<s>[INST] Quartic polynomials of a complex variable I want to answer the following:

The equation 
  $$
    a \bigg(z + \frac{1}{z}\bigg)^2
    + b \bigg(z + \frac{1}{z}\bigg)
    + c = 0
$$
  has four solutions. Find which quartics can be put in this form after apply a linear change of vars.

 [/INST] The crucial step is the substitution:
$u = z + \frac{1}{z}$
Then, we obtain the quadratic:
$au^{2} + bu + c = 0$
By the Quadratic Formula:
$u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Let the roots of the above quadratic be $s$ and $t$. Then, for $s$:
$z + \frac{1}{z} = s$
$z^{2} - sz + 1 = 0$
By the Quadratic Formula:
$z = \frac{s\pm\sqrt{s^{2} - 4}}{2}$
By a similar argument for $t$:
$z = \frac{t\pm\sqrt{t^{2} - 4}}{2}$
$\boxed{z = \frac{s+\sqrt{s^{2} - 4}}{2},\frac{s-\sqrt{s^{2} - 4}}{2},\frac{t+\sqrt{t^{2} - 4}}{2}, \frac{t-\sqrt{t^{2} - 4}}{2}}$
</s>