Let $f$ be a non-constant entire function satisfying the following conditions:
- $f(0)=0$
- For every positive real number $M$, the set $ \{z:|f(z)|<M\}$ is connected.
Prove that $f(z)=cz^n$ for some constant $c$ and positive integer $n$.
My attempt:
Since $\ f$ is entire so zeros of $\ f$ are isolated so we can choose an $r>0$ such that $B_r=\{z:|z| \le r\}$ contains no zero of $\ f$ other than $0$. Now if I consider $m=\inf_{|z|=r} |f(z)|$ then the set $S=\{z:|f(z)|<m\}$ is connected (by 2nd condition) and also $S \cap B_r \ne \phi$ which implies $S \subset B_r $. Therefore, $\ f$ has only zero at $z=0$ in $\mathbb{C}$.
Now I think $\ f$ has a pole at $\infty$....which would imply it is a polynomial which has only zero at $z=0$...And this should imply that $\ f$ is of the desired form.
So I try to prove: $\lim_{|z| \to \infty}|f(z)|=\infty$
For this we have to show that for any $R>0$ there exists a $P>0$ such that $|f(z)| \ge R$ whenever $|z|>P$.
But from our previous discussion we can say that outside $B_r, ~ |f(z)| \ge m$.....So for any $R>m$ we can choose $P$ to be $r$. But what if $R \le m$..?
How can I prove that $\ f$ has a pole at $\infty $ ?
If I'm going wrong then please let me know...
Please help. Thank you..
If $f$ is bounded at infinity, then $f \equiv 0$. Otherwise, suppose $f$ had an essential singularity there. Choose $\epsilon > 0$ such that there exists a circle $S_R$ of radius $R$ around the origin on which $|f(z)| \ge \epsilon$ for all $z \in S_R$.
By any number of theorems (e.g. big Picard or Casorati-Weierstrass), the image of $\{z : |z| > R\}$ under $f$ contains points arbitrarily close to the origin. Therefore, $\{z : |f(z)| < \epsilon\}$ contains both $0$ and a point outside $S_R$, which contradicts connectivity.
Hence, $f$ must have a pole at infinity, so it's a polynomial. The only zero is at the origin, so $f$ is a multiple of $z^n$.