Quartic polynomials of a complex variable

Question

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.

Answer 1

$a(z+\frac{1}{z})^2+b(z+\frac{1}{z})+c = 0$ This is a quartic polynomial, if we multipy by $z^2$ we obtain $az^4+bz^3+(2a+c)z^2+bz+a = 0$ This quartic can be related to the general quartic polynomial

$f(x) = Ax^4+Bx^3+Cx^2+Dx+E$ Now compare the coefficient of the two polynomials $az^4+bz^3+(2a+c)z^2+bz+a$ and $Ax^4+Bx^3+Cx^2+Dx+E$

Are you looking for a way to transform the general quartic to this form $az^4+bz^3+(2a+c)z^2+bz+a$ ?

Now we are looking for a quartic that can be put in that form, since it has $3$ variables $a,b,c$, then reduce the polynomials and compare with a depressed quartic $y^4+py^2+qy+r$, equate their coefficient then put $a,b,c$ in terms of the order coefficient

Answer 2

Let $\zeta =z+\frac 1 z$. Solve the quadratic $a\zeta^{2}+b\zeta +c=0$ to get two solution s $\zeta_1$ and $\zeta_2$. Then solve the equations $z+\frac 1 z=\zeta _1$ and $z+\frac 1 z=\zeta_2$. Each of these is a quadratic in $z$ so you get two soultions in each case.

Answer 3

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 $t$:

$z = \frac{t\pm\sqrt{t^{2} - 4}}{2}$

$\boxed{z = \frac{s+\sqrt{s^{2} - 4}}{2},\frac{s-\sqrt{s^{2} - 4}}{2},\frac{t+\sqrt{t^{2} - 4}}{2}, \frac{t-\sqrt{t^{2} - 4}}{2}}$