I want to find all entire function such that $|f(z)| \geq C/|z|^n$ for $|z| \geq R$ I want to find all entire function such that (for fixed R,C>0 and fixed integer n)
$|f(z)| \geq C/|z|^n$ for $|z| \geq R$
I know that f does not have any zeros for $|z| \geq R$, then define $g=1/f$ where g is analytic for $|z| \geq R$.
Then the above inequality changes to $|g| \leq |z|^n/C$ for $|z| \geq R$.
I also know that if g is entire, g is polynomial, but g is not...(it is just analytic at outer disk)
("If g is polynomial with greater than 1 degree, then f has a pole, so it is not entire.. then f is a constant" is what i want to say if i can show g is polynomial)
(However this is also wrong because i found f=z that for R=C=n=1 the equality is true for f=z)
then how can i find such $f$?
 A: I presume the quantities $C,n,R$ are fixed independently of $f$.
Then an entire function $f$ satisfies the condition in the question iff $f$ is a polynomial whose roots are in $|z| <R$ and $\min_t |R^n f(R e^{i t})| \ge C$.
Suppose $f$ is entire and satisfies the condition in the question. It is clear that we have $\min_t |R^n f(R e^{i t})| \ge C$.
Since $f$ is entire we see that $f$ has a finite number of zeros in $|z| < R$. Hence we can write $f(z) = p(z) g(z)$ for some polynomial $p$ (whose zeroes are the zeroes of $f$) and some entire $g$ that has no zeroes in $|z| < R$ (and hence no zeroes anywhere). 
In particular, ${1 \over g}$ is entire.
We have $|z^n p(z)| \ge {C \over |g(z)|}$ for $|z| \ge R$. It follows from this that ${1 \over g}$ is a polynomial and hence a non zero constant (otherwise it would have zeros). Hence $f$ is a polynomial whose roots
are in $|z|<R$.
For the other direction, suppose $f$ is a polynomial whose roots are in $|z|<R$ and $\min_t |R^n f(R e^{i t})| \ge C$. Since $f$ is entire we just need to show that $f$ satisfies the condition in the question.
Pick $w$ such that $|w| >R$ and let $w^*$ be the closest point in $|z|=R$ to $w$. We know that $f$ has the form  $f(z) = K\prod_k (z-z_k)$ with $|z_k| < R$. Since $w^*$ is the closest point to $w$ we see that
$|w-z_k| \ge |w^*-z_k|$ for all $k$ and hence $|f(w)| \ge |f(w^*)|$ and
so $|w^n f(w)| \ge |(w^*)^nf(w^*)| \ge C$ as required.
