Complex Analysis - Liouville's Theorem application and polynomial degree Exercise : 
Let $f:\mathbb C \to \mathbb C$ be an entire function.
(a) If $|f(z)| \geq 1$ $\forall z \in \mathbb C$, show that the function $f$ is constant in $\mathbb C.$
(b) If :
$$\lim_{|z|\to \infty} \frac{f(z)}{1 + |z|^{5/2}} = 0$$
show that $f$ is a polynomial of degree $\leq 2.$
Attempt :
(a)
Since $f: \mathbb C \to \mathbb C$ is an entire complex function and f :
It is : 
$$|e^f| \leq e^{|f|} \geq e^1 \Leftrightarrow |e^f| \leq e  $$ 
Thus we can say that $e^f$ is constant (by Liouville's Theorem). Now, I know I must show that this implies that $f$ is constant but I do not see how to do it, so I'd appreciate some help (I think I should try to see how $f$ could be constant and then prove that it contradicts the assumption that $f$ is entire).
For (b) I do not know how to grasp the work in such exercises and I would really like a thorough solution, because I just stepped on the Liouville's Theorem examples and exercises, so I'd like to figure out how such questions can be solved.
I would really appreciate any help !
Kind regars,
Charalampos Filippatos.
 A: (a) If $(\forall z\in\mathbb{C}):\bigl|f(z)\bigr|\geqslant1$, then $(\forall z\in\mathbb{C}):\bigl|\frac1{f(z)}\bigr|\leqslant1$ and therefore $\frac1f$ is constant, by Liouville's theorem.
(b) If $f(z)=a_0+a_1z+a_2z^2+\cdots$, then, by Cauchy's inequalities,$$(\forall R>0):|a_n|\leqslant\frac1{R^n}\sup_{|z|=R}\bigl|f(z)\bigr|$$It follows from your inequality that this converges to $0$ when $R$ goes to $+\infty$ and $n>2$. Therefore, $f(z)=a_0+a_1z+a_2z^2$.
A: For the first one use $g=1/f$. Clearly g is entire because $|f(z)| \geq 1$. Now $|g(z)| \leq 1$. Hence by Lioville's theorem , g is constant, hence f is constant.
(b) Here observe that due to the limiting condition 
$$ \Big{|} \frac{f(z)}{1+|z|^{\frac{5}{2}}} \Big{|} \leq M$$.
Now use the following Cauchy's estimate: if $|f(z)| \leq M$ on $C(0,R)$, then
$$ |f^{n}(0)| \leq \frac{n!M}{R^{n}}$$.
Use this to get from the above:
$$|f^{n}(0)| \leq \frac{n!M(1+R^{5/2})}{R^{n}}$$
Now, observe if $n\geq 3$, then as $R\to \infty$, the RHS of the above inequality approaches 0 , hence $f^{n}(0) =0$ for all $n \geq 3$. Now use the global power series representation of f around 0 to conclude the result. 
