Describe the locus in the complex plane of the zeros of a quartic polynomial as the constant term varies

(Diagram and setup from UCSMP Precaluclus and Discrete Mathematics, 3rd ed.)

Above is a partial plot of the zeros of $p_c(x)=4x^4+8x^3-3x^2-9x+c$. The text stops at showing the diagram and does not discuss the shape of the locus of the zeros or describe the resulting curves. Are the curves in the locus some specific (named) type of curve? Is there a simple way to describe the curves (equations)?

The question need not be limited to the specific polynomial given--a similar sort of locus is generated by the zeros of nearly any quartic polynomial as the constant term is varied.

I presume that you are restricting $c$ to be real, so your question is:

Which complex numbers $z$ can be roots of equations $$4z^4+8x^3 -3z^2 -9z+c=0$$ as $c$ varies through the reals?

This is the same question as

For which complex numbers is $$g(z)=4z^4+8x^3 -3z^2 -9z$$ real?

Certainly if $z$ is real then $g(z)$ is real, so we are really interested in non-real solutions for $z$. Now a complex number $w$ is real iff $w-\overline{w} =0$ where $\overline{w}$ is the complex conjugate of $w$. So we want to solve the equation $$g(z)-\overline{g(z)}=0,$$ equivalently $$4(z^4-\overline{z}^4)+6(z^3-\overline{z}^3) -3(z^2-\overline{z}^2)-9(z-\overline{z})=0.\qquad\qquad(*)$$

If we set $z=x+yi$ with $y$ real we can simplify this using $$z-\overline{z}=2yi,$$ $$z^2-\overline{z}^2=4xyi,$$ $$z^3-\overline{z}^3=2(3x^2-y^2)yi$$ and $$z^4-\overline{z}^4=2(4x^3-4xy^2)yi.$$ Thus $(*)$ becomes, on cancelling $2i$, $$y[16(x^3-xy^2)+16(3x^2-y^2)-6x-9]=0.$$

The factor of $y$ in the above equation accounts for the $x$-axis (so all real values of $z$) and so the other solutions lie on the curve defined by the equation $$16(x^3-xy^2)+16(3x^2-y^2)-6x-9=0.$$ I'm not going to analyse this equation further, but it looks a fairly typical cubic, so it probably defines an elliptic curve.

We need to find all $z:\exists c\in \mathbb{R}:p_c(z)=0$.

Some manipulation gives

$\begin{array} &&p_c(z)=0\\\ \Leftrightarrow&p_0(z)+c=0\\\ \Leftrightarrow&Re(p_0)+i Im(p_0)+c=0\\\ \Leftrightarrow&Re(p_0)+c=0\text{ and } Im(p_0)=0\\\ \end{array}$

Now $Im(p_0(z))=0$ is a fixed set of points on the complex plane.

$Re(p_0)+c$ defines a surface on on $\mathbb{C}$ which varies with $c$. The surfaces that we get varying $c$ are just the translation of the surface $z=Re(p_0)$, along $z$ axis.

Our required set of points are those that lie in intersection of these the sets

$\\{z\in \mathbb{C}:Im(p_0(z))=0\\}\cap\\{z\in \mathbb{C}:\exists c:Re(p_0(z))+c=0\\}$

$=\\{z\in \mathbb{C}:Im(p_0(z))=0\\}\cap\mathbb{C}$

$=\\{z\in \mathbb{C}:Im(p_0(z))=0\\}$

Example 1: As an example let's take $p_c(z)=z^2+c$.

Plot of $Im(p_0(z))=2 x y = 0$ is just the two axes, as given below.

This is the required locus.

A plot of $Re(p_0(x))$ is also given below. This doesn't help in finding the locus though, and this is attached only to see why the second term in the intersection above is entire $\mathbb{C}$. Evidently, the points which intersect with the complex plane as we move the surface up and down is the whole complex plane. This will hold true for any polynomial as polynomials are holomorphic functions.

Example 2: Let's take your example $p_c(z)=z^4+8z^3−3z^2−9z+c$.

We have

$\begin{array}\ Im(p_0(z))&=&Im(p_0(x+iy))\\\ &=&16 x^3 y+24 x^2 y-16 x y^3-6 x y-8 y^3-9 y\\\ &=&-y \left(-16 x^3-24 x^2+16 x y^2+6 x+8 y^2+9\right)\end{array}$

Plot of $p_0(z)=0$ (which gives your locus) is shown below.

Well, you're looking at $p_c(x) = 4x^4 + 8x^3 - 3x^2 - 9x + c=0$, which is just $c=-4x^4-8x^3+3x^2+9x$, so if you graph in the $xc$-plane, you'll get a quartic that opens down with the horizontal cross sections the zeros for given $c$. In fact, because we can solve for $c$ like this to make it a graph in the affine plane, it is a rational quartic curve, so you might want to start looking there.