Let $P(z) = \prod_{i=1}^n (z-\alpha_i) \in \mathbb{Z}[z]$, $n > 1$ be an integer monic irreducible polynomial, that is also reciprocal, i.e. $z^n P(\frac{1}{z}) = P(z)$, and suppose also that its distinct roots are symmetric about the unit circle, so $\alpha_i = \frac{1}{\overline{\alpha_{n-i+1}}}$ for all $i=1...n$ and also suppose no $\alpha_i$ lies on the unit circle. Note that this means $n$ is even.
Define for each $i=1...n$, a holomorphic branch of the square root $h_i^{j}(z) = \sqrt{(z-\alpha_i^{j})(z-\alpha_{n-i+1}^{j})}$, outside the radial segment $[\alpha_{i}^{j},\alpha_{n-i+1}^{j}] \subset \mathbb{C}$, that is asymptotic to $z$ as $z \rightarrow \infty$, for $j=2,4$(note that all these segments do not contain $0$), denote by $E$ the union of all of these radial segments.
Then define a holomorphic branch of the square root $\sqrt{P_j}$ outside of $E$ by setting $\sqrt{P_j}(z) = \prod_{i=1}^{\frac{n}{2}} h_i^j(z)$,where $P_j(z) = \prod_{i=1}^n (z - \alpha_i ^j)$ for $j=2,4$, and set $\sqrt{Q}(z) = \sqrt{P_2}(z) \sqrt{P_4}(z) = \sqrt{P_2P_4}(z)$, here $Q = P_2P_4$
I want to prove that $\sqrt{Q}(0) = 1$. An idea for an approach I have is to show that , due to $Q$ being reciprocal (and therefore the domain $\mathbb{C} \setminus E$ being conformally mapped onto itself by $z \rightarrow \frac{1}{z}$), $z^{n} \sqrt{Q}(\frac{1}{z}) = \sqrt{Q}(z)$ holds for all $z \in \mathbb{C} \setminus E$.
If I could show this, then since $\sqrt{Q}$ is asymptotic to $z^n$ for large $z$, the LHS is $z^n(\frac{\sqrt{Q}(\frac{1}{z})}{\frac{1}{z^{n}}}) z^{-n}$ which tends to $1$ as $z \rightarrow 0$.
Thank you in advance for any help.
Edit: Added crucial change to the question (as it currently stands I don't think its necessarily true)
Here is a tentative answer (leaving the question open in case there is a simpler way forward):
One can show that in order for each $h_i^j(z)$ to be asymptotic to $z$ as $z \rightarrow \infty$, we need to choose the branch of logarithm with cut given by the half-line opposite to $\alpha_i^j$ when defining each square root, in which case one calculates that $h_i^j(0) = -\alpha_{i}^{\frac{j}{2}} \alpha_{n-i+1}^{\frac{j}{2}}$. From this one gets that $\sqrt{P}(0) = 1$.