How to show a given bound For a polynomial $p(z)=\sum_{j=0}^{n} a_{j} z^{j},$ how to show that
$$|p(z)| \leq A_{\epsilon} e^{\epsilon|z|},$$
for any $\epsilon > 0$, for some $A_{\epsilon} > 0$.
I tried to bound each term of the polynomial by some exponential term and to get the maximum of all of them but wasn't successful. Any suggestion is much appreicated. 
 A: Let $M=\max\{|a_{j}|: j=0,...,n\}$, then $|p(z)|\leq M\displaystyle\sum_{j=0}^{n}|z|^{j}$. Fix a $j=0,...,n$, as $\dfrac{|z|^{j}}{e^{(\epsilon+\rho)|z|}}\rightarrow 0$ as $|z|\rightarrow\infty$, then we can find some $N>0$ such that for each $j=0,...,n$, for all $|z|\geq N$, $|z|^{j}\leq e^{(\epsilon+\rho)|z|}$.
Now $z\rightarrow\displaystyle\sum_{j=0}^{n}|z|^{j}$ is continuous on $\{|z|\leq N\}$, so for some $M'>0$, $\displaystyle\sum_{j=0}^{n}|z|^{j}\leq M'$ on $\{|z|\leq N\}$.
Now $z\rightarrow e^{(\epsilon+\rho)|z|}$ is continuous on $\{|z|\leq N\}$, so for some $m>0$, $e^{(\epsilon+\rho)|z|}\geq m$ on $\{|z|\leq N\}$.
So we conclude that on $\{|z|\leq N\}$, $|p(z)|\leq MM'=\dfrac{MM'}{m}\cdot m\leq\dfrac{MM'}{m}e^{(\epsilon+\rho)|z|}$.
On $\{|z|>N\}$, $|p(z)|\leq M(n+1)e^{(\epsilon+\rho)|z|}$, overall we have $|p(z)|\leq M\left(\dfrac{M'}{m}+n+1\right)e^{(\epsilon+\rho)|z|}$.
A: The $ρ$ is not necessary. Actually, given a fixed positive $\epsilon$, $\frac{|p(z)|}{e^z}$ tends to $0$ when $|z|$ tends to infinity. 
So it must have a maximum which can be chosen as $A_\epsilon$.
