Find all entire functions that are uniformly continuous on the complex plane.
I think the answer must be the the linear polynomials, but nor sure, since neither of polynomials of degree greater than 1 or transcendental entire function are uniformly continuous on $\mathbb{C}.$
Given $\epsilon >o$, there exists $\delta>0$ for any $z_1, z_2$ such that $|z_1-z_2|\leq \delta $ then we have $|f(z_1)-f(z_0)|\leq \epsilon.$
Let write the taylor expansion of $f(z)$ around point $z \in \mathbb{C}$. $$ f(z) =\sum_{n=0}^{\infty}\frac{f^{(n)}(0)z^n}{n!}. $$ so if for $z_1, z_2$ we have $|z_1-z_2|\leq \delta $, we get
$$ |f(z_1)-f(z_2)| \le \sum_{n=0}^{\infty} \frac{f^{(n)}|z_1^n-z^n_2|}{n!} $$
Intuitively, I can see that $n=1,0$ to have a uniformly continuous function, but I might be wrong. I appreciate any help.