I have been going through Ahlfors Complex Analysis and am currently studying the section about isolated singularities. I have tried some problems from this section but cannot make any progress at all including looking at previous answers posted about these questions. For instance:
- If an entire function has a nonessential singularity at $\infty$ it reduces to a polynomial.
Okay, well nonessential means we are either removable or a pole. Let us deal with the case for the removable singularity. Define $g(z)=f(1/z)$, then if $f$ has a removable singularity at $\infty$ then $g$ must have a removable singularity at $0$. From here we can extend the function $g$ so that it actually becomes analytic at the point $0$ simply by defining $g(0) = \lim_{z \rightarrow 0} g(0).$ Here is where I get stuck, where do I go from here? Of course we can substitute what $g$ is in the limit, but how does that help?
For the pole, we know that this means that $\lim_{z \rightarrow \infty}f(z) = \infty$. Therefore again define $g(z) = \frac{1}{f(1/z)}$, and $g$ will have a removable singularity at $0$. I do not know how to proceed from here using what Ahlfors covers so far.
It seems like I am stuck basically at the same part in both cases. Any advice on how I can understand this concept and solve this problem is much appreciated.
Thanks!
Best Answer
Please check if this is right.
Let $f$ be an entire function. Since it is entire, it has Taylor series $$\sum_{n = 0}^{+\infty} a_n z^n$$ at $0$ converging on $\mathbb{C}$.
We need to show $f$ has nonessential singularity at $\infty$ $\implies$ $f$ is a polynomial.
Taking contrapositive, it suffices to show infinitely many $a_n \neq 0 \implies$ $f$ has essential singularity at $\infty$.
Now $f\left(\frac{1}{z}\right)$ has Laurent series $$\sum_{n = 0}^{+\infty} \frac{a_n}{z^n}$$ at $0$ converging on $\mathbb{C} \setminus \{0\}$.
Infinitely many $a_n \neq 0$ $\implies$ $f\left(\frac{1}{z}\right)$ has essential singularity at $0$ $\implies$ $f(z)$ has essential singularity at $\infty$.