# How to prove f has degree at most n, with the caveat that |z|>R

Prove the following statement. Suppose that f is entire and that there exist$k > 0, R > 0, n \in \mathbb{N}$ such that $|f(z)| ≤ k|z|^n$ n for all $z$ such that $|z| > R$. Then f is a polynomial of degree at most n.

at first I thought this was a pretty easy proof until I realized the question said |z|>R.

Basically what I've done so far is;

if f is entire then it has a taylor series representation at $z_0=0$

$$f(z)=\sum_{n=0}^{\infty}a_nz^n$$

the coefficients are of the form $a_m=f^m(0)/m!.$ so if we can show that $f^m(0)=0 \forall m>n$ this will imply that $a_m=0 \forall m>n$ and hence show that f has degree at most n.

Originally I was going to use Cauchy's Inequality but then i realised the definition for this says we must have $0<z<R$ which means I can't use it right ?. I looked through my notes but I couldn't find any that allowed $|z|>R$ so I'm at a loss for what theorem to use.

Note: we are currently working on Laurent series and singularities. We have covered Taylors theorem , Cauchy's integral theorems, Liovilles theorem and laurents theorem and also there corollaries . just to give some context on the knowledge we're expected to have.

Quoting OP: "Originally I was going to use Cauchy's Inequality but then i realised the definition for this says we must have $0<z<R$ which means I can't use it right ?. I looked through my notes but I couldn't find any that allowed $|z|>R$ so I'm at a loss for what theorem to use."
You need to be able to see through those symbolisms. This exercise simply refers to a radius $R>0$. Your notes probably refer to the radius of a disc being included in the (open) domain of your holomorphic function. When reading a theorem or a lemma, try to understand the relations between the objects that are being discussed.
Let $z\in\mathbb{C}\setminus\overline{D(0,R)}$. Then, for $r>|z|$ Let $C_r$ denote the circumference of the circle centered at 0 and radius $r$, orientated counter-clockwise. By Cauchy's integral formula: $f^{(n)}(z)=\displaystyle{\frac{n!}{2\pi i}\int_{C_r}\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta}$. Applying the M-L inequality one gets $|f^{(n)}(z)|\leq n! r\displaystyle{\max_{\zeta\in C_r}|\frac{f(\zeta)}{(\zeta-z)^{n+1}}|}\leq \frac{n!\cdot r\cdot k\cdot r^n}{(r-|z|)^{n+1}}$. This inequality holds for all $r>|z|$, therefore by letting $r\to \infty$ at the inequality above, one gets $|f^{(n)}(z)|\leq n!\cdot k$, for all $z\in \mathbb{C}\setminus\overline{D(0,R)}$. The set $\overline{D(0,R)}$ is compact, so $f^{(n)}$ which is continuous on $\mathbb{C}$ is bounded on this compact disc. Hence $f^{(n)}$ is bounded on the whole plane, therefore is constant (Liouville's theorem). Thus $f$ is a polynomial