Complex Analysis Proof Application of Liouville's Theorem I am studying for my Complex Analysis and have been trying to answer this question for some time now.  My professor gave a hint to use the power series expansion of $f(z)$ (to manipulate $f(z)$) and then to use or apply Liouville's Theorem.  I have tried this, but am getting nowhere.  Any help, suggestions, or tips would be much appreciated.  Please note that this is for an elementary course on complex analysis so please do not give too advanced an explanation.  Thank you. I have attached two pictures, one of the Cauchy estimates and one of the actual question.


 A: I would do Jennifer's proof, but if you want it explicitly using Liouville's theorem, use the same as her but exactly with $f^{(m)}(z)$, where $F$ of your Liouville's theorem statement is $f^m$.
By the Cauchy inequality with $z_0 = 0$ and $r\geq R_0$,
\begin{equation}
|f^{(m)}(0)|\leq\frac{m!}{r^m}\max_{|z|=r}|f(z)| \leq \frac{n!}{r^n}A|z|^m = (\ast).
\end{equation}
As $|z| = r$,
\begin{equation}
(\ast) = n! A = M
\end{equation}
and by Liouville's theorem $f^{(m)}$ is constant, which implies that $f$ is a polynomial of degree $m$ or less (because it's entire and you can write it like a power serie as Jennifer said).
A: You don't have to use Liouville's theorem.
You can write $f(z)=\sum_{n=0}^\infty a_nz^n$. With Cauchy inequality, $\forall n,|a_n|\leq\frac{\max_{|z-z_0|=r}|f(z)|}{r^n}$. So with the hypoyhesis of the question : $\forall n,\forall r \geq R_0,|a_n|\leq\frac{Ar^m}{r^n}={Ar^{m-n}}$.
When $n>m$ if you make tend $r\rightarrow \infty$, you have $|a_n|=0$. So $f(z)=\sum_{n=0}^m a_nz^n$ so $f$ is a polynomial of degree $m$ or less.

We will show this result by induction on $m$.  
By Liouville's theorem, the statement is true for $m=0$, indeed in this case $f$ is bounded so $f$ is constant, so $f$ is a polynomial of degree at most $0$. 
Suppose true for $m-1$. Consider $$g(z)=\left\{\begin{array}{ll} \frac{f(z)-f(0)}{z}, & z\neq 0\\ f'(0), & z=0\end{array}\right.$$
Then $g$ is entire, and for sufficiently large $|z|$, $$|g(z)|\le \left|\frac{f(z)}{z}\right|+\left|\frac{f(0)}{z}\right|\le|Az^{m-1}|+|f(0)|\le C|z|^{m-1}$$ for some constant $C$. 
Hence by induction hypothesis, $g$ is a polynomial of degree at most $m-1$.
So $f$ is a polynomial of degree at most $m$. 
