Finding the critical points of a degree $5$ Blaschke product The derivative of a degree $5$ polynomial $p\in\mathbb{C}[z]$ is a degree four polynomial $p'\in\mathbb{C}[z]$, and as such, the zeros of $p'$ may be found explicitly using the quartic formulae.
One may think of a finite Blaschke product as playing the same role on the unit disk that a polynomial plays on the plane (in some ways!).  In particular, a degree $5$ finite Blaschke product has $5$ zeros and $4$ critical points in the disk.  However in general, a finite Blaschke product $B$ will have additional critical points outside the unit disk.
Suppose $B$ is a finite Blaschke product of degree $n=5$.  Let $w_1,w_2,w_3,w_4\in\mathbb{D}$ be the critical points of $B$ which lie in the disk.  Assume that no $w_i=0$.  Due to the conjugate symmetry of $B$, the critical points of $B$ outside the unit disk are exactly $1/\bar{w_1},1/\bar{w_2},1/\bar{w_3},1/\bar{w_4}\in\mathbb{D}\setminus\overline{\mathbb{C}}$. Thus if $B'=P/Q$ for relatively prime polynomials $P$ and $Q$, then $\deg(P)=8$.
Can the known conjugate symmetry of the degree $8$ polynomial $P$ across the unit circle allow us to find its zeros using the quartic formulae?
NOTE 1: The above is a narrowing of this question on MSE, wherein I also asked about whether a degree $8$ polynomial $p(z)$ whose set of zeros is known to consist of four conjugate pairs could be factored somehow using the quartic formula.
NOTE 2: The same question above goes for degree $4$ finite Blaschke products.  Of course for a degree $3$ or less Blaschke product, there are at most $4$ critical points anyway, so one may just use the quartic formula straightforwardly.
 A: Even for degree $4$ Blaschke products, the critical points need not be contained in a solvable Galois extension of the field of coefficients of $P$. For instance, let $t$ be another variable which is supposed to parametrize the purely imaginary numbers. Take e.g. $w_k=1/(k+1)+t$. Then the Blaschke product is, up to a constant factor, $B(z)=\prod\frac{z-1/(k+1)-t}{(1/(k+1)-t)z-1}$. The numerator $P(z)$ of the derivative of $B(z)$ is a degree $6$ polynomial in $z$ over the field $\mathbb Q(t)$. It is not hard to see that the Galois group of $P(z)$ over $\mathbb C(t)$ is the symmetric group $S_6$. So there is no quartic formula expressing the $w_k$'s in terms of the coefficients of $P(z)$.
In order to get an arithmetic example, note that the Galois group of $P(z)$ over the smaller field $\mathbb Q(\sqrt{-1},t)$ is $S_6$ even more. So generically (for instance use Hilbert's irreducibility theorem) upon specializing $t$ to $\sqrt{-1}a$ for most rationals $a$ we get that the Galois group of $P(z)$ is still $S_6$. Instead of using HIT, one may use explicit examples. For instance $a=1/2$ works.
The same kind of construction works for degree $5$ too.
A: For example, consider the Blaschke product
$$ B(z) = {\frac { \left( z-1/2 \right)  \left( z-1/3 \right)  \left( z-1/4
 \right)  \left( z-1/5 \right)  \left( z-1/6 \right) }{ \left( 1-z/2
 \right)  \left( 1-z/3 \right)  \left( 1-z/4 \right)  \left( 1-z/5
 \right)  \left( 1-z/6 \right) }}
$$
The numerator of $B'(z)$ is
$$13356\,{z}^{8}-222040\,{z}^{7}+1402555\,{z}^{6}-4211480\,{z}^{5}+
6172978\,{z}^{4}-4211480\,{z}^{3}+1402555\,{z}^{2}-222040\,z+13356
$$
which is an irreducible polynomial with galois group "8T44" (according to Maple).  This is a solvable group, so the polynomial should be solvable
by radicals.  
Indeed, if $w = z - 1/z$ for a root $z$, then $w$ is a root of 
$$ {w}^{8}-{\frac {26555231\,{w}^{6}}{455058}}+{\frac {1322220803\,{w}^{4
}}{1213488}}-{\frac {61878800\,{w}^{2}}{8427}}+{\frac{38769600}{2809}}
$$
This is a quartic in $w^2$, so can be solved using the quartic formula.  And then $$ z = \frac{w \pm \sqrt{w^2 +4}}{2}$$
