5

Let $f:\mathbb{C} \rightarrow \hat{\mathbb{C}}$ a meromorphic function with an essential singularity at infinity. Does $f'(z)$ have a zero? No, considering $z\mapsto e^z$. But if I assume that $f$ is surjective, is it true? and can i say something on the order of vanishing.

I am not a specialist of holomorphic function. This question arise in geomerty. More precisely my function satisfies $\frac{f'}{1+\vert f\vert^2}=O(\frac{1}{z})$, where the right-hand side is called the spherical derivative. Thanks to the work of Letho,The spherical derivative of meromorphic functions in the neighborhood of an isolated singularity, we know that $f$ must be surjective.

Added: The full question is: If $f$ has an essential singularity and satisfies $\frac{f'}{1+\vert f\vert^2}=O(\frac{1}{z})$, does $f'$ vanishes? and can we say something about the order of vanishing.

Ana
  • 155

2 Answers2

3

A simple counter-example is: $$f(z)=2\int_0^ze^{w^2}dw-\frac{e^{z^2}}{z}.$$ Obviously $f$ is a meromorphic function on $\Bbb C$ with an essential singularity at infinity and $$f'(z)=\frac{e^{z^2}}{z^2}$$ has no zero.

To show $f(\Bbb C)=\hat{\Bbb C}$, since $f(0)=\infty$, let us assume that there exists $c\in \Bbb C$ such that $c\notin f(\Bbb C)$ to get some contradiction. By our assumption, $$g(z):=z(f(z)-c)=2z\int_0^ze^{w^2}dw-e^{z^2}-cz$$ is an entire function with no zero. Moreover, it is easy to see that the order of $g$ at infinity is $2$. Then by Hadamard's theorem, there exists a polynomial $P$ of degree $2$, such that $$g(z)=e^{P(z)}\Longrightarrow g'(z)=P'(z)g(z),$$ which is absurd.

Hu Zhengtang
  • 3,757
  • It looks true. But as you say the order of $g$ at infinity is $2$ but as explain in my question I am interested in function satisfying $\frac{f'}{1+\vert f\vert^2}=O(\frac{1}{z})$. The bounty concerns this refinement but if no one found it will be yours. – Ana Nov 09 '13 at 19:55
  • 2
    @Ana: When I was answering your question, I thought your question was contained in the first paragraph, and the second paragraph was merely to explain your motivation, so I gave such an example. According to the paper you cited, if $f$ satisfies $\frac{f'}{1+|f|^2}=O(\frac{1}{z})$ around infinity, then at least $f$ maps any neighborhood of infinity onto $\hat{\Bbb C}$, and clearly my example fails to satisfy this property. I'm totally unfamiliar this topic, so let me stop here without modifying my answer. – Hu Zhengtang Nov 10 '13 at 11:32
2

Note that the answer is no if surjective meromorphic function $f:\mathbb{C} \to \hat{\mathbb{C}}$ is replaced by surjective entire function $f:\mathbb{C} \to \mathbb{C}$. In this case, an example is given by $$f(z)=z\cdot e^{\int_0^zg(t)dt}$$ where $g(z)=\frac{e^z-1}{z}.$

See this question

Community
  • 1
Malik Younsi
  • 3,868
  • 1
    Malik: The example needs a modification since Ana asked about a meromorphic function, whose image is the whole Riemann sphere, while I asked about a entire function whose image is the whole complex plane. Still, this is a very strange coincidence. – Moishe Kohan Nov 06 '13 at 21:14
  • Thanks for the example. But it doesn't feet the question as studiosus remarks it. – Ana Nov 07 '13 at 09:12
  • @studiosus : you are right, I didn't read carefully. Indeed, it is a strange coincidence. – Malik Younsi Nov 07 '13 at 13:46