$f$ is an entire function such that $f(0)=0$

Question

Let $f$ be a non-constant entire function satisfying the following conditions: $f(0)=0$ and for each $N \gt 0$ the set $\{z \mid \left| f(z)\right| < N\}$ is connected.

Prove that $f(z)=cz^n$ for some constant $c$ and positive integer $n$.

$f(0)=0$ implies there is $r>0$ such that $f(z)\neq 0$ for any $z\in \{z: 0<|z|\leq r\}$

All I can see it $0$ is the only root of $f(z)$. Suppose not, for small $M$, the set $\{z: |f(z)|<M\}$ contains disjoint open sets which contradicts connectedness of the set. So, there is only one root.

So, $0$ is the only root of $f(z)$.

Can we now say that we are forced to have $f(z)=cz^n$ for some $n$?

I could not see more than this.. please give some hints.

Answer 1

Let's start by assuming that $f(z)$ is a polynomial. Since we know it is an entire function, we know that it can at least be written as a power series, so we'll begin with this simplification.

Since it is a polynomial, we can factor it over $\mathbb{C}$ as $f(z) = z(z - a_1) \cdots (z - a_k)$ for some $a_i \in \mathbb{C}$. Suppose without loss of generality that $a_k \neq 0$. Locally around $a_k$ (and 0) the function looks like $a(z- a_k)^{b_k}$. In particular, we can choose $N > 0$ small enough so that $|f|^{-1}([0, N])$ is (at least) two disjoint neigbourhoods of $a_k$ and 0, respectively. So if it's a polynomial, you're fine.

So what if it is not a polynomial? Well, then it is an entire function. Since entire functions attain every single value (except perhaps one) infinitely often, it follows that there are infinitely many $\lambda \in \mathbb{C}$ such that $f(\lambda) = 0$. Using a similar argument to the polynomial case, you can then show that the set is disconnected for $N > 0$ sufficiently small.

Edit As pointed out in the comments, this last part doesn't quite work; for example, our function could be $f(z) = z e^z$ which attains the value 0 exactly once. However, one should be able to note that in such a case, it must attain the value $\varepsilon$ infinitely often, so choosing $N > \varepsilon > 0$ one should be able to make a similar argument (since the function attains the value $\varepsilon$ close to zero, and also infinitely often close to $\infty$).

Answer 2

Picard is a big gun, maybe even a WMD (weapon of math destruction), for this problem. We know the only zero of $f$ is $0$ from the connectivity condition. Thus $f(z) = z^ng(z)$ for some $n \in \mathbb N$ and some entire $g$ that never vanishes.

We also know there is $r>0$ such that $\min_{|z|=r} |f| = m > 0.$ It then follows, again from the connectivity condition, that $\{|f|<m/2\} \subset D(0,r).$ Thus $|f(z)| \ge m/2$ for $|z| \ge r.$

Putting this together shows

$$\frac{1}{|f(z)|} = \frac{1}{|z|^n|g(z)|}\le \frac{2}{m}, \,\,|z|\ge r.$$

Now $1/g$ is entire, and the above gives $|1/g(z)| \le (2/m)|z|^n$ for $|z|\ge r.$ So $1/g$ is an entire function that grows no faster than a polynomial. A well known result then implies $1/g$ is a polynomial. But $1/g$ never vanishes. The only polynomials that never vanish are the nonzero constants. This gives $f(z) = cz^n$ as desired.

Answer 3

Here is a minimalistic approach:

$f(0) = 0$. So unless $f$ is identically zero, for any small enough $r > 0$, $|f|$ has a minimum $m_r > 0$ on $|z| = r$. Therefore, as mentioned by zhw, by connectivity, $|f(z)| \ge m_r/2$ for $|z| \ge r$.

This proves two things at once:

a) $f(1/z)$ can not have an essential singularity at zero. Hence by using Laurent series, conclude that $f$ is a polynomial.

b) The only root of $f$ is zero.

Therefore, $f = c z^n$.