1. ## Complex Analysis

Suppose that $f$ is an entire functionand that there are positive constants $A$ and $m$ with $|f(z)| \leq A|z|^{m}$ if $|z| \geq R_{0}$. Show that $f$ is a polynomial of degree $m$ or less.

2. Originally Posted by chiph588@
Suppose that $f$ is an entire functionand that there are positive constants $A$ and $m$ with $|f(z)| \leq A|z|^{m}$ if $|z| \geq R_{0}$. Show that $f$ is a polynomial of degree $m$ or less.
isn't this just a slight twist on the (extended) Liouville's theorem?

3. Originally Posted by chiph588@
Suppose that $f$ is an entire functionand that there are positive constants $A$ and $m$ with $|f(z)| \leq A|z|^{m}$ if $|z| \geq R_{0}$. Show that $f$ is a polynomial of degree $m$ or less.
Let $R>R_0$.

If $f$ is an entire function then $f(z) = \sum_{n=0}^{\infty}a_n z^n$ and $a_n = \frac{1}{2\pi i}\oint \limits_{|z|=R} \frac{f(z)}{z^{n+1}} dz$.

If $|f| \leq M$ on $|z|=R$ then $\left| a_n \right| = \left| \frac{1}{2\pi i}\oint \limits_{|z|=R} \frac{f(z)}{z^{n+1}}dz \right| \leq \frac{M}{R^n}$

By hypothesis $|f(z)| \leq A |z|^m$ for $|z| \geq R_0$. Therefore, if $R > R_0$ then on $|z| = R$ we have $|f(z)| \leq A |z|^m = A\cdot R^m$

Thus, $|a_n| \leq \frac{A \cdot R^m}{R^n}$.
Thus, if $n>m$ then by making $R$ sufficiently large we have $|a_n| = 0 \implies a_n = 0$.

Thus, $f$ is a polynomial of degree at most $m$.

4. That's the proof I used, does that prove it for the disk $|z| < R_{0}$ also?