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complex plot of the zeros

(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.

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Wouldn't this be spec(Z[x,y]/(xP(x)+y))(C) for a cubic polynomial P(x)? (Where we view the scheme spec(...)(-) as a functor CRing^op->Sets). –  user126 Jul 23 '10 at 6:38
(I can't answer Harry's question as category theory is where abstract algebra stopped making any sense to me.) –  Isaac Jul 23 '10 at 18:05
Do you expect to do any better than what the quartic formula gets you? –  Qiaochu Yuan Jul 29 '10 at 7:39
@Qiaochu Yuan: Not expect, but hope. –  Isaac Jul 29 '10 at 7:44
@Isaac: Is it just me, or is the image that is supposed to appear in your question missing? –  Jonas Meyer Dec 11 '10 at 5:38

4 Answers 4

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.

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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.

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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.

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I see how the horizontal cross-sections you talk about correspond to the real zeros of the polynomial, but not how to get anything about non-real zeros from that graph in the xc-plane. –  Isaac Jul 27 '10 at 22:08
Well, over the complex numbers, things are simpler, this is a rational plane quartic, so you can say quite a lot. If nothing else, though, take "horizontal" cross sections in $\mathbb{C}^2$, and those will be the zeros, this is, of course, intersecting $y=c$ with $y=f(x)$, which will give you the four points. I'm not sure what more you're looking for (please be explicit) but here are equations for the zeros, a name for the sort of curve they cut out, and even a parameterization. –  Charles Siegel Jul 27 '10 at 22:20

One way to look at your problem is that you are trying to find the graph of an equation p(x,c)=0. In other words, you are trying to solve an implicit function equation.

In general the solutions to implicit polynomial equations in two variables are called Algebraic Varieties, and can be very complicated (http://en.wikipedia.org/wiki/Algebraic_variety). So you can't really say necessarily what kind of curve you will get. However, maybe for this specific polynomial it is a simple curve.

What can be said is that if you know p(x, c) = 0 for some given (x,c), then for all points d near enough to c one can find a root x(d) such that p(x(d), d) = 0. This is called the implicit function theorem, and another result of the theorem is that x(d) is even differentiable with respect to d.

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