Understanding Isolated Singularities. 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:

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*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!
 A: 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 $\infty$.
A: For the removable singularity:  You have extended $g$ to a continuous function on a compact set: the closed unit disk, so $g$ is bounded on the closed unit disk.  This says $f$ is bounded on the exterior of the open unit disk, say by $M$.  Since $f$ is entire, $f$ is continuous on the closed unit disk, so $f$ is bounded there, say by $m$.  So $f$ is a bounded (by $\max\{m,M\}$) entire function, to which we apply Liouville's theorem to discover $f$ is a constant function.
For the pole:  (Are you sure your limit isn't  $\lim_{|z| \rightarrow \infty} |f(z)| = \infty$?)  Suppose the pole has order $n$.  Let $g(z) = f(1/z)$ so that $g$ has a pole of order $n$ at $z = 0$.  Then $z^n g(z)$ has a removable singularity at $z = 0$ and is bounded on the closed unit disk.  So $(1/z)^n g(1/z) = z^{-n}f(z)$ is bounded on the exterior of the open unit disk.  (In fact, $\lim_{|z| \rightarrow \infty} z^{-n}f(z)$ is some nonzero number.)  Since $f$ is entire, it has a power series centered at $z = 0$,
$$  f(z) = \sum_{j=0}^\infty a_j z^j  \text{.}  $$
Then
$$  z^{-n} f(z) = \frac{a_0}{z^n} + \frac{a_1}{z^{n-1}} + \cdots + a_n + \sum_{j=1}^\infty a_{n+j} z^{j}  $$
Observe that $\sum_{j=1}^\infty a_{n+j} z^{j}$ is an entire function with a removable singularity at $z = \infty$, so by the first part of the solution, is a constant.  Since that sum has no constant term, that sum is zero.  Therefore, $f$ is the polynomial
$$  f(z) = \sum_{j=0}^n a_j z^j  \text{.}  $$
