Is there a clever way to factor this special degree four polynomial? Suppose that $f(z)=\alpha z^4+\beta z^3+\gamma z^2+\overline{\beta}z+\overline{\alpha}.$  Furthermore, suppose that two of the roots are complex and lie on a unit circle (and are conjugate to each other), but we don't know what they are.  Is there a clever way to find all four complex roots of $f(z)$?  This seems similar to a palindromic polynomial.  Here, $\alpha,\beta,$ and $\gamma$ are allowed to be complex.  The case $\alpha=0$ is easy, so we can assume $\alpha \neq 0$.  Also, $\gamma$ is real.
 A: If we have distinct zeros $z_1$ and $\overline z_1$ on the unit circle, then we
must have a factorisation
$$f(z)=(z^2+cz+1)(\alpha z^2+dz+\overline\alpha)$$
where $c$ is real (indeed $|c|\le2$). This gives
$$\beta=d+c\alpha$$
$$\overline\beta=d+c\overline\alpha$$
(so $d$ is real)
and
$$\gamma=\alpha+\overline\alpha+cd.\tag{*}$$
If we put $d=\beta-c\alpha$ into (*) we get a quadratic equation for $c$
which must have a real root in the interval $[-2,2]$. So we can solve this
quadratic, and then find $d$ also. Then we can solve two more quadratics
to factor $f(z)$.
A: If $r$ is a root whose conjugate is also a root, we have both
$$ \alpha r^4 + \beta r^3 + \gamma r^2 + \overline{\beta} r + \overline{\alpha} = 0$$
and (substituting $\overline{r}$ for $r$ and then conjugating) $$ \overline{\alpha} r^4 + \overline{\beta} r^3 +\overline{\gamma} r^2 + \beta r + \alpha = 0 $$
But if $r$ is on the unit circle, this conjugate is $1/r$, from which (substituting $1/r$ for $r$ in the original equation and multiplying by $r^4$)
$$ \overline{\alpha} r^4 + \overline{\beta} r^3 +{\gamma} r^2 + \beta r + \alpha = 0 $$
Comparing these two, we see that $\overline{\gamma} = \gamma$, i.e. $\gamma$ must be real.
Multiply the first equation by $\overline{\alpha}$, the third by $\alpha$, and subtract: we find that $r$ satisfies a cubic
$$ \left( \overline{\alpha}\beta-\overline{\beta}\alpha \right) {r}^{3}+
 \left( \overline{\alpha}\gamma-\gamma\,\alpha \right) {r}^{2}+
 \left( \overline{\alpha}\overline{\beta}-\alpha\,\beta \right) r+
 \left( \overline{\alpha} \right) ^{2}-{\alpha}^{2} = 0$$
Again, $1/r$ also satisfies this, so
$$ \left((\overline{\alpha})^2 - \alpha^2\right) r^3 +  \left( \overline{\alpha}\overline{\beta}-\alpha\,\beta \right) r^2 + \left( \overline{\alpha}\gamma-\gamma\,\alpha \right) r + \overline{\alpha}\beta-\overline{\beta}\alpha = 0$$
and by combining these last two we should get the quadratic satisfied by $r$ and $\overline{r}$.
A: We have $z_1 z_2 = 1$. From Vieta's formulas, we have
\begin{align}
z_1 + z_2 + z_3 + z_4 &= -\frac{\beta}{\alpha}, \tag{1}\\
z_1z_2 + z_3z_4 + (z_1 + z_2)(z_3 + z_4) &= \frac{\gamma}{\alpha}, \tag{2}\\
z_1z_2(z_3+z_4) + (z_1 + z_2)z_3z_4 &= -\frac{\bar{\beta}}{\alpha}, \tag{3}\\
z_1z_2z_3z_4 &= \frac{\bar{\alpha}}{\alpha}.\tag{4}
\end{align}
There are two possible cases:

*

*$\alpha = \bar{\alpha}$:

It is easy to get $\beta = \bar{\beta}$ and
\begin{align}
z_1 z_2 &= 1, \\
z_3 z_4 &= 1, \\
z_1 + z_2 + z_3 + z_4 &= -\frac{\beta}{\alpha} \tag{5}\\
(z_1+z_2)(z_3+z_4) &= \frac{\gamma}{\alpha} - 2.\tag{6}
\end{align}
From (5) and (6), we can get $z_1 + z_2$ and $z_3 + z_4$. Then it is easy to get $z_1, z_2, z_3, z_4$.


*$\alpha \ne \bar{\alpha}$:

It is easy to get
\begin{align}
z_1 z_2 &= 1, \\[6pt]
z_1 + z_2 &= \left(1 - \frac{\bar{\alpha}}{\alpha}\right)^{-1} 
\left(\frac{\bar{\beta}}{\alpha} - \frac{\beta}{\alpha}\right), \\[6pt]
z_3 z_4 &= 
\frac{\bar{\alpha}}{\alpha}, \\[6pt]
z_3 + z_4 &= \left(1 - \frac{\bar{\alpha}}{\alpha}\right)^{-1}
\left(\frac{\beta}{\alpha}\frac{\bar{\alpha}}{\alpha}
- \frac{\bar{\beta}}{\alpha}\right).
\end{align}
It is easy to obtain $z_1, z_2, z_3, z_4$.
