because if $\deg p>0,$ then it is elementary that as $|z|\to \infty$ within $E,$ $|p(z)|\to \infty.$ Thus we must have $\deg p=0,$ which implies $p$ is constant.
We are given $p_n\to f$ uniformly on $E.$ This implies there exists $N$ such that $n\ge N$ implies $|p_n-p_N|<1$ on $E.$ By the above, $p_n-p_N$ is a constant $c_n$ for $n\ge N.$ Evaluate at any $z\in E$ and let $n\to \infty$ to see that $c_n$ converges to some $c.$ So letting $n\to \infty$ in the equation $p_n=c_n +p_N$ gives $f= c+P_N.$ Thus $f$ is a polynomial as desired.
