Proof that a holomorphic square root of function exists Problem: Let $f(z)$ be a polynomial of even degree. We want to prove that it has a holomorphic square root in the annulus ouside all the zeros of the polynomial.

I asked this question before and got this answer from Lord Shark The Unknown. To summarise the bit I want to understand better, I will write it out here.

Firstly some definitions are given:

$$f(z)=b_0 z^{2n}+b_{1}z^{2n-1}+\cdots+b_{2n}
=b_0z^{2n}F(1/z)$$where$$F(z)=1+\frac{b_1}{b_0}z+\cdots+\frac{b_{2n}}{b_0}z^{2n}
=\prod_{k=1}^{2n}(1-\alpha_k z).$$

Then the following concluding sentence is given:

Now $F(z)$ will be nonzero for $|z|<1/a$ so each $|\alpha_k|<a$.
  Each $(1-\alpha_kz)$ will have a holomorphic square root on the disc
  $\{z:|z|<1/a\}$.


I do not understand fully why such a conclusion can be made. Firstly, why must $F(1/z)$ be used? What is the problem with using $F(z)$ instead? Secondly, why can we say each $(1-\alpha_k z)$ has a holomorphi square root in that disc? Is there some theorem which allows us to make this conclusion?
 A: Basic theorem: If $U$ is simply connected, $f$ is holomorphic in $U,$ and $f$ is never $0$ in $U,$ then $f$ has a holomorphic logarithm in $U.$ I. e., $f=e^g$ for some $g$ holomorphic in $U.$
Once we have $f=e^g,$ then $f$ has holomorphic $n$th roots, because $(e^{g/n})^n = e^g = f.$
Suppose now $U= \{|z|>R\}$ for some $R>0.$ Assume $f$ is holomorphic in $U,$ and $f$ is never $0$ in $U.$ Assume futher that $\lim_{z\to \infty} f(z) = L,$ where $L \in \mathbb C\setminus \{0\}.$ Then $f$ has a holomorphic logarithm in $U.$
Proof: $U$ is not simply connected, so we can't use the theorem above. However, $f(1/z)$ is holomorphic in $\{0<|z|<1/R\}$ and extends nicely to the disc $\{|z|<1/R\}.$ I'll leave the rest as an exercise.
Now to our problem: Suppose
$$p(z) = (z-a_1)\cdot (z-a_2) \cdots (z-a_{2n})$$
is an even polynomial. We can write this, for $z\ne 0,$ as 
$$p(z) = z^{2n}(1-a_1/z)\cdot (1-a_2/z) \cdots (1-a_{2n}/z)=z^{2n}q(z).$$
Suppose $R=\max\{|a_1|, \dots ,|a_{2n}|\}.$ Then $q(z)$ is holomorphic and never $0$ in $U=\{|z|>R\},$  and has limit $1$ at $\infty.$ Therefore $q$ has a holomorphic logarithm in $U.$ Thus $q$ has a holomorphic square root in $U$, and $z^n$ times this square root is a holomorphic square root of $p$ in $U.$
