From: TJ Evert <t_evert@hotmail.com>
To: Teacher2Teacher Service
Date: Nov 26, 2003 at 20:00:29
Subject: The geometry of the complex roots of graphs

Dear Math Forum,

I have stumbled onto an interesting way of graphically presenting the 
complex roots of a qudratic equation and I was hoping you could steer me 
to more ideas along this line of reasoning. There's a big build-up, so I 
thank you for your patience in advance.

=======================================================================

The roots to the quadratic equation ax^2 + bx + c = 0 are  x = -b/2a ± 
d/2a, where d = sqrt(b^2 - 4ac). For simplicity's sake, let's say b/2a = A 
and d/2a = B. Thus (for real-valued coefficients) x = A ± B, where A is 
strictly a real number and B is real if b^2 > 4ac and imaginary if b^2 < 
4ac.

It's not groundbreaking to notice A = -b/2a is the vertex of the parabola. 
Since the graphic representation the real roots of a qudratic equation are 
simply where the graph of the quadratic equation crosses the x-axis, then 
B can be thought of as the *distance* the two root are from the vertex.

Let's assume that "a" is positive and the parabola opens "up". Let's 
further assume this parabola has two distinct real roots. If we think 
of "c" as an additive constant, as it becomes more positive the graph 
is "raised" and the roots get closer together, or the graph
  +               +
---+-------------+---
    +           +
     +         +
      +       +
        +   +
          v

becomes

   +             +
    +           +
     +         + 
------+-------+------
        +   +
          v

Algebraically, this makes sense. If "a" is positive and "c" becomes more 
positive, then B = sqrt(b^2 - 4ac)/2a (the "distance" for the roots from 
the vertex) becomes smaller. When b^2 = 4ac, then the distance between the 
roots is zero and we can say that there are two root at the same point, or

   +             +
    +           +
     +         + 
      +       +
        +   +
----------v----------

But what if b^2 < 4ac? Then B is imaginary and our graph looks like

   +             +
    +           +
     +         + 
      +       +
        +   +
          v

---------------------

But I thought one way to "imagine" the roots is using a "phantom 
parabola". This parabola shares the vertex of the "real" parabola, but is 
a vertical mirror image, or

   +             +
    +           +
     +         + 
      +       +
        +   +
          v
        *   *
------*-------*------
     *         *
    *           *
   *             *

Now the solution x = -b/2a ± d/2a = A ± B still has a geometric 
intrepretaion. "A" is still the vertex, but the imaginary "B" can be 
thought of as the *distance* of the two *imaginary* roots on the "phantom" 
parabola from the vertex.

=======================================================================

Whew! Now the questions:
1) As far as you know, is this a novel idea? (and please, not beacause of 
its lengthy description...)

2) If so, is this an idea worthy of publication? How would I proceed? 
Where do ideas like this "get into circulation"?

3) If not, has any thought been given to using this idea for higher 
powered polynomials. For instance, the inflection point of the graph of a 
a cubic has many of the nice geometric "properties" of a parabola's vertex 
(a single point of symmetry). Perhaps this idea of a "phantom" graph could 
be extended.

Thank you again for your time and consideration. I think you all do a 
great job. From the few contributions I've made in alt.algebra.help, I 
know intimately how hard you job is. I look forward to your reply.

Sincerely,
TJ Evert

Post a public discussion message
Ask Teacher2Teacher a new question